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SCHOLAR Study Guide
CfE Advanced Higher MathematicsCourse materialsTopic 1: Partial fractions
Authored by:Fiona Withey (Stirling High School)
Karen Withey (Stirling High School)
Reviewed by:Margaret Ferguson
Previously authored by:Jane S Paterson
Dorothy A Watson
Heriot-Watt University
Edinburgh EH14 4AS, United Kingdom.
First published 2016 by Heriot-Watt University.
This edition published in 2017 by Heriot-Watt University SCHOLAR.
Copyright © 2017 SCHOLAR Forum.
Members of the SCHOLAR Forum may reproduce this publication in whole or in part for educationalpurposes within their establishment providing that no profit accrues at any stage, Any other use of thematerials is governed by the general copyright statement that follows.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmittedin any form or by any means, without written permission from the publisher.
Heriot-Watt University accepts no responsibility or liability whatsoever with regard to the informationcontained in this study guide.
Distributed by the SCHOLAR Forum.
SCHOLAR Study Guide Course Materials: CfE Advanced Higher Mathematics
1. CfE Advanced Higher Mathematics Course Code: C747 77
AcknowledgementsThanks are due to the members of Heriot-Watt University's SCHOLAR team who planned and created thesematerials, and to the many colleagues who reviewed the content.
We would like to acknowledge the assistance of the education authorities, colleges, teachers and studentswho contributed to the SCHOLAR programme and who evaluated these materials.
Grateful acknowledgement is made for permission to use the following material in the SCHOLARprogramme:
The Scottish Qualifications Authority for permission to use Past Papers assessments.
The Scottish Government for financial support.
The content of this Study Guide is aligned to the Scottish Qualifications Authority (SQA) curriculum.
All brand names, product names, logos and related devices are used for identification purposes only and aretrademarks, registered trademarks or service marks of their respective holders.
1
Topic 1
Partial fractions
Contents1.1 Looking back . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.1.1 Division by (x - a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.1.2 Factor theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.1.3 Factorising polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.2 Introduction to partial fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.3 Linear factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.4 Repeated factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.5 Irreducible quadratic factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
1.6 Algebraic long division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
1.7 Reduce improper rational functions by division . . . . . . . . . . . . . . . . . . . . . . . . . . 29
1.8 Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
1.9 Extended information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
1.10 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
2 TOPIC 1. PARTIAL FRACTIONS
Learning objective
By the end of this topic, you should be able to:
• use the method of partial fractions to express proper rational functions as a sum ofpartial fractions;
• use algebraic long division and factorise polynomials of up to degree 3.
© HERIOT-WATT UNIVERSITY
TOPIC 1. PARTIAL FRACTIONS 3
1.1 Looking back
SummaryPolynomials
• A polynomial is an expression containing the sum or difference of algebraic terms with powersor the equivalent in factorised form.e.g. 2x3 − 4x2 − 3x + 10 and (x + 5)(x2 − 5x − 3)
• The degree of a polynomial is the value of the highest power.e.g. 2x3 − 4x + 10 has degree 3.
• Synthetic division is a method for factorising a polynomial.
◦ (x − a) is the divisor e.g. (3x3 + 5x + 4) ÷ (x − 1)
• If a polynomial divided by (x − a) has remainder 0 then (x − a) is a factor.
• If (x − a) is a factor then the remainder under division by (x − a) is 0.
• When trying a divisor of the form (x − a) it is usually a good idea to start with (x − 1).
◦ If that does not work then think of the possible factors of the constant on the end of yourpolynomial.
◦ Be systematic and don't rub out any attempts that do not work.
• Solving a polynomial is best done in factorised form and allows you to identify the roots (i.e.the places where the graph of the polynomial crosses the x-axis).
• To determine the equation of a polynomial from its graph:
◦ use the roots to determine the factors e.g. roots a, b, cgive factors (x − a)(x − b)(x − c);
◦ remember the polynomial may have a scalar ke.g. y = k(x − a)(x − b)(x − c);
◦ substitute the coordinates of the y-intercept to determine k;
© HERIOT-WATT UNIVERSITY
4 TOPIC 1. PARTIAL FRACTIONS
1.1.1 Division by (x - a)
We can divide polynomials using the method of synthetic division.
Key point
The Remainder Theorem
If a polynomial f(x) is divided by (x− a) the remainder is f(a).
Example
Problem:
f(x) = 3x3 + 5x + 4Use synthetic division to find (3x3 + 5x + 4) ÷ (x − 1).
Solution:
(x − a) is the divisor and in this example a = 1. Using synthetic division we will be able tofind the quotient and the remainder.
Also, notice that we do not have a squared term so we must interpret this in a polynomial as0x2. The polynomial should be interpreted as 3x3 + 0x2 + 5x + 4.
So under division by (x − 1) the quotient is 3x2 + 3x + 8 and the remainder is 12.
We can express the function as f(x) = (x − 1)( 3x2 + 3x + 8 ) + 12.
Key point
When f(x) is divided by (x − a) we can say
f(x) = (x − a)Q(x) + R
where Q(x) is the quotient and R is the remainder.
Examples
1. Problem:
If f(x) = 2x4 + x2 − x + 1, divide f(x) by (x + 1).
Solution:
We must interpret the function as 2x4 + 0x3 + x2 − x + 1.
We can only divide by (x − a) so we interpret the divisor as (x − (−1)).
© HERIOT-WATT UNIVERSITY
TOPIC 1. PARTIAL FRACTIONS 5
This gives,
So the quotient is 2x3 − 2x2 + 3x − 4 and the remainder is 5.
Hence f(x) = (x + 1)( 2x3 − 2x2 + 3x − 4) + 5.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2. Problem:
If f(x) = 2x3 + 5x2 − x − 1, divide f(x) by (2x − 1).
Solution:
We can only divide by (x − a) so we interpret the divisor as 2(x − 1/2).
This gives,
So the quotient is 2x2 + 6x + 2 and the remainder is 0
So f(x) = (x − 1/2)( 2x2 + 6x + 2) but we have to take the common factor of 2 out of thequotient and put it back into the divisor.(x − 1
2
)(2x2 + 6x + 2
)=
(x − 1
2
)× 2
(x2 + 3x + 1
)
= 2
(x − 1
2
)(x2 + 3x + 1
)
giving f(x) = (2x − 1)(x2 + 3x + 1).
Go onlineDivision by (x - a) practice
Q1: What is the remainder when (4x2 − 10x + 2) ÷ (x − 3)?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q2: What is the remainder when (5x3 − 7x2 + 14) ÷ (x − 3)?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q3: What is the remainder when (3x2 − 8x + 4) ÷ (x + 3)?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q4: Express (6x3 + 7x2 − 1) ÷ (3x − 1) in the form (3x− 1)Q(x) + R, where Q(x) isthe quotient and R is the remainder.
© HERIOT-WATT UNIVERSITY
6 TOPIC 1. PARTIAL FRACTIONS
1.1.2 Factor theorem
The Factor Theorem states that if f(a) = 0 then (x − a) is a factor of f(x) and if (x − a) is a factorof f(x) then f(a) = 0.
Key point
If a polynomial divided by (x − a) has remainder 0 then (x − a) is a factor and if (x − a) isa factor then the remainder under division by (x − a) is equal to 0.
Example
Problem:
Is (x − 4) a factor of f(x) where f(x) = x3 + 7x2 − 26x − 72?
Solution:
Since the remainder is 0, (x − 4) is a factor of f(x) and f(x) = (x − 4)(x2 + 11x + 18).
Notice that we can factorise the quotient and f(x) = (x − 4)(x + 2)(x + 9) which givesthe function in its fully factorised form.
Go onlineFactor theorem practice
Q5: Is (x − 1) a factor of f(x) where f(x) = x4 + 2x3 − 7x2 − 8x + 12?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q6: Is (x − 2) a factor of f(x) where f(x) = x3 + 3x2 − 4x − 12?
1.1.3 Factorising polynomials
Key point
Synthetic division can be used to help fully factorise a polynomial.
Example
Problem:
Factorise fully x4 + 2x3 − 7x2 − 8x + 12.
Solution:
We are not given a divisor for this example so we have to use trial and error to find the firstfactor. The simplest divisor to try is (x − 1) giving,
© HERIOT-WATT UNIVERSITY
TOPIC 1. PARTIAL FRACTIONS 7
Since the remainder is 0, (x − 1) is a factor ofx4 + 2x3 − 7x2 − 8x + 12 = (x − 1)(x3 + 3x2 − 4x − 12).Now we need to factorise the quotient x3 + 3x2 − 4x − 12. We could try (x − 1) again butlet's try (x − 2) giving,
Since the remainder is 0, (x − 2) is a factor ofx3 + 3x2 − 4x − 12 = (x − 2)(x2 + 5x − 6).
So it follows that x4 + 2x3 − 7x2 − 8x + 12 = (x − 1)(x − 2)( x2 + 5x − 6).
Now all we have to do is try to factorise x2 + 5x − 6. . . and so,x2 + 5x − 6 = (x + 2) (x + 3)
x4 + 2x3 − 7x2 − 8x + 12 = (x − 1) (x − 2) (x + 2) (x + 3)
= (x − 1) (x − 2) (x + 2) (x + 3)
Key point
When trying a divisor of the form (x − a) it is usually a good idea to start with (x − 1). If thatdoes not work then think of the possible factors of the constant on the end of your polynomial.
For the constant -12 the factors and divisors are:
1 and -12 (x − 1) and (x + 12)
2 and -6 (x − 2) and (x + 6)
3 and -4 (x − 3) and (x + 4)
4 and -3 (x − 4) and (x + 3)
6 and -2 (x − 6) and (x + 2)
12 and -1 (x − 12) and (x + 1)
Be systematic. Don't rub out any attempts that do not work just put a line through them. Thiswill allow you to see what you have tried.
Example
Problem:
Fully factorise x3 + 2x2 − 5x − 6.
Solution:
For the constant -6 the factors and divisors are:
© HERIOT-WATT UNIVERSITY
8 TOPIC 1. PARTIAL FRACTIONS
1 and -6 (x − 1) and (x + 6)
2 and -3 (x − 2) and (x + 3)
3 and -2 (x − 3) and (x + 2)
6 and -1 (x − 6) and (x + 1)
Try (x − 1).
Evaluate f(1):
Since the remainder �= 0, (x − 1) is not a factor.
Try (x + 1).
Evaluate f(−1):
Since the remainder = 0, (x + 1) is a factor andx3 + 2x2 − 5x − 6 = (x + 1)
(x2 + x − 6
)= (x + 1) (x + 3) (x − 2)
Go onlineFactorising polynomials practice
Q7: Fully factorise x3 − 3x + 2.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q8: Fully factorise 2x3 − 3x2 − 11x + 6.
1.2 Introduction to partial fractions
There are some very complex looking algebraic equations. To try to differentiate or integrate themas they stand would be very difficult. In this section methods for splitting them into manageableterms are investigated.
The following definitions will help to make this section clearer.
Key point
If P (x) = anxn + an − 1x
n - 1 + an − 2xn - 2 + . . . + a2x
2 + a1x1 + a0
© HERIOT-WATT UNIVERSITY
TOPIC 1. PARTIAL FRACTIONS 9
Key point continued
where a0, . . . , an ∈ R then P is a polynomial of degree n.
Example
x + 3 has degree 1;
x2 − 2x + 3 has degree 2;
4x3 + 2x2 − 5 has degree 3;
a constant such as 7 has degree 0
Key point
If P (x) and Q (x) are polynomials then P (x)Q(x) is called a rational function.
Key point
Let P (x) be a polynomial of degree n and Q (x) be a polynomial of degree m.
If n < m then P (x)Q(x) is a proper rational function.
For example,x2 + 2x − 1x3 − 3x + 4
Key point
Let P (x) be a polynomial of degree n and Q (x) be a polynomial of degree m.
If n ≥ m then P (x)Q(x) is an improper rational function.
For example,x3 + 2x − 1
x2 − 4or x2 + 2x − 1
x2 − 4
An improper rational function can always be expressed as a polynomial plus a proper rationalfunction (by using algebraic long division).
Note: The process of algebraic long division is covered in a later section in this topic.
© HERIOT-WATT UNIVERSITY
10 TOPIC 1. PARTIAL FRACTIONS
We already know how to simplify algebraic fractions by finding a common denominator e.g.1
1 + x + 1x + 2 = (x + 2) + (1 + x)
(1 + x)(x + 2) = 2x + 3(1 + x)(x + 2)
The opposite process, for example, expressing 2x + 3(1 + x)(x + 2) as 1
1 + x + 1x + 2 , is called putting proper
rational functions into partial fractions.
Key point
The process of taking a proper rational function and splitting it into separate terms each witha factor of the original denominator as its denominator is called expressing the function aspartial fractions.
The way in which the rational function splits up depends on whether the denominator is a quadraticequation or a cubic equation.
It also depends on whether this denominator has linear, repeated linear or quadratic factors (with noreal roots).
The different ways in which a proper rational function with a denominator of degree at most threecan be split into partial fractions are now explained.
Examples
1. Linear factors
Regardless of whether the denominator is a quadratic or a cubic, if it can be factorised intodistinct factors that are linear (have a degree of 1) then its partial fractions take the followingform:
This has a denominator of a quadratic with two distinct factors.···
(x + a)(x + b) ≡ A(x + a) + B
(x + b)
This has a denominator of a cubic with three distinct factors.···
(x + a)(x + b)(x + c) ≡ A(x + a) + B
(x + b) + C(x + c)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2. Repeated linear factors
This has a denominator of a quadratic with two repeated factors.···
(x + a)2≡ A
(x + a) + B(x + a)2
This has a denominator of a cubic with three repeated factors.···
(x + a)3≡ A
(x + a) + B(x + a)2
+ C(x + a)3
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3. Irreducible quadratic factors
This has an irreducible quadratic denominator (i.e. the denominator has no real roots).···
x2 + bx + c ≡ Ax + Bx2 + bx + c
This has a cubic denominator. One factor is linear and the other is an irreducible quadratic.···
(x + a)(x2 + bx + c) ≡ Ax + a + Bx + C
x2 + bx + c
© HERIOT-WATT UNIVERSITY
TOPIC 1. PARTIAL FRACTIONS 11
1.3 Linear factors
Before we begin the process of expressing a function as a sum of partial fractions we must ensurethat we are working with a proper rational function, where the degree of the numerator is lessthan the degree of the denominator. If this is not the case, algebraic long division must be carriedout before the process of partial fraction is applied.
Note: The process of algebraic long division is covered in a later section in this topic.
Key point
A linear polynomial is one which has at most a degree of 1, graphically this would berepresented by a straight line.
The following examples will examine proper rational functions taking the general form shown belowwith linear factors and constants A, B . . . N to be determined. Each constant is written over onefactor from the original denominator and added to the next term. Resulting in a sum of terms on theRHS.
···(x + a)(x + b)···(x + n) ≡ A
(x + a) + B(x + b) + · · · + N
(x + n)
Note that you will be expected to factorise both quadratic and cubic polynomials.
Examples
1. Problem:
Express in partial fraction form 7x + 1x2 + x − 6
Solution:
Step 1: Factorise the denominator.
x2 + x − 6 = (x − 2)(x + 3)
Step 2: Identify the type of partial fraction. In this case it has linear factors.7x + 1
(x − 2)(x + 3) = A(x − 2) + B
(x + 3)
Step 3: Obtain the fractions with a common denominator.7x + 1
(x − 2)(x + 3) = A(x + 3)(x − 2)(x + 3) + B(x − 2)
(x − 2)(x + 3)
Step 4: Equate numerators since the denominators are equal.
7x + 1 ≡ A(x + 3) + B(x − 2)
Step 5: Choose two values of x to find the values of the two constants.
Whilst any two values for x will work, we can make the calculations easier by choosing x = 2and x = − 3 because either A or B will be multiplied by 0.
When x = 2:7 × 2 + 1 = A(2 + 3) + B(2 − 2)
15 = 5A
A = 3
© HERIOT-WATT UNIVERSITY
12 TOPIC 1. PARTIAL FRACTIONS
When x = − 3:7 × ( - 3) = A( - 3 + 3) + B( - 3 − 2)
- 20 = - 5B
B = 4
So our answer is,7x + 1
x2 + x − 6≡ 3
(x − 2) + 4(x + 3)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.
This example will demonstrate an alternative method to selecting values of x called equatingcoefficients. This method of working out the constants A, B, C, ... has advantages in certaincircumstances and this will be explored in further examples.
Problem:
Express x + 4(x2 − 7x + 10) in partial fractions.
Solution:
Step 1: Factorise the denominator.
x + 4(x2 − 7x + 10) ≡ x + 4
(x − 2)(x − 5)
Step 2: Identify the type of partial fraction. In this case it has linear factors.
Let x + 4(x2 − 7x + 10)
≡ A(x − 2) + B
(x − 5)
Step 3: Obtain the fractions with a common denominator.
x + 4(x2 − 7x + 10)
≡ A(x − 5)(x − 2)(x − 5) + B(x − 2)
(x − 2)(x − 5)
Step 4: Equate numerators since the denominators are equal.
x + 4 = A(x − 5) + B(x − 2) (*)
Step 5: Select values of x and substitute into (*)
Whilst we can choose any value of x we can make the calculations easier by choosing it suchthat (x − 5) = 0 and (x − 2) = 0 since then A and B will be multiplied by zero, respectively.
Hence the values taken are x = 5 and x = 2
Let x = 5:x+ 4 = A(x− 5) +B(x− 2)
5 + 4 = A(5− 5) +B(5− 2)
9 = 3B
B = 3
Let x = 2:x+ 4 = A(x− 5) +B(x− 2)
2 + 4 = A(2− 5) +B(2− 2)
6 = −3A
A = −2
© HERIOT-WATT UNIVERSITY
TOPIC 1. PARTIAL FRACTIONS 13
By either method the result follows x + 4(x2 − 7x + 10)
≡ −2(x − 2) + 3
(x − 5)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3. Problem:
Express x2 − 13x3 − 7x + 6
in partial fractions.
Solution:
Step 1: The denominator can be factorised using synthetic division.x2 − 13
x3 − 7x + 6 ≡ x2 − 13(x − 1)(x − 2)(x + 3)
Step 2: Identify the type of partial fraction. In this case again it has linear factors.
Let x2 − 13(x − 1)(x − 2)(x + 3) ≡ A
(x − 1) + B(x − 2) + C
(x + 3)
Step 3: Obtain the fractions with a common denominator.x2 − 13
(x − 1)(x − 2)(x + 3) ≡ A(x − 2)(x + 3)(x − 1)(x − 2)(x + 3) + B(x − 1)(x + 3)
(x − 1)(x − 2)(x + 3) + C(x − 1)(x − 2)(x − 1)(x − 2)(x + 3)
Step 4: Equate numerators since the denominators are equal.
x2 − 13 = A(x − 2)(x + 3) + B(x − 1)(x + 3) + C(x − 1)(x − 2) (*)
Step 5: Select values of x and substitute into (*).
Whilst we can choose any value of x we can make the calculations easier by choosing it suchthat (x − 2) = 0, (x + 3) = 0 and (x − 1) = 0 since then A, B and C will be multipliedby zero respectively.
x2 − 13 = A(x − 2)(x + 3) + B(x − 1)(x + 3) + C(x − 1)(x − 2)
Let x = 2 then,−9 = 0 + 5B + 0
−9 = 5B
B = −9
5Let x = 1 then,−12 = −4A
A = 3
Let x = − 3 then,−4 = 20C
C = −1
5
Therefore, x2 − 13x3 − 7x + 6
≡ 3(x − 1) − 9
5(x − 2) − 15(x + 3)
Go onlineLinear factors practice
Q9: Express x + 7x2 − x − 2
into partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q10: Express 7x − 42x2 − 3x − 2
into partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
© HERIOT-WATT UNIVERSITY
14 TOPIC 1. PARTIAL FRACTIONS
Q11: Express 8xx3 + 3x2 − x − 3
into partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q12: Express −11x − 34x3 + 5x2 + 2x − 8
into partial fractions.
Go onlineLinear factors exercise
Q13: Express 2x + 18x2 + 2x − 15 in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q14: Express −7x + 11−x2 + x + 6 in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q15: Express −x − 5x2 − 1
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q16: Express −4x − 10x2 + 2x − 8
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q17: Express 5x − 13x2 − 5x + 6
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q18: Express 5x2 + 6x + 7x3 − 2x2 − x + 2
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q19: Express 6x2 + 4x − 6x3 − 7x − 6
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q20: Express −7x + 9x3 − 2x2 − 9x + 18
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q21: Express 2x2 − 3x3 + 7x2 + 14x + 8
in partial fractions.
1.4 Repeated factors
When factorising quadratic and cubic functions factors may be repeated. This section will examineproper rational functions taking the form below with repeated factors on the denominator. ConstantsA, B, . . . , N are to be determined.
···(x + a)n ≡ A
(x + a) + B(x + a)2
+ · · · + N(x + a)n
Note that you will be expected to factorise both quadratic and cubic polynomials.
© HERIOT-WATT UNIVERSITY
TOPIC 1. PARTIAL FRACTIONS 15
Examples
1. Problem:
Express in partial fraction form 5x + 18x2 + 8x + 16
Solution:
Step 1: Factorise the denominator.
x2 + 8x + 16 = (x + 4)(x + 4)
Step 2: Identify the type of partial fraction. In this case it has repeated linear factors.5x + 18(x + 4)2
= Ax + 4 + B
(x + 4)2
Step 3: Obtain the fractions with a common denominator.5x + 18(x + 4)2
= A(x + 4)
(x + 4)2+ B
(x + 4)2
Step 4: Equate numerators since the denominators are equal.
5x + 18 = A (x + 4) + B (*)
Step 5: Select values of x and substitute into (*).
Whilst any value for x will work, we can make the calculations easier by choosing x = − 4as A will be multiplied by zero.
Let x = − 4:
5x + 18 = A (x + 4) + B
5(−4) + 18 = A (−4 + 4) + B
−2 = B
B = −2
Now substitute B = − 2 back into (*) and select a value of x to simplify the calculation.
Let x = 0:5x + 18 = A (x + 4) + B
5(0) + 18 = A (0 + 4) − 2
18 + 2 = 4A
A = 5
We have A = 5 and B = − 2.
Therefore: 5x + 18x2 + 8x + 16
= 5x + 4 − 2
(x + 4)2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2. Problem:
Express in partial fraction form 2x − 10x3 + 6x2 + 12x + 8
Solution:
Step 1: Factorise the denominator. Synthetic division can be used to do this.
x3 + 6x2 + 12x + 8 = (x + 2)3
Step 2: Identify the type of partial fraction. In this case it has repeated linear factors.2x − 10(x + 2)3
= A(x + 2) + B
(x + 2)2+ C
(x + 2)3
© HERIOT-WATT UNIVERSITY
16 TOPIC 1. PARTIAL FRACTIONS
Step 3: Obtain the fractions with a common denominator.
2x − 10(x + 2)3
= A(x + 2)2
(x + 2)3+ B(x + 2)
(x + 2)3+ C
(x + 2)3
Step 4: Equate numerators since the denominators are equal.
2x − 10 = A(x + 2)2 + B (x + 2) + C (*)
Step 5: Select values of x and equate coefficients.
First we will select a value for x and substitute into (*).
Whilst any value for x will work, we can make the calculations easier by choosing x = − 2as A and B will be multiplied by zero.
Let x = − 2:
2x − 10 = A(x + 2)2 + B (x + 2) + C
2(−2) − 10 = A(−2 + 2)2 + B (−2 + 2) + C
−14 = C
C = −14
In this case choosing another value of x would not eliminate A or B. Instead we use themethod of equating coefficients. This is done in the following way.
Substitute C = − 14 back into (*), expand the brackets and collect like terms.
2x − 10 = A(x + 2)2 + B (x + 2) + C
2x − 10 = A(x2 + 4x + 4
)+ Bx + 2B − 14
2x − 10 = Ax2 + 4Ax + 4A + Bx + 2B − 14
2x − 10 = Ax2 + (4A + B)x + 4A + 2B − 14
Set the coefficient in front of x2 on the LHS equal to the coefficient in front of x2 on the RHS.
In this case there is no x2 on the LHS. Its coefficient is therefore zero. The correspondingterm on the RHS is Ax2. The coefficient of x2 is A. We therefore equate zero and A.
0 = A
Set the coefficients in front of x on the LHS equal to the coefficients in front of x on the RHS.
2 = 4A + B
2 = 4(0) + B
B = 2
We have A = 0, B = 2 and C = − 14.
Therefore: 2x − 10x3 + 6x2 + 12x + 8
≡ 2(x + 2)2
− 14(x + 2)3
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3. Problem:
Express in partial fraction form 5x + 2x3 + x2
Solution:
Step 1: Factorise the denominator.
x3 + x2 = x2(x + 1)
Step 2: Identify the type of partial fraction. In this case it has distinct linear and repeatedlinear factors.
© HERIOT-WATT UNIVERSITY
TOPIC 1. PARTIAL FRACTIONS 17
5x + 2x2(x + 1)
= Ax + B
x2 + Cx + 1
Step 3: Obtain the fractions with a common denominator.5x + 2
x2(x + 1)= Ax(x + 1)
x2(x+1)+ B(x + 1)
x2(x+1)+ Cx2
x2(x+1)
Step 4: Equate numerators since the denominators are equal.
5x + 2 = Ax (x + 1) + B (x + 1) + Cx2 (*)
Step 5: Select values of x and substitute into (*).
Whilst any values for x will work, we can make the calculations easier by choosing x = − 1,x = 0 and x = 1.
Let x = − 1:
5x + 2 = Ax (x + 1) + B (x + 1) + Cx2
5(−1) + 2 = A(−1) (−1 + 1) + B (−1 + 1) + C(−1)2
−3 = C
C = −3
Let x = 0:5x + 2 = Ax (x + 1) + B (x + 1) + Cx2
5(0) + 2 = A(0) (0 + 1) + B (0 + 1) + C(0)2
2 = B
B = 2
Now substitute B = 2 and C = − 3 back into (*).
Let x = 1:5x + 2 = Ax (x + 1) + B (x + 1) + Cx2
5(1) + 2 = A(1) (1 + 1) + 2 (1 + 1) − 3(1)2
7 = 2A + 1
A = 3
We have A = 3, B = 2 and C = − 3.
Therefore: 5x + 2x3 + x2 = 3
x + 2x2 − 3
x + 1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4. Problem:
Express x + 3x2 − 4x + 4
in partial fractions.
© HERIOT-WATT UNIVERSITY
18 TOPIC 1. PARTIAL FRACTIONS
Solution:
Step 1: Factorise the denominator.x + 3
x2 − 4x + 4≡ x + 3
(x − 2)2
Step 2: Identify the type of partial fraction. In this case it has repeated linear factors.x + 3
(x − 2)2≡ A
(x − 2) + B(x − 2)2
Step 3: Obtain the fractions with a common denominator.x + 3
(x − 2)2≡ A(x − 2)
(x − 2)2+ B
(x − 2)2
Step 4: Equate numerators since the denominators are equal.
x + 3 = A(x − 2) + B (*)
Step 5: Select values of x and substitute into (*).
Let x = 2 :
x + 3 = A (x − 2) + B
2 + 3 = A (2 − 2) + B
5 = B
B = 5
Since we know B and have only A left to work out we can choose any value of x and substituteit along with the value of B into (*). Choose x to make the calculation easy. Here we havechosen x = 0.
Let x = 0 :
x + 3 = A (x − 2) + B
0 + 3 = A (0 − 2) + 5
3 = −2A + 5
−2 = −2A
A = 1
Therefore: x + 3(x − 2)2
≡ 1(x − 2) + 5
(x − 2)2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5. Problem:
Express 2x2 + 5x + 3x3 + 6x2 + 12x + 8 in partial fractions.
Solution:
Step 1: Use synthetic division to factorise the denominator.2x2 + 5x + 3
x3 + 6x2 + 12x + 8 = 2x2 + 5x + 3(x + 2)3
Step 2: Identify the type of partial fraction. In this case it has repeated linear factors.2x2 + 5x + 3
(x + 2)3= A
(x + 2) + B(x + 2)2
+ C(x + 2)3
Step 3: Obtain the fractions with a common denominator.
2x2 + 5x + 3(x + 2)3
= A(x + 2)2
(x + 2)3+ B(x + 2)
(x + 2)3+ C
(x + 2)3
Step 4: Equate numerators since the denominators are equal.
2x2 + 5x + 3 = A(x + 2)2 + B (x + 2) + C (*)
© HERIOT-WATT UNIVERSITY
TOPIC 1. PARTIAL FRACTIONS 19
Step 5: Select values of x and equate coefficients.
First we will select a value for x and substitute into (*).
Let x = − 2:
2x2 + 5x + 3 = A(x + 2)2 + B (x + 2) + C
2(−2)2 + 5(−2) + 3 = A(−2 + 2)2 + B (−2 + 2) + C
8 − 10 + 3 = C
1 = C
C = 1
Now substitute C = 1 back into (*), expand the brackets and collect like terms.
2x2 + 5x + 3 = A(x + 2)2 + B (x + 2) + C
2x2 + 5x + 3 = Ax2 + 4Ax + 4A + Bx + 2B + 1
2x2 + 5x + 3 = Ax2 + (4A + B)x + (4A + 2B + 1)
Set the coefficient in front of x2 on the LHS equal to the coefficient in front of x2 on the RHS.
x2 : 2 = A
A = 2
Now, set the coefficient in front of x on the LHS equal to the coefficient in front of x on theRHS.
x : 5 = 4A + B
Now, substitute A = 2 and evaluate for B.
5 = 4(2) + B
5 = 8 + B
B = −3
We have the values of A, B and C so we do not need to equate the constants.
We have A = 2, B = − 3 and C = 1.
Therefore: 2x2 + 5x + 3x3 + 6x2 + 12x + 8
= 2(x + 2) − 3
(x + 2)2+ 1
(x + 2)3
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6. Problem:
Express 4x2 + 9x3 + 4x2 − 3x − 18 in partial fractions.
Solution:
Step 1: Use synthetic division to factorise the denominator.4x2 + 9
x3 + 4x2 − 3x − 18 = 4x2 + 9(x − 2)(x + 3)2
Step 2: Identify the type of partial fraction. In this case it has a distinct linear factor andrepeated linear factors.
4x2 + 9(x − 2)(x + 3)2
= A(x − 2) + B
(x + 3) + C(x + 3)2
Step 3: Obtain the fractions with a common denominator.
4x2 + 9(x − 2)(x + 3)2
= A(x + 3)2
(x − 2)(x + 3)2+ B(x − 2)(x + 3)
(x − 2)(x + 3)2+ C(x − 2)
(x − 2)(x + 3)2
© HERIOT-WATT UNIVERSITY
20 TOPIC 1. PARTIAL FRACTIONS
Step 4: Equate numerators since the denominators are equal.
4x2 + 9 = A(x + 3)2 + B (x − 2) (x + 3) + C (x − 2) (*)
Step 5: Select values of x and equate coefficients.
First we will select a value for x and substitute into (*).
Let x = − 3:
4x2 + 9 = A(x + 3)2 + B (x − 2) (x + 3) + C (x − 2)
4(−3)2 + 9 = A(−3 + 3)2 + B (−3 − 2) (−3 + 3) + C (−3 − 2)
36 + 9 = C (−5)
45 = −5C
C = −9
Let x = 2:4x2 + 9 = A(x + 3)2 + B (x − 2) (x + 3) + C (x − 2)
4(2)2 + 9 = A(2 + 3)2 + B (2 − 2) (2 + 3) + C (2 − 2)
16 + 9 = A(5)2
25 = 25A
A = 1
Now substitute A = 1 and C = −9 back into (*), expand the brackets and collect like terms.
4x2 + 9 = A(x + 3)2 + B (x − 2) (x + 3) + C (x − 2)
4x2 + 9 = (x + 3)2 + B (x − 2) (x + 3) − 9 (x − 2)
4x2 + 9 = x2 + 6x + 9 + B(x2 + x − 6
) − 9x + 18
4x2 + 9 = x2 + 6x + 9 + Bx2 + Bx − 6B − 9x + 18
4x2 + 9 = x2 + Bx2 + Bx − 3x − 6B + 27
4x2 + 9 = (1 + B) x2 + (B − 3)x + (−6B + 27)
Set the coefficient in front of x2 on the LHS equal to the coefficient in front of x2 on the RHS.
x2 : 4 = 1 +B
B = 3
We have A = 1, B = 3 and C = − 9.
Therefore: 4x2 + 9(x − 2)(x + 3)2
= 1(x − 2) + 3
(x + 3) − 9(x + 3)2
Key point
When working out the values of A, B, C... always substitute values for x first. When this isexhausted equate coefficients.
Go onlineRepeated factors practice
Q22: Express 2x + 2(x + 3)(x + 3) in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
© HERIOT-WATT UNIVERSITY
TOPIC 1. PARTIAL FRACTIONS 21
Q23: Express x2 − 7x + 2x3 − 3x2 + 3x − 1 in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q24: Express −3x2 + 6x + 20x3 − x2 − 8x + 12
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q25: Express x2 + 11x + 15x3 + 3x2 − 4
in partial fractions.
Go onlineRepeated factors exercise
Q26: Express 3x + 11(x + 1)2
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q27: Express 2x + 11x2 + 8x + 16
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q28: Express 3x + 5(x + 2)(x + 2) in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q29: Express x2 − 2x + 3(x + 1)3
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q30: Express 2x2 + 7x + 8x3 + 6x2 + 12x + 8
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q31: Express 6x2 + 5x + 2x3 − 3x − 2
in partial fractions.
1.5 Irreducible quadratic factors
Not all quadratic functions can be factorised into linear factors. When the resulting quadratic cannotbe factorised to give real roots it is called irreducible. This section will examine proper rationalfunctions taking the form below with irreducible quadratic factors on the denominator. Constants Aand B are to be determined.
···ax2 + bx + c ≡ Ax + B
ax2 + bx + c
Note that you will be expected to factorise both quadratic and cubic polynomials.
Examples
1. Problem:
Express in partial fraction form x − 12x3 + 4x
© HERIOT-WATT UNIVERSITY
22 TOPIC 1. PARTIAL FRACTIONS
Solution:
Step 1: Factorise the denominator.x − 12x3 + 4x
= x − 12x(x2 + 4)
Step 2: Identify the type of partial fraction. In this case it has a distinct linear factor and anirreducible quadratic factor. This can be checked by working out b2 − 4ac.
b2 − 4ac = 02 − 4 × 1× 4 = −16 which is less than zero therefore x2 + 4 is a irreduciblequadratic factor.
The partial fraction form is therefore x − 12x(x2 + 4)
= Ax + Bx + C
x2 + 4.
Step 3: Obtain the fractions with a common denominator.
x − 12x(x2 + 4) =
A(x2 + 4)x + (Bx + C)x
x2 + 4
Step 4: Equate numerators since the denominators are equal.
x − 12 = A(x2 + 4
)+ (Bx + C) x (*)
Step 5: Select values of x and equate coefficients.
First we will select a value for x and substitute into (*).
Let x = 0:0 − 12 = A (0 + 4) + (0B + C) 0
−12 = 4A
A = −3
Now substitute A = − 3 back into (*), expand the brackets and collect like terms.
x − 12 = −3(x2 + 4
)+ (Bx + C)x
x − 12 = −3x2 − 12 + Bx2 + Cx
x − 12 = (−3 + B) x2 + Cx − 12
Set the coefficients in front of x2 on the LHS equal to the coefficients in front of x2 on theRHS.0 = −3 + B
B = 3
Set the coefficients in front of x on the LHS equal to the coefficients in front of x on the RHS.
1 = C
C = 1
We have A = − 3, B = 3 and C = 1.
Therefore: x − 12x3 + 4x = −3
x + 3x + 1x2 + 4
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2. Problem:
Express 4x + 1x3 − x2 + x − 6
in partial fractions.
Solution:
Step 1: Factorise the denominator.4x + 1
x3 − x2 + x − 6 = 4x + 1(x − 2)(x2 + x + 3)
© HERIOT-WATT UNIVERSITY
TOPIC 1. PARTIAL FRACTIONS 23
Step 2: Identify the type of partial fraction. In this case it has a distinct linear factor and anirreducible quadratic factor.
4x + 1(x − 2)(x2 + x + 3)
= Ax − 2 + Bx + C
x2 + x + 3
Step 3: Obtain the fractions with a common denominator.
4x + 1(x − 2)(x2 + x + 3)
=A(x2 + x + 3)
(x − 2)(x2 + x + 3)+ (Bx + C)(x − 2)
(x − 2)(x2 + x + 3)
Step 4: Equate numerators since the denominators are equal.
4x + 1 = A(x2 + x + 3
)+ (Bx + C) (x − 2) (*)
Step 5: Select values of x and substitute into (*).
Let x = 2:4x + 1 = A
(x2 + x + 3
)+ (Bx + C) (x − 2)
4(2) + 1 = A(22 + 2 + 3
)+ (2B + C) (2 − 2)
9 = 9A
A = 1
Now substitute A = 1 back into (*) and select a value of x to simplify the calculation.
Let x = 0:4x + 1 = A
(x2 + x + 3
)+ (Bx + C) (x − 2)
4(0) + 1 = 1(02 + 0 + 3
)+ (0B + C) (0 − 2)
1 = 3 − 2C
2C = 2
C = 1
Now substitute A = 1 and C = 1 back into (*) and select another value of x to simplify thecalculation.
Let x = 1:4x + 1 = A
(x2 + x + 3
)+ (Bx + C) (x − 2)
4(1) + 1 = 1(12 + 1 + 3
)+ (1B + 1) (1 − 2)
5 = 5 − B − 1
B = −1
We have A = 1, B = − 1 and C = 1.
Therefore: 4x + 1(x − 2)(x2 + x + 3) = 1
x − 2 + −x + 1x2 + x + 3
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3. Problem:
Express −7x + 5x3 − x2 + x + 3 in partial fractions.
Solution:
Step 1: Use synthetic division to factorise the denominator.−7x + 5
x3 − x2 + x + 3 = −7x + 5(x + 1)(x2 − 2x + 3)
Step 2: Identify the type of partial fraction. In this case it has a distinct linear factor and anirreducible quadratic factor.
−7x + 5(x + 1)(x2 − 2x + 3) = A
x + 1 + Bx + Cx2 − 2x + 3
© HERIOT-WATT UNIVERSITY
24 TOPIC 1. PARTIAL FRACTIONS
Step 3: Obtain the fractions with a common denominator.
−7x + 5(x + 1)(x2 − 2x + 3)
=A(x2 − 2x + 3)
(x + 1)(x2 − 2x + 3)+ (Bx + C)(x + 1)
(x + 1)(x2 − 2x + 3)
Step 4: Equate numerators since the denominators are equal.
−7x + 5 = A(x2 − 2x + 3
)+ (Bx + C) (x + 1) (*)
Step 5: Select values of x and substitute into (*).
Let x = − 1:
−7x + 5 = A(x2 − 2x + 3
)+ (Bx + C) (x + 1)
−7(−1) + 5 = A((−1)2 − 2(−1) + 3
)+ (B(−1) + C) (−1 + 1)
12 = 6A
A = 2
Now substitute A = 2 back into (*) and select another value of x to simplify the calculation.
Let x = 0:−7x + 5 = A
(x2 − 2x + 3
)+ (Bx + C) (x + 1)
−7(0) + 5 = 2(02 − 2(0) + 3
)+ (B(0) + C) (0 + 1)
5 = 6 + C
C = −1
Now substitute A = 2 and C = − 1 back into (*) and select another value of x to simplifythe calculation.
Let x = 1:−7x + 5 = A
(x2 − 2x + 3
)+ (Bx + C) (x + 1)
−7(1) + 5 = 2(12 − 2(1) + 3
)+ (B(1) − 1) (1 + 1)
−2 = 4 + 2B − 2
2B = −4
B = −2
We have A = 2, B = − 2 and C = − 1.
Therefore: −7x + 5(x + 1)(x2 − 2x + 3)
= 2x + 1 + −2x − 1
x2 − 2x + 3
Go onlineIrreducible quadratic factors practice
Q32: Express x2 − 4x + 14x3 − x2 +2x − 8
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q33: Express 10x2 + 16x + 92x3 + 7x2 + 5x + 6
in partial fractions.
Go onlineIrreducible quadratic factors exercise
Q34: Express 5x2 − 4x + 1x3 − 2x2 − 2x + 1 in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
© HERIOT-WATT UNIVERSITY
TOPIC 1. PARTIAL FRACTIONS 25
Q35: Express x + 4x3 − 4x2 + 7x − 6
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q36: Express 2x2 + 11x + 22x3 + x − 10
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q37: Express x2 + 9x2x3 − 3x2 − x − 2
in partial fractions.
Go onlinePartial fraction formation
Given 5x − 1x2 − x − 2
, this has a denominator of a quadratic with two distinct factors.
Factorise the denominator:5x − 1
x2 − x − 2→ 5x − 1
(x − 2)(x + 1)
Equate with the general form ofthis type:
5x − 1(x − 2)(x + 1) → A
(x − 2) + B(x + 1)
Obtain a common denominatorfor both fractions:
A(x − 2) + B
(x + 1) → A(x + 1)(x − 2)(x + 1) + B(x − 2)
(x − 2)(x + 1)
Equate numerator: 5x − 1 = A (x + 1) + B (x − 2)
Select values of x and substituteto solve for A and B:
Let x = −1 then −6 = A (0) +B (−3)
Let x = 2 then 9 = A (3) +B (0)
Hence A = 3 and B = 2 which gives5x − 1
x2 − x − 2= 3
x − 2 +2
x + 1
Given 2x − 5x2 − 6x + 9 , this has a denominator of a quadratic with two repeated factors.
Factorise the denominator:2x − 5
x2 − 6x + 9 → 2x − 5(x − 3)2
Equate with the general form ofthis type:
2x − 5(x − 3)2
→ A(x − 3) + B
(x − 3)2
Obtain a common denominatorfor both fractions:
A(x − 3) + B
(x − 3)2→ A(x − 3)
(x − 3)2+ B
(x − 3)2
Equate numerator: 2x − 5 = A (x − 3) + B
Select values of x and substituteto solve for A and B:
Let x = 3 then −6 = A (0) +B (−3)
Let x = 0 and B = 1 then −5 = A (−3) + 1
Hence A = 2 and B = 1 which gives2x − 5
x2 − 6x + 9 = 2x−3 +
1(x−3)2
Given 5x2 + 6x + 4x3 + 2x2 − 2x − 1 , this has a cubic denominator. One factor is linear and the other is an
irreducible quadratic.
© HERIOT-WATT UNIVERSITY
26 TOPIC 1. PARTIAL FRACTIONS
Factorise the denominator: 5x2 + 6x + 4x3 + 2x2 − 2x − 1
→ 5x2 + 6x + 4(x − 1)(x2 + 3x + 1)
Equate with the general form ofthis type:
5x2 + 6x + 4(x − 1)(x2 + 3x + 1)
→ A(x − 1) + Bx + C
(x2 + 3x + 1)
Obtain a common denominatorfor both fractions:
A
(x − 1)+
Bx + C
(x2 + 3x + 1)
↓A(x2 + 3x + 1
)(x − 1) (x2 + 3x + 1)
+Bx + C (x − 1)
(x − 1) (x2 + 3x + 1)
Equate numerator:5x2 + 6x + 4 =A(x2 + 3x + 1
)+ (Bx + C) (x − 1)
Select values of x and substituteto solve for A and B:
Let x = 1 then 12 = A (5) + (B (1) + C) (0)
Let x = 0 andA = 3
then 4 = 3 (1) + (B (0) + C) (−1)
Let x = −1, A = 3 and
C = −1
then 3 = 3 (−1) + (B (−1) + (−1)) (−2)
Hence A = 3, B = 2 and C = − 1 which gives5x2 + 6x + 4
x3 + 2x2 − 2x − 1= 3
x − 1 + 2x − 1x2 + 3x + 1
1.6 Algebraic long division
Key point
The dividend in a long division calculation is the expression which is being divided. As afraction it is the numerator.
Key point
The divisor is the expression which is doing the dividing.
It is the expression outside the division sign. As a fraction it is the denominator.
Key point
The quotient is the answer to the division not including the remainder.
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TOPIC 1. PARTIAL FRACTIONS 27
Key point
The remainder is what is left over after dividing.
Before performing long division with polynomials a numerical example will be used to illustrate themethod.
Examples
1. 351 ÷ 8 = 43 r 7
The dividend is 351.The divisor is 8.The quotient is 43 and the remainder is 7.
In long division style this is written as:
84 3
)3 5 1
3 2 ↓3 1
2 4
7
This would be written in fraction terms as 43 78 .
The same technique can be used for dividing polynomials.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2. Problem:
Divide x3 − 2x + 5 by x2 + 2x − 3
Solution:
Step 1:Lay out the long division taking account of 'missing terms'.
Step 2:Divide the first term of the divisor (x2) into the first of the dividend (x3) and write the answerat the top (x).
Step 3:Multiply each of the terms in the divisor by the first term of the quotient and write underneaththe dividend.
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28 TOPIC 1. PARTIAL FRACTIONS
Step 4:Subtract to give a new last line in the dividend.
Step 5:Divide the first term of the divisor (x2) into the first term of the last line (−2x2) and write theanswer at the top.
Step 6:Multiply each of the terms in the divisor by the 2nd term of the quotient (−2) and writeunderneath the divisor.
Step 7:Subtract to give a new last line in the dividend (5x − 1).The division stops here in this case as the degree of the divisor (2) is greater than the degreeof the last line (1).
Therefore x3 − 2x + 5x2 + 2x − 3 = x − 2 + 5x − 1
x2 + 2x − 3
(x − 2) is the quotient and (5x − 1) is the remainder.
© HERIOT-WATT UNIVERSITY
TOPIC 1. PARTIAL FRACTIONS 29
Go onlineAlgebraic long division practice
Q38: Divide 3x3 − 2x2 + 6 by x2 + 4
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q39: Divide x3 + 8x2 + 13x − 10 by x + 5.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q40: Evaluate x4 + 7x2 + 13x − 4x2 + 4x
.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q41: What is x5 − x3
x5 + 1?
Go onlineAlgebraic long division exercise
Q42: Divide 3x6 − 4x3 + 9 by x3 + 4.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q43: Divide x5 − 2x4 + 5x + 3 by x2 − 2.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q44: Divide x4 − 2x + 5 by x2 + 4.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q45: Find the remainder when x5 − 4x3 is divided by x − 4.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q46: Divide x3 + 3x2 + 7 by x2 − 2x.
1.7 Reduce improper rational functions by division
A variation on the previous problems occurs when the initial expression is an improper rationalfunction. In these circumstances it is necessary to divide through and obtain a polynomial anda proper rational function first. This rational function can then be expressed as a sum of partialfractions using the appropriate method from the types given in the previous sections.
Summary of partial fraction forms
Before we looking at reducing improper rational functions by division. Let's look back at thetypes of partial fractions from the previous sections. This activity provides a summary of thedifferent partial fraction forms and when they should used.
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30 TOPIC 1. PARTIAL FRACTIONS
Example : Partial fraction forms
In the denominator each linear factor e.g. 2x + 1, requires one partial fraction of the form:A
2x + 1
each repeated linear factor e.g. (2x + 1)2 requires two partial fractions: A2x + 1 and B
(2x + 1)2
each irreducible quadratic factor e.g. x2 + 5, requires a partial fraction of the form: Ax + Bx2 + 5
For the following fractions where the denominator has already been factorised identify thecorrect form of partial fractions.
Q47:
1) 1(x + 1)(x − 2) a) A
(x + 1) + B(x − 2) + C
(x − 2)2
2) 1(x2 + 3) b) A
(x + 1) + B(x + 1)2
+ C(x − 2)
3) 1(x + 1)2 c) Ax + B
(x2 − 6) + C(x + 1)
4) 1(x + 3)(x − 1) d) Ax + B
x2 + 3
5) 1(x2 − 6)(x + 1) e) A
(x + 1) + B(x − 2)
6) 1(x + 1)(x − 2)2 f) Ax + B
(x2 + 1) + C(x − 2) + D
(x − 2)2
7) 1(x + 1)2(x − 2) g) A
(x + 1) + B(x + 1)2
8) 1(x2 + 1)(x − 2)2 h) A
(x + 3) + B(x − 1)
Examples
1. Improper rational functions
Problem:
Express in partial fraction form x2 + 8x − 5x2 + x − 6
Solution:
It is clear that we are working with an improper rational function as the degree of thenumerator = degree of the denominator. We must therefore use algebraic long division beforewriting it as the sum of partial fractions.
Step 1: Divide using algebraic long division.
This gives x2 + 8x − 5x2 + x − 6
= 1 + 7x + 1x2 + x − 6
.
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TOPIC 1. PARTIAL FRACTIONS 31
Step 2: Now express 7x + 1x2 + x − 6 in partial fractions by first factorising the denominator.
7x + 1x2 + x − 6 = 7x + 1
(x + 3)(x − 2)
Step 3: Now express 7x + 1(x + 3)(x − 2) in partial fractions. In this case it has distinct linear factors.
7x + 1(x + 3)(x − 2) = A
x + 3 + Bx − 2
Step 4: Obtain the fractions with a common denominator.7x + 1
(x + 3)(x − 2) = A(x − 2)(x + 3)(x − 2) + B(x + 3)
(x + 3)(x − 2)
Step 5: Equate numerators since the denominators are equal.
7x + 1 = A (x − 2) + B (x + 3) (*)
Step 6: Select values of x and substitute into (*).
Let x = − 3:
7x + 1 = A (x − 2) + B (x + 3)
7(−3) + 1 = A (−3 − 2) + B (−3 + 3)
−20 = −5A
A = 4
Let x = 2:7x + 1 = A (x − 2) + B (x + 3)
7(2) + 1 = A (2 − 2) + B (2 + 3)
15 = 5B
B = 3
Therefore:7x + 1
(x + 3)(x − 2) = 4x + 3 + 3
x − 2
Step 7: Substitute partial fractions back into the expression.
x2 + 8x − 5
x2 + x − 6= 1 +
7x + 1
x2 + x − 6
= 1 +4
x + 3+
3
x − 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2. Problem:
Express 2x3 − 7x2 − 11x + 10(x + 2)(x − 4) in partial fractions.
Solution:
If we expand the denominator it becomes x2 − 2x − 8. It is then clear that we are working withan improper rational function as the degree of the numerator > degree of the denominator.
Step 1: Divide using algebraic long division.
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32 TOPIC 1. PARTIAL FRACTIONS
This gives 2x3 − 7x2 − 11x + 10(x + 2)(x − 4) = 2x − 3 − x + 14
(x + 2)(x − 4) .
Step 2: Now express x + 14(x + 2)(x − 4) in partial fractions. In this case it has a distinct linear
factors.x + 14
(x + 2)(x − 4) =A
(x + 2) +B
(x − 4)
Step 3: Obtain the fractions with a common denominator.x + 14
(x + 2)(x − 4) =A(x − 4)
(x + 2)(x − 4) +B(x + 2)
(x + 2)(x − 4)
Step 4: Equate numerators since the denominators are equal.
x + 14 = A (x − 4) + B (x + 2) (*)
Step 5: Select values of x and substitute into (*).
Let x = − 2:
x + 14 = A (x − 4) + B (x + 2)
−2 + 14 = A (−2 − 4) + B (−2 + 2)
12 = −6A
A = −2
Let x = 4:x + 14 = A (x − 4) + B (x + 2)
4 + 14 = A (4 − 4) + B (4 + 2)
18 = 6B
B = 3
Therefore:x + 14
(x + 2)(x − 4) =−2
x + 2 + 3x − 4
Step 6: Substitute partial fractions back into the expression.
2x3 − 7x2 − 11x + 10
(x + 2) (x − 4)= 2x − 3 − x + 14
(x + 2) (x − 4)
= 2x − 3 −[ −2
x + 2+
3
x − 4
]
= 2x − 3 +2
x + 2− 3
x − 4
Note that the sum of partial fraction is being subtracted and so the signs of the partial fractionschanges.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3. Problem:
Express x3 − 5x2 + 6x − 8x2 − 2x − 3
in partial fractions.
Solution:
It is clear that we are working with an improper rational function as the degree of thenumerator > degree of the denominator.
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TOPIC 1. PARTIAL FRACTIONS 33
Step 1: Divide using algebraic long division.
This gives x3 − 5x2 + 6x − 8x2 − 2x − 3
= x − 3 + 3x − 17x2 − 2x − 3
.
Step 2: Now express 3x − 17x2 − 2x − 3
in partial fractions by first factorising the denominator.3x − 17
x2 − 2x − 3= 3x − 17
(x + 1)(x − 3)
Step 3: Identify the type of partial fraction. In this case it has a distinct linear factors.3x − 17
(x + 1)(x − 3) =A
(x + 1) +B
(x − 3)
Step 4: Obtain the fractions with a common denominator.3x − 17
(x + 1)(x − 3) =A(x − 3)
(x + 1)(x − 3) +B(x + 1)
(x + 1)(x − 3)
Step 5: Equate numerators since the denominators are equal.
3x − 17 = A (x − 3) + B (x + 1) (*)
Step 6: Select values of x and substitute into (*).
Let x = − 1:
3x − 17 = A (x − 3) + B (x + 1)
3(−1) − 17 = A (−1 − 3) + B (−1 + 1)
−20 = −4A
A = 5
Let x = 3:3x − 17 = A (x − 3) + B (x + 1)
3(3) − 17 = A (3 − 3) + B (3 + 1)
−8 = 4B
B = −2
Therefore:3x − 17
(x + 1)(x − 3) =5
(x + 1) − 2(x − 3)
Step 7: Substitute partial fractions back into the expression.
x3 − 5x2 + 6x − 8
x2 − 2x − 3= x − 3 +
3x − 17
x2 − 2x − 3
= x − 3 +5
x + 1− 2
x − 3
© HERIOT-WATT UNIVERSITY
34 TOPIC 1. PARTIAL FRACTIONS
Go onlineReduce improper rational functions by division practice
Q48: Express x3
(x + 1)(x + 2) in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q49: Express x3 + 3x2 + 4x + 3(x + 1)(x + 2) in partial fractions.
Go onlineReduce improper rational functions by division exercise
Q50: Express 5x2 − 6x − 1(x − 1)2
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q51: Express 2x3 − 5x2 − x + 3(x − 1)(x − 2) in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q52: Express −7x2 − 63x − 110−x2 − 8x − 12
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q53: Express 3x4+12x3+16x2 + 7x + 1x3 + 4x2 + 5x + 2
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q54: Express x5 − 2x4 + x3 + x2 − 10x + 5x3 − 2x2 − x + 2
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q55: Express x3 − 5x2 + 9x − 11(x − 2)3
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q56: Express 2x4 + 9x3 + 22x2 + 27x + 152x3 + 7x2 + 5x + 6
in partial fractions.
© HERIOT-WATT UNIVERSITY
TOPIC 1. PARTIAL FRACTIONS 35
1.8 Learning points
Partial fractionsRational functions
• A proper rational function of two polynomials is when the degree of the numerator < thedegree of the denominator.x2 + 2x − 1x3 − 3x + 4
• An improper rational function of two polynomials is when the degree of the numerator ≥ thedegree of the denominator.x3 + 2x − 1
x2 + 4or x2 + 2x − 1
x2 + 4
Algebraic long division
• The method of partial fractions is applied to proper rational functions only. If the function isimproper, algebraic long division must be carried out first.
• When setting up algebraic long division each power of the variable must be accounted for inthe dividend.In the example above 0x2 has been written in the dividend to retain the position of the x2
terms.
Linear factors
• Any rational functions that have distinct linear factors on the denominator take the followingform:
···(x + a)(x + b)(x + c) ≡ A
(x + a) + B(x + b) + C
(x + c)
• Note that this rule can be extended to any number of distinct linear factors on the denominator.
Repeated linear factors
• Any rational functions that have repeated linear factors on the denominator take the followingform:
···(x + a)3
≡ A(x + a) + B
(x + a)2+ C
(x + a)3
• Note that this rule can be extended to any number of distinct linear factors on the denominator.
© HERIOT-WATT UNIVERSITY
36 TOPIC 1. PARTIAL FRACTIONS
Irreducible quadratic factors
• An Irreducible Quadratic is a quadratic polynomial that cannot be factorised to give real rootsi.e. b2 − 4ac < 0
• Any rational functions that have irreducible quadratic factors on the denominator take thefollowing form:
···x2 + bx + c
≡ Ax + Bx2 + bx + c
• Note that this rule can be extended to any irreducible polynomial factor on the denominator.
1.9 Extended information
Learning objective
To encourage an interest in related topics
The following web links should serve as an insight to the wealth of information and encouragereaders to explore the subject further.The authors do not maintain these web links and no guarantee can be given as to theireffectiveness at a particular date.
http://www.quickmath.com/webMathematica3/quickmath/page.jsp?s1=algebra&s2=partialfractions&s3=basicAn extremely useful site for checking answers and solving problems.
© HERIOT-WATT UNIVERSITY
TOPIC 1. PARTIAL FRACTIONS 37
1.10 End of topic test
Go onlineEnd of topic 1 test
Linear Factors
Q57: Express 5x − 8x2 − 3x + 2
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q58: Express −6x − 16x2 + 4x + 3
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q59: Express −4x − 10x2 + 2x − 8
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q60: Express 4x2 + x + 1x3 − 7x − 6
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q61: Express 5x2 + 6x + 7x3 − 2x2 − x + 2
in partial fractions.
Repeated Linear Factors
Q62: Express 2x − 7x2 − 4x + 4
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q63: Express 2x + 8x2 + 6x + 9 in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q64: Express x2 + 7x + 19x3 + 3x2 − 4 in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q65: Express 2x2 − 5x + 6x3 − 6x2 +12x − 8 in partial fractions.
Irreducible Factors
Q66: Express 4x2 + 15x + 23x3 + 5x2 + 11x + 7 in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q67: Express 8x − 15x3 − 3x2 + 5x − 3
in partial fractions.
Improper Fractions
Q68: Use algebraic division to express x3
x2 + 3x + 2as the sum of a polynomial and a proper
fraction.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
© HERIOT-WATT UNIVERSITY
38 TOPIC 1. PARTIAL FRACTIONS
Q69: Now express x3
x2 + 3x + 2 in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q70: Use algebraic division to express 2x4 − 2x3 − 2x2 − 15x + 20x2 − 3x + 2
as the sum of a polynomialand a proper fraction.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q71: Now express 2x4 − 2x3 − 2x2 − 15x + 20x2 − 3x + 2
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q72: Use algebraic division to express 4x3 − 16x2 + 20x − 11x2 − 4x + 4
as the sum of a polynomial and aproper fraction.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q73: Now express 4x3 − 16x2 + 20x − 11x2 − 4x + 4
in partial fractions.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q74: Use algebraic division to express −3x4 + 11x3 − 17x2 − 4x + 18x3 − 2x2 + 2x + 5
as the sum of a polynomialand a proper fraction.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q75: Now express −3x4 + 11x3 − 17x2 − 4x + 18x3 − 2x2 + 2x + 5 in partial fractions.
© HERIOT-WATT UNIVERSITY
GLOSSARY 39
Glossary
dividend
the dividend in a long division calculation is the expression which is being divided; as afraction it is the numerator
divisor
the divisor is the expression which is doing the dividing; it is the expression outside the divisionsign; as a fraction it is the denominator
improper rational function
let P (x) be a polynomial of degree n and Q (x) be a polynomial of degree m
If n ≥ m then P (x)Q(x) is an improper rational function.
For examplex3 + 2x − 1
x2 − 4 or x2 + 2x − 1x2 − 4
linear polynomial
a linear polynomial is one which has at most a degree of 1, graphically this would berepresented by a straight line
partial fractions
the process of taking a proper rational function and splitting it into separate terms each with afactor of the original denominator as its denominator is called expressing the function inpartial fractions
polynomial of degree n
if P (x) = anxn + an − 1x
n - 1 + an − 2xn - 2 + . . . + a2x
2 + a1x1 + a0
where a0, . . . , an ∈ R
then P is a polynomial of degree n.
proper rational function
let P (x) be a polynomial of degree n and Q (x) be a polynomial of degree m
If n < m then P (x)Q(x) is a proper rational function.
For example,x2 + 2x − 1x3 − 3x + 4
quotient
the quotient is the answer to the division but not including the remainder
rational function
if P (x) and Q (x) are polynomials then P (x)Q(x) is called a rational function
remainder
the remainder is what is left over after dividing
© HERIOT-WATT UNIVERSITY
40 ANSWERS: UNIT 1 TOPIC 1
Answers to questions and activities
Topic 1: Partial fractions
Division by (x - a) practice (page 5)
Q1: 8
Q2: 86
Q3: 55
Q4:
Remember we do not have an x term so 6x3 + 7x2 − 1 must be interpreted as 6x3 + 7x2 + 0x − 1and the divisor is factorised to give 3
(x − 1
3
).
So f(x) = (x − 1/3)( 6x2 + 9x + 3) but we have to take the common factor of 3 out of the
quotient and put it back into the divisor.(x − 1
3
)(6x2 + 9x + 3
)=
(x − 1
3
)× 3
(2x2 + 3x + 1
)
= 3
(x − 1
3
)(2x2 + 3x + 1
)
giving, f(x) = (3x − 1)(2x2 + 3x + 1).
So the quotient is 2x2 + 3x + 1 and the remainder = 0.
Factor theorem practice (page 6)
Q5: Yes
Q6: Yes
Factorising polynomials practice (page 8)
Q7:
Evaluate f(1):
Since the remainder = 0, (x − 1) is a factor and,x3 − 3x + 2 = (x − 1)
(x2 + x − 2
)= (x − 1) (x + 2) (x − 1)
= (x + 2) (x − 1)2
© HERIOT-WATT UNIVERSITY
ANSWERS: UNIT 1 TOPIC 1 41
Q8:
Evaluate f(1):
Since the remainder �= 0, (x − 1) is not a factor.
© HERIOT-WATT UNIVERSITY
42 ANSWERS: UNIT 1 TOPIC 1
Evaluate f(−1):
Since the remainder �= 0, (x + 1) is not a factor.
Evaluate f(2):
Since the remainder �= 0, (x − 2) is not a factor.
Evaluate f(−2):
Since the remainder = 0, (x + 2) is a factor and,2x3 − 3x2 − 11x + 6 = (x + 2)
(2x2 − 7x + 3
)= (x + 2) (2x − 1) (x − 3)
Linear factors practice (page 13)
Q9:
Step 1: Factorise the denominator.x + 7
x2 − x − 2= x + 7
(x + 1)(x − 2)
Step 2: Identify the type of partial fraction. In this case it has linear factors.x + 7
(x + 1)(x − 2) = A(x + 1) + B
(x − 2)
Step 3: Obtain the fractions with a common denominator.x + 7
(x + 1)(x − 2) = A(x − 2)(x + 1)(x − 2) + B(x + 1)
(x + 1)(x − 2)
Step 4: Equate numerators since the denominators are equal.
x + 7 = A(x − 2) + B(x + 1) (*)
Step 5: Select values of x and substitute into (*).
Let x = 2:x + 7 = A(x − 2) + B(x + 1)
2 + 7 = A(2 − 2) + B(2 + 1)
9 = 3B
B = 3
Let x = − 1:
© HERIOT-WATT UNIVERSITY
ANSWERS: UNIT 1 TOPIC 1 43
x + 7 = A(x − 2) + B(x + 1)
−1 + 7 = A(−1 − 2) + B(−1 + 1)
6 = −3A
A = −2
Therefore: x + 7x2 − x − 2
= 3x − 2 − 2
x + 1
Q10:
Step 1: Factorise the denominator.7x − 4
2x2 − 3x − 2 = 7x − 4(2x + 1)(x − 2)
Step 2: Identify the type of partial fraction. In this case it has linear factors.7x − 4
(2x + 1)(x − 2) = A(2x + 1) + B
(x − 2)
Step 3: Obtain the fractions with a common denominator.7x − 4
(2x + 1)(x − 2) = A(x − 2)(2x + 1)(x − 2) + B(2x + 1)
(2x + 1)(x − 2)
Step 4: Equate numerators since the denominators are equal.
7x − 4 = A(x − 2) + B(2x + 1) (*)
Step 5: Select values of x and substitute into (*).
Let x = 2:7x − 4 = A(x − 2) + B(2x + 1)
7(2) − 4 = A(2 − 2) + B(2(2) + 1)
10 = 5B
B = 2
Let x = − 12 :
7x − 4 = A(x − 2) + B(2x + 1)
7(−1
2) − 4 = A(−1
2− 2) + B(2(−1
2) + 1)
−7
2− 4 = −2
1
2A
−15
2= −5
2A
A =15
5A = 3
Therefore: 7x − 42x2 − 3x − 2
= 32x + 1 + 2
x − 2
Q11:
Step 1: Factorise the denominator using synthetic division.8x
x3 + 3x2 − x − 3= 8x
(x + 1)(x − 1)(x + 3)
Step 2: Identify the type of partial fraction. In this case it has linear factors.8x
(x + 1)(x − 1)(x + 3) = A(x + 1) +
B(x − 1) +
C(x + 3)
Step 3: Obtain the fractions with a common denominator.
8x
(x+ 1) (x− 1) (x+ 3)=
A (x− 1) (x+ 3)
(x+ 1) (x− 1) (x+ 3)+
B (x+ 1) (x+ 3)
(x+ 1) (x − 1) (x+ 3)+
C (x+ 1) (x− 1)
(x+ 1) (x− 1) (x+ 3)
© HERIOT-WATT UNIVERSITY
44 ANSWERS: UNIT 1 TOPIC 1
Step 4: Equate numerators since the denominators are equal.
8x = A(x − 1)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x − 1) (*)
Step 5: Select values of x and substitute into (*).
Let x = 1:8x = A (x− 1) (x+ 3) +B (x+ 1) (x+ 3) + C (x+ 1) (x− 1)
8(1) = A (1− 1) (1 + 3) +B (1 + 1) (1 + 3) + C (1 + 1) (1− 1)
8 = 8B
B = 1
Let x = − 1:
8x = A (x− 1) (x+ 3) +B (x+ 1) (x+ 3) + C (x+ 1) (x− 1)
8(−1) = A (−1− 1) (−1 + 3) +B (−1 + 1) (−1 + 3) + C (−1 + 1) (−1− 1)
−8 = −4A
A = 2
Let x = − 3:
8x = A (x− 1) (x+ 3) +B (x+ 1) (x+ 3) + C (x+ 1) (x− 1)
8(−3) = A (−3− 1) (−3 + 3) +B (−3 + 1) (−3 + 3) + C (−3 + 1) (−3− 1)
−24 = 8C
C = −3
Therefore: 8x2x3 + 3x2 − x − 3 = 2
x + 1 + 1x − 1 − 3
x + 3
Q12:
Step 1: Factorise the denominator using synthetic division.−11x − 34
x3 + 5x2 + 2x − 8= −11x − 34
(x + 2)(x − 1)(x + 4)
Step 2: Identify the type of partial fraction. In this case it has linear factors.−11x − 34
(x + 2)(x − 1)(x + 4) = A(x + 2) +
B(x − 1) +
C(x + 4)
Step 3: Obtain the fractions with a common denominator.−11x−34
(x+2)(x−1)(x+4) =A(x−1)(x+4)
(x+2)(x−1)(x+4) +B(x+2)(x+4)
(x+2)(x−1)(x+4) +C(x+2)(x− 1)
(x+2)(x−1)(x+4)
Step 4: Equate numerators since the denominators are equal.
−11x − 34 = A(x − 1)(x + 4) + B(x + 2)(x + 4) + C(x + 2)(x − 1) (*)
Step 5: Select values of x and substitute into (*).
Let x = 1:−11x− 34 = A (x− 1) (x+ 4) +B (x+ 2) (x+ 4) + C (x+ 2) (x− 1)
−11(1) − 34 = A (1− 1) (1 + 4) +B (1 + 2) (1 + 4) + C (1 + 2) (1− 1)
−45 = 15B
B = −3
Let x = − 2:
−11x− 34 = A (x− 1) (x+ 4) +B (x+ 2) (x+ 4) + C (x+ 2) (x− 1)
−11(−2) − 34 = A (−2− 1) (−2 + 4) +B (−2 + 2) (−2 + 4) + C (−2 + 2) (−2− 1)
−12 = −6A
A = 2
© HERIOT-WATT UNIVERSITY
ANSWERS: UNIT 1 TOPIC 1 45
Let x = − 4:−11x− 34 = A (x− 1) (x+ 4) +B (x+ 2) (x+ 4) + C (x+ 2) (x− 1)
−11(−4) − 34 = A (−4− 1) (−4 + 4) +B (−4 + 2) (−4 + 4) + C (−4 + 2) (−4− 1)
10 = 10C
C = 1
Therefore: −11x − 34x3 + 5x2 + 2x − 8
= 2x + 2 − 3
x − 1 + 1x + 4
Linear factors exercise (page 14)
Q13:
Step 1: Factorise the denominator.2x + 18
x2 + 2x − 15 = 2x + 18(x + 5)(x − 3)
Step 2: Identify the type of partial fraction. In this case it has linear factors.2x + 18
(x + 5)(x − 3) =A
(x + 5) + B(x − 3)
Step 3: Obtain the fractions with a common denominator.2x + 18
(x + 5)(x − 3) =A(x − 3)
(x + 5)(x − 3) + B(x + 5)(x + 5)(x − 3)
Step 4: Equate numerators since the denominators are equal.
2x + 18 = A (x − 3) + B (x + 5) (*)
Step 5: Select values of x and substitute into (*).
Let x = − 5:2x + 18 = A (x − 3) + B (x + 5)
2(−5) + 18 = A (−5 − 3) + B (−5 + 5)
8 = −8A
A = −1
© HERIOT-WATT UNIVERSITY
46 ANSWERS: UNIT 1 TOPIC 1
Let x = 3:2x + 18 = A (x − 3) + B (x + 5)
2(3) + 18 = A (3 − 3) + B (3 + 5)
24 = 8B
B = 3
Therefore: 2x + 18x2 + 2x − 15
= −1x + 5 + 3
x − 3
Q14:
Step 1: Factorise the denominator.−7x + 11
−x2 + x + 6 = −7x + 11(−x + 3)(x + 2)
Step 2: Identify the type of partial fraction. In this case it has linear factors.−7x + 11
(−x + 3)(x + 2) =A
(−x + 3) +B
(x + 2)
Step 3: Obtain the fractions with a common denominator.−7x + 11
(−x + 3)(x + 2) =A(x + 2)
(−x + 3)(x + 2) +B(−x + 3)
(−x + 3)(x + 2)
Step 4: Equate numerators since the denominators are equal.
−7x + 11 = A (x + 2) +B (−x + 3) (*)
Step 5: Select values of x and substitute into (*).
Let x = − 2:−7x + 11 = A (x + 2) +B (−x + 3)
−7(−2) + 11 = A (−2 + 2) +B (−(−2) + 3)
25 = 5B
B = 5
Let x = 3:−7x + 11 = A (x + 2) +B (−x + 3)
−7(3) + 11 = A (3 + 2) +B (−3 + 3)
−10 = 5A
A = −2
Therefore: −7x + 11−x2 + x + 6
−2−x + 3 + 5
x + 2
Q15:
Step 1: Factorise the denominator.−x − 5x2 − 1
= −x − 5(x + 1)(x − 1)
Step 2: Identify the type of partial fraction. In this case it has linear factors.−x − 5
(x + 1)(x − 1) =A
(x + 1) +B
(x − 1)
Step 3: Obtain the fractions with a common denominator.−x − 5
(x + 1)(x − 1) =A(x − 1)
(x + 1)(x − 1) +B(x + 1)
(x + 1)(x − 1)
© HERIOT-WATT UNIVERSITY
ANSWERS: UNIT 1 TOPIC 1 47
Step 4: Equate numerators since the denominators are equal.
−x − 5 = A (x − 1) +B (x + 1) (*)
Step 5: Select values of x and substitute into (*).
Let x = − 1:
−x − 5 = A (x − 1) +B (x + 1)
− (−1) − 5 = A (−1 − 1) +B (−1 + 1)
−4 = −2A
A = 2
Let x = 1:−x − 5 = A (x − 1) +B (x + 1)
− (1) − 5 = A (1 − 1) +B (1 + 1)
−6 = 2B
B = −3
Therefore: −x − 5x2 − 1
= 2x + 1 − 3
x − 1
Q16:
Step 1: Factorise the denominator.−4x − 10
x2 + 2x − 8= −4x − 10
(x + 4)(x − 2)
Step 2: Identify the type of partial fraction. In this case it has linear factors.−4x − 10
(x + 4)(x − 2) =A
(x + 4) +B
(x − 2)
Step 3: Obtain the fractions with a common denominator.−4x − 10
(x + 4)(x − 2) =A(x − 2)
(x + 4)(x − 2) +B(x + 4)
(x + 4)(x − 2)
Step 4: Equate numerators since the denominators are equal.
−4x − 10 = A (x − 2) +B (x + 4) (*)
Step 5: Select values of x and substitute into (*).
Let x = − 4:
−4x − 10 = A (x − 2) +B (x + 4)
−4 (−4) − 10 = A (−4 − 2) +B (−4 + 4)
6 = −6A
A = −1
Let x = 2:−4x − 10 = A (x − 2) +B (x + 4)
−4 (2) − 10 = A (2 − 2) +B (2 + 4)
−18 = 6B
B = −3
Therefore: −4x − 10x2 + 2x − 8
= −1x + 4 − 3
x − 2
© HERIOT-WATT UNIVERSITY
48 ANSWERS: UNIT 1 TOPIC 1
Q17:
Step 1: Factorise the denominator.5x − 13
x2 − 5x + 6= 5x − 13
(x − 3)(x − 2)
Step 2: Identify the type of partial fraction. In this case it has linear factors.5x − 13
(x − 3)(x − 2) =A
(x − 3) +B
(x − 2)
Step 3: Obtain the fractions with a common denominator.5x − 13
(x − 3)(x − 2) =A(x − 2)
(x − 3)(x − 2) +B(x − 3)
(x − 3)(x − 2)
Step 4: Equate numerators since the denominators are equal.
5x − 13 = A (x − 2) +B (x − 3) (*)
Step 5: Select values of x and substitute into (*).
Let x = 2:5x − 13 = A (x − 2) +B (x − 3)
5(2) − 13 = A (2 − 2) +B (2 − 3)
−3 = −B
B = 3
Let x = 3:5x − 13 = A (x − 2) +B (x − 3)
5(3) − 13 = A (3 − 2) +B (3 − 3)
2 = A
A = 2
Therefore: 5x − 13x2 − 5x + 6 = 2
x − 3 + 3x − 2
Q18:
Step 1: Factorise the denominator.5x2 + 6x + 7
x3 − 2x2 − x + 2= 5x2 + 6x + 7
(x − 1)(x − 2)(x + 1)
Step 2: Identify the type of partial fraction. In this case it has linear factors.5x2 + 6x + 7
(x − 1)(x − 2)(x + 1) =A
(x − 1) +B
(x − 2) +C
(x + 1)
Step 3: Obtain the fractions with a common denominator.5x2 + 6x + 7
(x − 1)(x − 2)(x + 1) =A(x − 2)(x + 1)
(x − 1)(x − 2)(x + 1) +B(x − 1)(x + 1)
(x − 1)(x − 2)(x + 1) +C(x − 1)(x − 2)
(x − 1)(x − 2)(x + 1)
Step 4: Equate numerators since the denominators are equal.
5x2 + 6x + 7 = A (x − 2) (x + 1) +B (x − 1) (x + 1) + C (x − 1) (x − 2) (*)
© HERIOT-WATT UNIVERSITY
ANSWERS: UNIT 1 TOPIC 1 49
Step 5: Select values of x and substitute into (*).
Let x = − 1:
5x2 + 6x + 7 = A (x − 2) (x + 1) +B (x − 1) (x + 1) + C (x − 1) (x − 2)
5(−1)2 + 6 (−1) + 7 = A (−1 − 2) (−1 + 1) +B (−1 − 1) (−1 + 1) + C (−1 − 1) (−1 − 2)
5 − 6 + 7 = C (−2) (−3)
6 = 6C
C = 1
Let x = 1:5x2 + 6x + 7 = A (x − 2) (x + 1) +B (x − 1) (x + 1) + C (x − 1) (x − 2)
5(1)2 + 6 (1) + 7 = A (1 − 2) (1 + 1) +B (1 − 1) (1 + 1) + C (1 − 1) (1 − 2)
18 = A (−1) (2)
18 = −2A
A = −9
Let x = 2:5x2 + 6x + 7 = A (x − 2) (x + 1) +B (x − 1) (x + 1) + C (x − 1) (x − 2)
5(2)2 + 6 (2) + 7 = A (2 − 2) (2 + 1) +B (2 − 1) (2 + 1) + C (2 − 1) (2 − 2)
39 = B (1) (3)
39 = 3B
B = 13
Therefore: 5x2 + 6x + 7x3 − 2x2 − x + 2
= −9x − 1 + 13
x − 2 + 1x + 1
Q19:
Step 1: Factorise the denominator.6x2 + 4x − 6x3 − 7x − 6
= 6x2 + 4x − 6(x + 1)(x + 2)(x − 3)
Step 2: Identify the type of partial fraction. In this case it has linear factors.6x2 + 4x − 6
(x + 1)(x + 2)(x − 3) =A
(x + 1) +B
(x + 2) +C
(x − 3)
Step 3: Obtain the fractions with a common denominator.6x2 + 4x − 6
(x + 1)(x + 2)(x − 3) =A(x + 2)(x − 3)
(x + 1)(x + 2)(x − 3) +B(x + 1)(x − 3)
(x + 1)(x + 2)(x − 3) +C(x + 1)(x + 2)
(x + 1)(x + 2)(x − 3)
Step 4: Equate numerators since the denominators are equal.
6x2 + 4x − 6 = A (x + 2) (x − 3) +B (x + 1) (x − 3) + C (x + 1) (x + 2) (*)
Step 5: Select values of x and substitute into (*).
Let x = − 2:
6x2 + 4x − 6 = A (x + 2) (x − 3) +B (x + 1) (x − 3) + C (x + 1) (x + 2)
6(−2)2 + 4 (−2) − 6 = A (−2 + 2) (−2 − 3) +B (−2 + 1) (−2 − 3) + C (−2 + 1) (−2 + 2)
24 − 8 − 6 = B (−1) (−5)
10 = 5B
B = 2
© HERIOT-WATT UNIVERSITY
50 ANSWERS: UNIT 1 TOPIC 1
Let x = − 1:6x2 + 4x − 6 = A (x + 2) (x − 3) +B (x + 1) (x − 3) + C (x + 1) (x + 2)
6(−1)2 + 4 (−1) − 6 = A (−1 + 2) (−1 − 3) +B (−1 + 1) (−1 − 3) + C (−1 + 1) (−1 + 2)
6 − 4 − 6 = A (1) (−4)
−4 = −4A
A = 1
Let x = 3:6x2 + 4x − 6 = A (x + 2) (x − 3) +B (x + 1) (x − 3) + C (x + 1) (x + 2)
6(3)2 + 4 (3) − 6 = A (3 + 2) (3 − 3) +B (3 + 1) (3 − 3) + C (3 + 1) (3 + 2)
54 + 12 − 6 = C (4) (5)
60 = 20C
C = 3
Therefore: 6x2 + 4x − 6x3 − 7x − 6
= 1x + 1 + 2
x + 2 + 3x − 3
Q20:
Step 1: Factorise the denominator.−7x + 9
x3 − 2x2 − 9x + 18 = −7x + 9(x + 3)(x − 2)(x − 3)
Step 2: Identify the type of partial fraction. In this case it has linear factors.−7x + 9
(x + 3)(x − 2)(x − 3) =A
(x + 3) +B
(x − 2) +C
(x − 3)
Step 3: Obtain the fractions with a common denominator.−7x + 9
(x + 3)(x − 2)(x − 3) =A(x − 2)(x − 3)
(x + 3)(x − 2)(x − 3) +B(x + 3)(x − 3)
(x + 3)(x − 2)(x − 3) +C(x + 3)(x − 2)
(x + 3)(x − 2)(x − 3)
Step 4: Equate numerators since the denominators are equal.
−7x + 9 = A (x − 2) (x − 3) +B (x + 3) (x − 3) + C (x + 3) (x − 2) (*)
Step 5: Select values of x and substitute into (*).
Let x = − 3:−7x + 9 = A (x − 2) (x − 3) +B (x + 3) (x − 3) + C (x + 3) (x − 2)
−7 (−3) + 9 = A (−3 − 2) (−3 − 3) +B (−3 + 3) (−3 − 3) + C (−3 + 3) (−3 − 2)
21 + 9 = A (−5) (−6)
30 = 30A
A = 1
Let x = 2:−7x + 9 = A (x − 2) (x − 3) +B (x + 3) (x − 3) + C (x + 3) (x − 2)
−7 (2) + 9 = A (2 − 2) (2 − 3) +B (2 + 3) (2 − 3) + C (2 + 3) (2 − 2)
−14 + 9 = B (5) (−1)
−5 = −5B
B = 1
© HERIOT-WATT UNIVERSITY
ANSWERS: UNIT 1 TOPIC 1 51
Let x = 3:−7x + 9 = A (x − 2) (x − 3) +B (x + 3) (x − 3) + C (x + 3) (x − 2)
−7 (3) + 9 = A (3 − 2) (3 − 3) +B (3 + 3) (3 − 3) + C (3 + 3) (3 − 2)
−21 + 9 = C (6) (1)
−12 = 6C
C = −2
Therefore: −7x + 9x3 − 2x2 − 9x + 18
= 1x + 3 + 1
x − 2 − 2x − 3
Q21:
Step 1: Factorise the denominator.2x2 − 3
x3 + 7x2 + 14x + 8= 2x2 − 3
(x + 1)(x + 2)(x + 4)
Step 2: Identify the type of partial fraction. In this case it has linear factors.2x2 − 3
(x + 1)(x + 2)(x + 4) =A
(x + 1) +B
(x + 2) +C
(x + 4)
Step 3: Obtain the fractions with a common denominator.2x2 − 3
(x + 1)(x + 2)(x + 4) =A(x + 2)(x + 4)
(x + 1)(x + 2)(x + 4) +B(x + 1)(x + 4)
(x + 1)(x + 2)(x + 4) +C(x + 1)(x + 2)
(x + 1)(x + 2)(x + 4)
Step 4: Equate numerators since the denominators are equal.
2x2 − 3 = A (x + 2) (x + 4) +B (x + 1) (x + 4) + C (x + 1) (x + 2) (*)
Step 5: Select values of x and substitute into (*).
Let x = − 4:2x2 − 3 = A (x + 2) (x + 4) +B (x + 1) (x + 4) + C (x + 1) (x + 2)
2(−4)2 − 3 = A (−4 + 2) (−4 + 4) +B (−4 + 1) (−4 + 4) + C (−4 + 1) (−4 + 2)
32 − 3 = C (−3) (−2)
29 = 6C
C =29
6Let x = − 2:
2x2 − 3 = A (x + 2) (x + 4) +B (x + 1) (x + 4) + C (x + 1) (x + 2)
2(−2)2 − 3 = A (−2 + 2) (−2 + 4) +B (−2 + 1) (−2 + 4) + C (−2 + 1) (−2 + 2)
8 − 3 = B (−1) (2)
5 = −2B
B = −5
2Let x = − 1:
2x2 − 3 = A (x + 2) (x + 4) +B (x + 1) (x + 4) + C (x + 1) (x + 2)
2(−1)2 − 3 = A (−1 + 2) (−1 + 4) +B (−1 + 1) (−1 + 4) + C (−1 + 1) (−1 + 2)
2 − 3 = A (1) (3)
−1 = 3A
A = −1
3
Therefore: 2x2 − 3x3 + 7x2 + 14x + 8
= −13(x + 1) − 5
2(x + 2) + 296(x + 4)
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52 ANSWERS: UNIT 1 TOPIC 1
Repeated factors practice (page 20)
Q22:
Step 1: The denominator has already been factorised and can be recognised as a repeated linearfactor.
2x + 2(x + 3)(x + 3) =
2x + 2(x + 3)2
Step 2: Identify the type of partial fraction. In this case it has repeated linear factors.2x + 2(x + 3)2
= A(x + 3) +
B(x + 3)2
Step 3: Obtain the fractions with a common denominator.2x + 2(x + 3)2
= A(x + 3)
(x + 3)2+ B
(x + 3)2
Step 4: Equate numerators since the denominators are equal.
2x + 2 = A (x + 3) + B (*)
Step 5: Select values of x and substitute into (*).
Let x = − 3:
2x + 2 = A (x + 3) + B
2(−3) + 2 = A (−3 + 3) + B
−6 + 2 = B
B = −4
Now substitute B = − 4 back into (*) and select another value of x to make solving for A easy.
Let x = 0:2x + 2 = A (x + 3) + B
2(0) + 2 = A (0 + 3) − 4
2 = 3A − 4
6 = 3A
A = 2
We have A = 2 and B = − 4.
Therefore: 2x + 2(x + 3)(x + 3) = 2
x + 3 − 4(x + 3)2
Q23:
Step 1: Use synthetic division to factorise the denominator.x2 − 7x + 2
x3 − 3x2 + 3x − 1 = x2 − 7x + 2(x−1)3
Step 2: Identify the type of partial fraction. In this case it has repeated linear factors.x2 − 7x + 2(x − 1)3
= Ax − 1 + B
(x − 1)2+ C
(x − 1)3
Step 3: Obtain the fractions with a common denominator.
x2 − 7x + 2(x − 1)3
= A(x − 1)2
(x − 1)3+ B(x − 1)
(x − 1)3+ C
(x − 1)3
Step 4: Equate numerators since the denominators are equal.
x2 − 7x + 2 = A(x − 1)2 + B (x − 1) + C (*)
Step 5: Select values of x and equate coefficients.
First we will select a value for x and substitute into (*).
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ANSWERS: UNIT 1 TOPIC 1 53
Let x = 1:
x2 − 7x + 2 = A(x − 1)2 + B (x − 1) + C
12 − 7(1) + 2 = A(1 − 1)2 + B (1 − 1) + C
−4 = C
C = −4
Now substitute C = − 4 back into (*), expand the brackets and collect like terms.
x2 − 7x + 2 = A(x − 1)2 + B (x − 1) − 4
x2 − 7x + 2 = A(x2 − 2x + 1
)+ B (x − 1) − 4
x2 − 7x + 2 = Ax2 − 2Ax + A + Bx − B − 4
x2 − 7x + 2 = Ax2 − 2Ax + Bx + A − B − 4
x2 − 7x + 2 = Ax2 + (−2A + B)x + A − B − 4
Set the coefficient in front of x2 on the LHS equal to the coefficient in front of x2 on the RHS.
x2 : 1 = A
A = 1
Now set the coefficient in front of x on the LHS equal to the coefficient in front of x on the RHS.
x : −7 = −2A+B
−7 = −2(1) +B
−5 = B
B = −5
We have A = 1, B = − 5 and C = − 4.
Therefore: x2 − 7x + 2x3 − 3x2 + 3x − 1
= 1x − 1 − 5
(x − 1)2− 4
(x − 1)3
Q24:
Step 1: Use synthetic division to factorise the denominator.−3x2 + 6x + 20
x3 − x2 − 8x + 12= −3x2 + 6x + 20
(x + 3)(x − 2)2
Step 2: Identify the type of partial fraction. In this case it has distinct linear factors and repeatedlinear factors.−3x2 + 6x + 20(x + 3)(x − 2)2
= A(x + 3) +
B(x − 2) +
C(x − 2)2
Step 3: Obtain the fractions with a common denominator.
−3x2 + 6x + 20(x + 3)(x − 2)2
= A(x − 2)2
(x + 3)(x − 2)2+ B(x + 3)(x − 2)
(x + 3)(x − 2)2+ C(x + 3)
(x + 3)(x − 2)2
Step 4: Equate numerators since the denominators are equal.
−3x2 + 6x + 20 = A(x − 2)2 +B (x + 3) (x − 2) + C (x + 3) (*)
Step 5: Select values of x and substitute into (*).
Let x = 2:−3x2 + 6x + 20 = A(x − 2)2 +B (x + 3) (x − 2) + C (x + 3)
−3(2)2 + 6(2) + 20 = A(2 − 2)2 +B (2 + 3) (2 − 2) + C (2 + 3)
−12 + 12 + 20 = 5C
20 = 5C
C = 4
© HERIOT-WATT UNIVERSITY
54 ANSWERS: UNIT 1 TOPIC 1
Let x = − 3:−3x2 + 6x + 20 = A(x − 2)2 +B (x + 3) (x − 2) + C (x + 3)
−3(−3)2 + 6(−3) + 20 = A(−3 − 2)2 +B (−3 + 3) (−3 − 2) + C (−3 + 3)
−27 − 18 + 20 = 25A
−25 = 25A
A = −1
Now substitute A = − 1 and C = 4 back into (*) and select a value of x to simplify the calculation.
Let x = 0:−3x2 + 6x + 20 = A(x − 2)2 +B (x + 3) (x − 2) + C (x + 3)
−3(0)2 + 6(0) + 20 = −1(0 − 2)2 +B (0 + 3) (0 − 2) + 4 (0 + 3)
20 = −4− 6B + 12
12 = −6B
B = −2
We have A = − 1, B = − 2 and C = 4.
Therefore: −3x2 + 6x + 20x3 − x2 − 8x + 12
= −1x + 3 − 2
x − 2 + 4(x − 2)2
Q25:
Step 1: Use synthetic division to factorise the denominator.x2 + 11x + 15x3 + 3x2 − 4
= x2 + 11x + 15(x − 1)(x + 2)2
Step 2: Identify the type of partial fraction. In this case it has distinct linear factors and repeatedlinear factors.x2 + 11x + 15(x − 1)(x + 2)2
= A(x − 1) +
B(x + 2) +
C(x + 2)2
Step 3: Obtain the fractions with a common denominator.x2 + 11x + 15(x − 1)(x + 2)2
= A(x + 2)2
(x − 1)(x + 2)2+ B(x − 1)(x + 2)
(x − 1)(x + 2)2+ C(x − 1)
(x − 1)(x + 2)2
Step 4: Equate numerators since the denominators are equal.
x2 + 11x + 15 = A(x + 2)2 +B (x − 1) (x + 2) + C (x − 1) (*)
Step 5: Select values of x and substitute into (*).
Let x = − 2:x2 + 11x + 15 = A(x + 2)2 +B (x − 1) (x + 2) + C (x − 1)
(−2)2 + 11(−2) + 15 = A(−2 + 2)2 +B (−2 − 1) (−2 + 2) + C (−2 − 1)
4 − 22 + 15 = −3C
−3 = −3C
C = 1
Let x = 1:x2 + 11x + 15 = A(x + 2)2 +B (x − 1) (x + 2) + C (x − 1)
12 + 11(1) + 15 = A(1 + 2)2 +B (1 − 1) (1 + 2) + C (1 − 1)
27 = 9A
A = 3
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ANSWERS: UNIT 1 TOPIC 1 55
Now substitute A = 3 and C = 1 back into (*) and select a value of x to simplify the calculation.
Let x = 0:x2 + 11x + 15 = A(x + 2)2 +B (x − 1) (x + 2) + C (x − 1)
02 + 11(0) + 15 = 3(0 + 2)2 +B (0 − 1) (0 + 2) + 1 (0 − 1)
15 = 12− 2B − 1
4 = −2B
B = −2
We have A = 3, B = − 2 and C = 1.
Therefore: x2 + 11x + 15x3 + 3x2 − 4
= 3x − 1 − 2
x + 2 + 1(x + 2)2
Repeated factors exercise (page 21)
Q26:
Step 1: Identify the type of partial fraction. In this case it has repeated factors.3x + 11(x + 1)2
= A(x + 1) +
B(x + 1)2
Step 2: Obtain the fractions with a common denominator.3x + 11(x + 1)2
= A(x + 1)
(x + 1)2+ B
(x + 1)2
Step 3: Equate numerators since the denominators are equal.
3x + 11 = A (x + 1) + B (*)
Step 4: Select values of x and substitute into (*).
Let x = − 1:
3x + 11 = A (x + 1) + B
3 (−1) + 11 = A (−1 + 1) + B
−3 + 11 = B
B = 8
Substitute B = 8 back into (*) and select another value of x.
Let x = 0:3x + 11 = A (x + 1) + B
3 (0) + 11 = A (0 + 1) + 8
11 = A + 8
A = 3
Therefore: 3x + 11(x + 1)2
= 3x + 1 + 8
(x + 1)2
Q27:
Step 1: Factorise the denominator.2x + 11
x2 + 8x + 16 = 2x + 11(x + 4)2
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56 ANSWERS: UNIT 1 TOPIC 1
Step 2: Identify the type of partial fraction. In this case it has repeated factors.2x + 11(x + 4)2
= A(x + 4) +
B(x + 4)2
Step 3: Obtain the fractions with a common denominator.2x + 11(x + 4)2
= A(x + 4)
(x + 4)2+ B
(x + 4)2
Step 4: Equate numerators since the denominators are equal.
2x + 11 = A (x + 4) +B (*)
Step 5: Select values of x and substitute into (*).
Let x = − 4:2x + 11 = A (x + 4) +B
2 (−4) + 11 = A (−4 + 4) +B
−8 + 11 = B
B = 3
Substitute B = 3 back into (*) and select another value of x.
Let x = 0:2x + 11 = A (x + 4) +B
2 (0) + 11 = A (0 + 4) + 3
11 = 4A+ 3
A = 2
Therefore: 2x + 11x2 + 8x + 16 = 2
x + 4 + 3(x + 4)2
Q28:
Step 1: Identify the type of partial fraction. In this case it has repeated factors.3x + 5
(x + 2)(x + 2) =A
(x + 2) +B
(x + 2)2
Step 2: Obtain the fractions with a common denominator.3x + 5
(x + 2)(x + 2) =A(x + 2)
(x + 2)2+ B
(x + 2)2
Step 3: Equate numerators since the denominators are equal.
3x + 5 = A (x + 2) +B (*)
Step 4: Select values of x and substitute into (*).
Let x = − 2:3x + 5 = A (x + 2) +B
3 (−2) + 5 = A (−2 + 2) +B
−6 + 5 = B
B = −1
Substitute B = − 1 back into (*) and select another value of x.
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ANSWERS: UNIT 1 TOPIC 1 57
Let x = 0:3x + 5 = A (x + 2) + B
3 (0) + 5 = A (0 + 2) − 1
5 = 2A − 1
A = 3
Therefore: 3x + 5(x + 2)(x + 2) =
3x + 2 − 1
(x + 2)2
Q29:
Step 1: Identify the type of partial fraction. In this case it has repeated factors.x2 − 2x + 3(x + 1)3
= A(x + 1) + B
(x + 1)2+ C
(x + 1)3
Step 2: Obtain the fractions with a common denominator.
x2 − 2x + 3(x + 1)3
= A(x + 1)2
(x + 1)3+ B(x + 1)
(x + 1)3+ C
(x + 1)3
Step 3: Equate numerators since the denominators are equal.
x2 − 2x + 3 = A(x + 1)2 + B (x + 1) + C (*)
Step 4: Select values of x and equate coefficients.
First we will select a value for x and substitute into (*).
Let x = − 1:
x2 − 2x + 3 = A(x + 1)2 + B (x + 1) + C
(−1)2 − 2 (−1) + 3 = A(−1 + 1)2 + B (−1 + 1) + C
1 + 2 + 3 = C
C = 6
Now substitute C = 6 back into (*), expand the brackets and collect like terms.
x2 − 2x + 3 = A(x + 1)2 + B (x + 1) + C
= A(x2 + 2x + 1
)+ B (x + 1) + 6
= Ax2 + 2Ax + A + Bx + B + 6
= Ax2 + (2A + B)x + A + B + 6
Set the coefficient in front of x2 on the LHS equal to the coefficient in front of x2 on the RHS.
x2: 1 = A
Set the coefficient in front of x on the LHS equal to the coefficient in front of x on the RHS.
x: −2 = 2A + B
Now, substitute A = 1 and evaluate for B.
−2 = 2A + B
−2 = 2 (1) + B
B = −4
We have the values of A, B and C so we do not need to equate the constants.
We have A = 1, B = − 4 and C = 6.
Therefore: x2 − 2x + 3(x + 1)3
= 1x + 1 − 4
(x + 1)2+ 6
(x + 1)3
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58 ANSWERS: UNIT 1 TOPIC 1
Q30:
Step 1: Factorise the denominator.2x2 + 7x + 8
x3 + 6x2 + 12x + 8= 2x2 + 7x + 8
(x + 2)3
Step 2: Identify the type of partial fraction. In this case it has repeated factors.2x2 + 7x + 8
(x + 2)3= A
(x + 2) + B(x + 2)2
+ C(x + 2)3
Step 3: Obtain the fractions with a common denominator.
2x2 + 7x + 8(x + 2)3
= A(x + 2)2
(x + 2)3+ B(x + 2)
(x + 2)3+ C
(x + 2)3
Step 4: Equate numerators since the denominators are equal.
2x2 + 7x + 8 = A(x + 2)2 + B (x + 2) + C (*)
Step 5: Select values of x and equate coefficients.
First we will select a value for x and substitute into (*).
Let x = − 2:
2x2 + 7x + 8 = A(x + 2)2 + B (x + 2) + C
2(−2)2 + 7 (−2) + 8 = A(−2 + 2)2 + B (−2 + 2) + C
8 − 14 + 8 = C
C = 2
Now substitute C = 2 back into (*), expand the brackets and collect like terms.
2x2 + 7x + 8 = A(x + 2)2 + B (x + 2) + C
= A(x2 + 4x + 4
)+ B (x + 2) + 2
= Ax2 + 4Ax + 4A + Bx + 2B + 2
= Ax2 + (4A + B)x + 4A + 2B + 2
Set the coefficient in front of x2 on the LHS equal to the coefficient in front of x2 on the RHS.
x2: 2 = A
Set the coefficient in front of x on the LHS equal to the coefficient in front of x on the RHS.
x: 7 = 4A + B
Now, substitute A = 2 and evaluate for B.
7 = 4A + B
7 = 4 (2) + B
B = −1
We have the values of A, B and C so we do not need to equate the constants.
We have A = 2, B = − 1 and C = 2.
Therefore: 2x2 + 7x + 8x3 + 6x2 + 12x + 8
= 2x + 2 − 1
(x + 2)2+ 2
(x + 2)3
Q31:
Step 1: Factorise the denominator.6x2 + 5x + 2x3 − 3x − 2 = 6x2 + 5x + 2
(x − 2)(x + 1)2
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ANSWERS: UNIT 1 TOPIC 1 59
Step 2: Identify the type of partial fraction. In this case it has a distinct linear factor and repeatedlinear factors.6x2 + 5x + 2
(x − 2)(x + 1)2= A
(x − 2) +B
(x + 1) +C
(x + 1)2
Step 3: Obtain the fractions with a common denominator.
6x2 + 5x + 2(x − 2)(x + 1)2
= A(x + 1)2
(x − 2)(x + 1)2+ B(x − 2)(x + 1)
(x − 2)(x + 1)2+ C(x − 2)
(x − 2)(x + 1)2
Step 4: Equate numerators since the denominators are equal.
6x2 + 5x + 2 = A(x + 1)2 + B (x − 2) (x + 1) + C (x − 2) (*)
Step 5: Select values of x and equate coefficients.
First we will select a value for x and substitute into (*).
Let x = − 1:
6x2 + 5x + 2 = A(x + 1)2 + B (x − 2) (x + 1) + C (x − 2)
6(−1)2 + 5 (−1) + 2 = A(−1 + 1)2 + B (−1 − 2) (−1 + 1) + C (−1 − 2)
6 − 5 + 2 = C (−3)
3 = −3C
C = −1
Let x = 2:6x2 + 5x + 2 = A(x + 1)2 + B (x − 2) (x + 1) + C (x − 2)
6(2)2 + 5 (2) + 2 = A(2 + 1)2 + B (2 − 2) (2 + 1) + C (2 − 2)
24 + 10 + 2 = A(3)2
36 = 9A
A = 4
Now substitute A = 4 and C = − 1 back into (*), expand the brackets and collect like terms.
6x2 + 5x + 2 = A(x + 1)2 + B (x − 2) (x + 1) + C (x − 2)
= A(x2 + 2x + 1
)+ B
(x2 − x − 2
)+ C (x − 2)
= 4(x2 + 2x + 1
)+ B
(x2 − x − 2
) − 1 (x − 2)
= 4x2 + 8x + 4 + Bx2 − Bx − 2B − x + 2
= (4 + B) x2 + (7 − B) x + (6 − 2B)
Set the coefficient in front of x2 on the LHS equal to the coefficient in front of x2 on the RHS.
x2: 6 = 4 + B so B = 2
We have A = 4, B = 2 and C = − 1.
Therefore: 6x2 + 5x + 2x3 − 3x − 2 = 4
x − 2 + 2x + 1 − 1
(x + 1)2
Irreducible quadratic factors practice (page 24)
Q32:
Step 1: Use synthetic division to factorise the denominator.x2 − 4x + 14
x3 − x2 +2x − 8= x2 − 4x + 14
(x − 2)(x2 + x + 4)
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60 ANSWERS: UNIT 1 TOPIC 1
Step 2: Identify the type of partial fraction. In this case it has a distinct linear factor and an irreduciblequadratic factor.
x2 − 4x + 14(x − 2)(x2 + x + 4)
= Ax − 2 + Bx+C
x2 + x + 4
Step 3: Obtain the fractions with a common denominator.
x2 − 4x + 14(x − 2)(x2 + x + 4)
=A(x2 + x + 4)
(x − 2)(x2 + x + 4)+ (Bx+C)(x − 2)
(x − 2)(x2 + x + 4)
Step 4: Equate numerators since the denominators are equal.
x2 − 4x + 14 = A(x2 + x + 4
)+ (Bx+ C) (x − 2) (*)
Step 5: Select values of x and substitute into (*).
Let x = 2:x2 − 4x + 14 = A
(x2 + x + 4
)+ (Bx+C) (x − 2)
22 − 4(2) + 14 = A(22 + 2 + 4
)+ (B(2) + C) (2 − 2)
10 = 10A
A = 1
Now substitute A = 1 back into (*) and select another value of x to simplify the calculation.
Let x = 0:x2 − 4x + 14 = A
(x2 + x + 4
)+ (Bx+C) (x − 2)
02 − 4(0) + 14 = 1(02 + 0 + 4
)+ (B(0) + C) (0 − 2)
14 = 4 − 2C
10 = −2C
C = −5
Now substitute A = 1 and C = − 5 back into (*) and select another value of x to simplify thecalculation.
Let x = 1:x2 − 4x + 14 = A
(x2 + x + 4
)+ (Bx + C) (x − 2)
12 − 4(1) + 14 = 1(12 + 1 + 4
)+ (B(1) − 5) (1 − 2)
11 = 6 − B + 5
B = 0
We have A = 1, B = 0 and C = − 5.
Therefore: x2 − 4x + 14x3 − x2 +2x − 8 = 1
x − 2 − 5x2 + x + 4
Q33:
Step 1: Use synthetic division to factorise the denominator.10x2 + 16x + 9
2x3 + 7x2 + 5x + 6 = 10x2 + 16x + 9(x + 3)(2x2 + x + 2)
Step 2: Identify the type of partial fraction. In this case it has a distinct linear factor and an irreduciblequadratic factor.
10x2 + 16x + 9(x + 3)(2x2 + x + 2) =
A(x + 3) +
Bx + C(2x2 + x + 2)
Step 3: Obtain the fractions with a common denominator.
10x2 + 16x + 9(x + 3)(2x2 + x + 2)
=A(2x2 + x + 2)
(x + 3)(2x2 + x + 2)+ (Bx + C)(x + 3)
(x + 3)(2x2 + x + 2)
Step 4: Equate numerators since the denominators are equal.
10x2 + 16x + 9 = A(2x2 + x + 2
)+ (Bx + C) (x + 3) (*)
© HERIOT-WATT UNIVERSITY
ANSWERS: UNIT 1 TOPIC 1 61
Step 5: Select values of x and substitute into (*).
Let x = − 3:10x2 + 16x + 9 = A
(2x2 + x + 2
)+ (Bx + C) (x + 3)
10(−3)2 + 16(−3) + 9 = A(2(−3)2 − 3 + 2
)+ (B(−3) + C) (−3 + 3)
51 = 17A
A = 3
Now substitute A = 3 back into (*) and select another value of x to simplify the calculation.
Let x = 0:10x2 + 16x + 9 = A
(2x2 + x + 2
)+ (Bx + C) (x + 3)
10(0)2 + 16(0) + 9 = 3(2(0)2 + 0 + 2
)+ (B(0) + C) (0 + 3)
9 = 6 + 3C
C = 1
Now substitute A = 3 and C = 1 back into (*) and select another value of x to simplify thecalculation.
Let x = 1:10x2 + 16x + 9 = A
(2x2 + x + 2
)+ (Bx + C) (x + 3)
10(1)2 + 16(1) + 9 = 3(2(1)2 + 1 + 2
)+ (B(1) + 1) (1 + 3)
35 = 15 + 4B + 4
16 = 4B
B = 4
We have A = 3, B = 4 and C = 1.
Therefore: 10x2 + 16x + 92x3 + 7x2 + 5x + 6 = 3
x + 3 + 4x + 12x2 + x + 2
Irreducible quadratic factors exercise (page 24)
Q34:
Step 1: Use synthetic division to factorise the denominator.5x2 − 4x + 1
x3 − 2x2 − 2x + 1 = 5x2 − 4x + 1(x + 1)(x2 − 3x + 1)
Step 2: Identify the type of partial fraction. In this case it has a distinct linear factor and anirreducible quadratic factor.
5x2 − 4x + 1(x + 1)(x2 − 3x + 1)
= A(x + 1) + Bx + C
(x2 − 3x + 1)
Step 3: Obtain the fractions with a common denominator.
5x2 − 4x + 1(x + 1)(x2 − 3x + 1)
=A(x2 − 3x + 1)
(x + 1)(x2 − 3x + 1)+ (Bx + C)(x + 1)
(x + 1)(x2 − 3x + 1)
Step 4: Equate numerators since the denominators are equal.
5x2 − 4x + 1 = A(x2 − 3x + 1
)+ (Bx + C) (x + 1) (*)
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62 ANSWERS: UNIT 1 TOPIC 1
Step 5: Select values of x and substitute into (*).
Let x = − 1:
5x2 − 4x + 1 = A(x2 − 3x + 1
)+ (Bx + C) (x + 1)
5(−1)2 − 4 (−1) + 1 = A((−1)2 − 3 (−1) + 1
)+ (B (−1) + C) (−1 + 1)
10 = 5A
A = 2
Now substitute A = 2 back into (*) and select another value of x to simplify the calculation.
Let x = 0:5x2 − 4x + 1 = A
(x2 − 3x + 1
)+ (Bx + C) (x + 1)
5(0)2 − 4 (0) + 1 = 2((0)2 − 3 (0) + 1
)+ (B (0) + C) (0 + 1)
1 = 2 + C
C = −1
Now substitute A = 2 and C = − 1 back into (*) and select another value of x to simplify thecalculation.
5x2 − 4x + 1 = A(x2 − 3x + 1
)+ (Bx + C) (x + 1)
5(1)2 − 4 (1) + 1 = 2((1)2 − 3 (1) + 1
)+ (B (1) − 1) (1 + 1)
2 = 2B − 4
B = 3
We have A = 2, B = 3 and C = − 1.
Therefore: 5x2 − 4x + 1x3 − 2x2 − 2x + 1
= 2x + 1 + 3x − 1
x2 − 3x + 1
Q35:
Step 1: Use synthetic division to factorise the denominator.x + 4
x3 − 4x2 + 7x − 6= x + 4
(x − 2)(x2 − 2x + 3)
Step 2: Identify the type of partial fraction. In this case it has a distinct linear factor and anirreducible quadratic factor.
x + 4(x − 2)(x2 − 2x + 3)
= A(x − 2) + Bx + C
(x2 − 2x + 3)
Step 3: Obtain the fractions with a common denominator.
x + 4(x − 2)(x2 − 2x + 3)
=A(x2 − 2x + 3)
(x − 2)(x2 − 2x + 3)+ (Bx + C)(x − 2)
(x − 2)(x2 − 2x + 3)
Step 4: Equate numerators since the denominators are equal.
x + 4 = A(x2 − 2x + 3
)+ (Bx + C) (x − 2) (*)
Step 5: Select values of x and substitute into (*).
Let x = 2:x + 4 = A
(x2 − 2x + 3
)+ (Bx + C) (x − 2)
2 + 4 = A(22 − 2 (2) + 3
)+ (B (2) + C) (2 − 2)
6 = 3A
A = 2
© HERIOT-WATT UNIVERSITY
ANSWERS: UNIT 1 TOPIC 1 63
Now substitute A = 2 back into (*) and select another value of x to simplify the calculation.
Let x = 0:x + 4 = A
(x2 − 2x + 3
)+ (Bx + C) (x − 2)
0 + 4 = 2(02 − 2 (0) + 3
)+ (B (0) + C) (0 − 2)
3 = 6 − 2C
C = 1
Now substitute A = 2 and C = 1 back into (*) and select another value of x to simplify thecalculation.
Let x = 1:x + 4 = A
(x2 − 2x + 3
)+ (Bx + C) (x − 2)
1 + 4 = 2(12 − 2 (1) + 3
)+ (B (1) + 1) (1 − 2)
5 = 3 − B
B = −2
We have A = 2, B = − 2 and C = 1.
Therefore: x + 4x3 − 4x2 + 7x − 6
= 2x − 2 + −2x + 1
x2 − 2x + 3
Q36:
Step 1: Use synthetic division to factorise the denominator.2x2 + 11x + 22x3 + x − 10 = 2x2 + 11x + 22
(x − 2)(x2 + 2x + 5)
Step 2: Identify the type of partial fraction. In this case it has a distinct linear factor and anirreducible quadratic factor.
2x2 + 11x + 22(x − 2)(x2 + 2x + 5)
= A(x − 2) + Bx + C
(x2 + 2x + 5)
Step 3: Obtain the fractions with a common denominator.
2x2 + 11x + 22(x − 2)(x2 + 2x + 5)
=A(x2 + 2x + 5)
(x − 2)(x2 + 2x + 5)+ (Bx + C)(x − 2)
(x − 2)(x2 + 2x + 5)
Step 4: Equate numerators since the denominators are equal.
2x2 + 11x + 22 = A(x2 + 2x + 5
)+ (Bx + C) (x − 2) (*)
Step 5: Select values of x and substitute into (*).
Let x = 2:2x2 + 11x + 22 = A
(x2 + 2x + 5
)+ (Bx + C) (x − 2)
2(2)2 + 11 (2) + 22 = A(22 + 2 (2) + 5
)+ (B (2) + C) ((2) − 2)
52 = 13A
A = 4
Now substitute A = 4 back into (*) and select another value of x to simplify the calculation.
Let x = 0:2x2 + 11x + 22 = A
(x2 + 2x + 5
)+ (Bx + C) (x − 2)
2(0)2 + 11 (0) + 22 = 4(02 + 2 (0) + 5
)+ (B (0) + C) (0 − 2)
22 = 20 − 2C
C = −1
Now substitute A = 4 and C = − 1 back into (*) and select another value of x to simplify thecalculation.
© HERIOT-WATT UNIVERSITY
64 ANSWERS: UNIT 1 TOPIC 1
Let x = 1:2x2 + 11x + 22 = A
(x2 + 2x + 5
)+ (Bx + C) (x − 2)
2(1)2 + 11 (1) + 22 = 4(12 + 2 (1) + 5
)+ (B (1) − 1) (1 − 2)
35 = 33 − B
B = −2
We have A = 4, B = − 2 and C = − 1.
Therefore: 2x2 + 11x + 22x3 + x − 10
= 4x − 2 + −2x − 1
x2 + 2x + 5
Q37:
Step 1: Use synthetic division to factorise the denominator.x2 + 9x
2x3 − 3x2 − x − 2= x2 + 9x
(x − 2)(2x2 + x + 1)
Step 2: Identify the type of partial fraction. In this case it has a distinct linear factor and anirreducible quadratic factor.
x2 + 9x(x − 2)(2x2 + x + 1)
= A(x − 2) + Bx + C
(2x2 + x + 1)
Step 3: Obtain the fractions with a common denominator.
x2 + 9x(x − 2)(2x2 + x + 1)
=A(2x2 + x + 1)
(x − 2)(2x2 + x + 1)+ (Bx + C)(x − 2)
(x − 2)(2x2 + x + 1)
Step 4: Equate numerators since the denominators are equal.
x2 + 9x = A(2x2 + x + 1
)+ (Bx + C) (x − 2) (*)
Step 5: Select values of x and substitute into (*).
Let x = 2:x2 + 9x = A
(2x2 + x + 1
)+ (Bx + C) (x − 2)
22 + 9 (2) = A(2(2)2 + 2 + 1
)+ (B (2) + C) (2 − 2)
22 = 11A
A = 2
Now substitute A = 2 back into (*) and select another value of x to simplify the calculation.
Let x = 0:x2 + 9x = A
(2x2 + x + 1
)+ (Bx + C) (x − 2)
02 + 9 (0) = 2(2(0)2 + 0 + 1
)+ (B (0) + C) (0 − 2)
0 = 2 − 2C
C = 1
Now substitute A = 2 and C = 1 back into (*) and select another value of x to simplify thecalculation.
Let x = 1:x2 + 9x = A
(2x2 + x + 1
)+ (Bx + C) (x − 2)
12 + 9 (1) = 2(2(1)2 + 1 + 1
)+ (B (1) + 1) (1 − 2)
10 = 7 − B
B = −3
We have A = 2, B = − 3 and C = 1.
Therefore: x2 + 9x2x3 − 3x2 − x − 2
= 2x − 2 + −3x + 1
2x2 + x + 1
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Algebraic long division practice (page 29)
Q38:
a) Set up the division correctly.
b) Divide x2 into 3x3 and type in the first term of the quotient.
c) Multiply 3x by the divisor and write the answer below the dividend.
d) Subtract to give a new last line in the dividend.
e) Divide x2 into −2x2 and type in the second term of the quotient.
f) Multiply −2 by the divisor and write the answer below new the dividend.
g) Subtract to give a new last line in the dividend.
h) Give the final solution.3x3 − 2x2 + 6
x2 + 4 = 3x − 2 + 14 − 12xx2 + 4
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Q39:
Therefore: x3 + 8x2 + 13x − 10x + 5 = x2 + 3x − 2
Q40:
Therefore: x4 + 7x2 + 13x − 4x2 + 4x
= x2 − 4x + 23 − 79x + 4x2 + 4x
Q41:
Therefore: x5 − x3
x5 + 1= 1 − x3 + 1
x5 + 1
Algebraic long division exercise (page 29)
Q42:
Therefore: 3x6 − 4x3 + 9x3 + 4 = 3x3 − 16 + 73
x3 + 4
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ANSWERS: UNIT 1 TOPIC 1 67
Q43:
Therefore: x5 − 2x4 + 5x + 3x2 − 2
= x3 − 24x2 + 2x − 48 + 9x − 93x2 − 2
Q44:
Therefore: x4 − 2x + 5x2 + 4
= x2 − 4 − 2x − 21x2 + 4
Q45:
Therefore: x5 − 4x3
x − 4 = x4 + 4x3 + 12x2 + 48x + 192 + 768x − 4
Q46:
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68 ANSWERS: UNIT 1 TOPIC 1
Therefore: x3 + 3x2 + 7x2 − 2x
= x + 5 + 10x + 7x2 − 2x
Summary of partial fraction forms (page 29)
Q47: 1-e, 2-d, 3-g, 4-h, 5-c, 6-a, 7-b, 8-f
Reduce improper rational functions by division practice (page ??)
Q48:
If we expand the denominator it becomes x2 + 3x + 2. It is then clear that we are working with animproper rational function as the degree of the numerator > degree of the denominator.
Step 1: Divide using algebraic long division.
This gives x3
(x + 1)(x + 2) = x − 3 + 7x + 6(x + 1)(x + 2) .
Step 2: Now express 7x + 6(x + 1)(x + 2) in partial fractions. In this case it has a distinct linear factors.
7x + 6(x + 1)(x + 2) =
A(x + 1) + B
(x + 2)
Step 3: Obtain the fractions with a common denominator.7x + 6
(x + 1)(x + 2) =A(x + 2)
(x + 1)(x + 2) + B(x + 1)(x + 1)(x + 2)
Step 4: Equate numerators since the denominators are equal.
7x + 6 = A (x + 2) + B (x + 1) (*)
Step 5: Select values of x and substitute into (*).
Let x = − 1:7x + 6 = A (x + 2) + B (x + 1)
7(−1) + 6 = A (−1 + 2) + B (−1 + 1)
−1 = A
A = −1
Let x = − 2:7x + 6 = A (x + 2) + B (x + 1)
7(−2) + 6 = A (−2 + 2) + B (−2 + 1)
−8 = −B
B = 8
Therefore: 7x + 6(x + 1)(x + 2) =
−1(x + 1) + 8
(x + 2)
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ANSWERS: UNIT 1 TOPIC 1 69
Step 6: Substitute partial fractions back into the expression.
x3
(x + 1) (x + 2)= x − 3 +
7x + 6
(x + 1) (x + 2)
= x − 3 +
[ −1
(x + 1)+
8
(x + 2)
]
= x − 3 − 1
(x + 1)+
8
(x + 2)
Q49:
It is clear that we are working with an improper rational function as the degree of the numerator >degree of the denominator.
Step 1: Divide using algebraic long division.
This gives x3 + 3x2 + 4x + 3(x + 1)(x + 2) = x + 2x + 3
(x + 1)(x + 2) .
Step 2: Now express 2x + 3(x + 1)(x + 2) in partial fractions. In this case it has a distinct linear factors.
2x + 3(x + 1)(x + 2) =
A(x + 1) +
B(x + 2)
Step 3: Obtain the fractions with a common denominator.2x + 3
(x + 1)(x + 2) =A(x + 2)
(x + 1)(x + 2) + B(x + 1)(x + 1)(x + 2)
Step 4: Equate numerators since the denominators are equal.
2x + 3 = A (x + 2) + B (x + 1) (*)
Step 5: Select values of x and substitute into (*).
Let x = − 1:
2x + 3 = A (x + 2) + B (x + 1)
2(−1) + 3 = A (−1 + 2) + B (−1 + 1)
1 = A
A = 1
Let x = − 2:
2x + 3 = A (x + 2) + B (x + 1)
2(−2) + 3 = A (−2 + 2) + B (−2 + 1)
−1 = −B
B = 1
Step 6: Substitute partial fractions back into the expression.
x3 + 3x2 + 4x + 3
(x + 1) (x + 2)= x +
2x + 3
(x + 1) (x + 2)
= x +1
(x + 1)+
1
(x + 2)
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Reduce improper rational functions by division exercise (page 34)
Q50:
If we expand the denominator it becomes x2 − 2x + 1. It is then clear that we are working with animproper rational function as the degree of the numerator = degree of the denominator. We musttherefore use algebraic long division before writing it as the sum of partial fractions.
Step 1: Divide using algebraic long division.
This gives 5x2 − 6x − 1(x − 1)2
= 5 + 4x − 6(x − 1)2
.
Step 2: Now express 4x − 6(x − 1)2
in partial fractions. In this case it has repeated linear factors.
4x − 6(x − 1)2
= Ax − 1 + B
(x − 1)2
Step 3: Obtain the fractions with a common denominator.4x − 6(x − 1)2
= A(x − 1)
(x − 1)2+ B
(x − 1)2
Step 4: Equate numerators since the denominators are equal.
4x − 6 = A (x − 1) + B (*)
Step 5: Select values of x and substitute into (*).
Let x = 1:4x − 6 = A (x − 1) + B
4 (1) − 6 = A (1 − 1) + B
B = −2
Now substitute B = − 2 back into (*) and select a value of x to simplify the calculation.
Let x = 0:4x − 6 = A (x − 1) + B
4 (0) − 6 = A (0 − 1) − 2
−6 = A (−1) − 2
A = 4
Therefore: 4x − 6(x − 1)2
= 4x − 1 − 2
(x − 1)2
Step 6: Substitute partial fractions back into the expression.5x2 − 6x − 1
(x − 1)2= 5 + 4
x − 1 − 2(x − 1)2
Q51:
If we expand the denominator it becomes x2 − 3x + 2. It is then clear that we are working with animproper rational function as the degree of the numerator > degree of the denominator. We musttherefore use algebraic long division before writing it as the sum of partial fractions.
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Step 1: Divide using algebraic long division.
This gives 2x3 − 5x2 − x + 3(x − 1)(x − 2) = 2x + 1 + −2x + 1
(x − 1)(x − 2) .
Step 2: Now express −2x + 1(x − 1)(x − 2) in partial fractions. In this case it has a distinct linear factors.
−2x + 1(x − 1)(x − 2) =
A(x − 1) + B
(x − 2)
Step 3: Obtain the fractions with a common denominator.−2x + 1
(x − 1)(x − 2) =A(x − 2)
(x − 1)(x − 2) + B(x − 1)(x − 1)(x − 2)
Step 4: Equate numerators since the denominators are equal.
−2x + 1 = A (x − 2) + B (x − 1) (*)
Step 5: Select values of x and substitute into (*).
Let x = 1:−2x + 1 = A (x − 2) + B (x − 1)
−2 (1) + 1 = A (1 − 2) + B (1 − 1)
−1 = A (−1)
A = 1
Let x = 2:−2x + 1 = A (x − 2) + B (x − 1)
−2 (2) + 1 = A (2 − 2) + B (2 − 1)
−3 = B (1)
B = −3
Therefore: −2x + 1(x − 1)(x − 2) =
1x − 1 − 3
x − 2
Step 6: Substitute partial fractions back into the expression.2x3 − 5x2 − x + 3
(x − 1)(x − 2) = 2x + 1 + 1x − 1 − 3
x − 2
Q52:
It is clear that we are working with an improper rational function as the degree of the numerator =degree of the denominator. We must therefore use algebraic long division before writing it as thesum of partial fractions.
Step 1: Divide using algebraic long division.
This gives −7x2 − 63x − 110−x2 − 8x − 12 = 7 + −7x − 26
−x2 − 8x − 12 .
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Step 2: Now express −7x − 26−x2 − 8x − 12
in partial fractions by first factorising the denominator.−7x − 26
−x2 − 8x − 12= −7x − 26
(−x − 2)(x + 6)
Step 3: Now express −7x − 26(−x − 2)(x + 6) in partial fractions. In this case it has a distinct linear factors.
−7x − 26(−x − 2)(x + 6) =
A−x − 2 + B
x + 6
Step 4: Obtain the fractions with a common denominator.−7x − 26
(−x − 2)(x + 6) =A(x + 6)
(−x − 2)(x + 6) + B(−x − 2)(−x − 2)(x + 6)
Step 5: Equate numerators since the denominators are equal.
−7x − 26 = A (x + 6) + B (−x − 2) (*)
Step 6: Select values of x and substitute into (*).
Let x = − 2:−7x − 26 = A (x + 6) + B (−x − 2)
−7 (−2) − 26 = A (−2 + 6) + B (− (−2) − 2)
−12 = 4A
A = −3
Let x = − 6:−7x − 26 = A (x + 6) + B (−x − 2)
−7 (−6) − 26 = A (−6 + 6) + B (− (−6) − 2)
16 = 4B
B = 4
Therefore: −7x − 26−x2 − 8x − 12
= −3−x − 2 + 4
x + 6
Step 7: Substitute partial fractions back into the expression.−7x2 − 63x − 110−x2 − 8x − 12
= 7 + −3−x − 2 + 4
x + 6
Q53:
It is clear that we are working with an improper rational function as the degree of the numerator >degree of the denominator. We must therefore use algebraic long division before writing it as thesum of partial fractions.
Step 1: Divide using algebraic long division.
This gives 3x4+12x3+16x2 + 7x + 1x3 + 4x2 + 5x + 2
= 3x + x2 + x + 1x3 + 4x2 + 5x + 2
.
Step 2: Now express x2 + x + 1x3 + 4x2 + 5x + 2
in partial fractions by first factorising the denominator.
x2 + x + 1x3 + 4x2 + 5x + 2
= x2 + x + 1(x + 2)(x + 1)2
Step 3: Now express x2 + x + 1(x + 2)(x + 1)2
in partial fractions. In this case it has a distinct linear factor and
repeated linear factors.x2 + x + 1
(x + 2)(x + 1)2= A
(x + 2) + B(x + 1) + C
(x + 1)2
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Step 4: Obtain the fractions with a common denominator.
x2 + x + 1(x + 2)(x + 1)2
= A(x + 1)2
(x + 2)(x + 1)2+ B(x + 2)(x + 1)
(x + 2)(x + 1)2+ C(x + 2)
(x + 2)(x + 1)2
Step 5: Equate numerators since the denominators are equal.
x2 + x + 1 = A(x + 1)2 + B (x + 2) (x + 1) + C (x + 2) (*)
Step 6: Select values of x and substitute into (*).
Let x = − 2:
x2 + x + 1 = A(x + 1)2 + B (x + 2) (x + 1) + C (x + 2)
(−2)2 + (−2) + 1 = A(−2 + 1)2 + B (−2 + 2) (−2 + 1) + C (−2 + 2)
4 − 2 + 1 = A(−1)2
A = 3
Let x = − 1:
x2 + x + 1 = A(x + 1)2 + B (x + 2) (x + 1) + C (x + 2)
(−1)2 + (−1) + 1 = A(−1 + 1)2 + B (−1 + 2) (−1 + 1) + C (−1 + 2)
1 = C (1)
C = 1
Now substitute A = 3 and C = 1 back into (*).
Let x = 0:x2 + x + 1 = A(x + 1)2 + B (x + 2) (x + 1) + C (x + 2)
(0)2 + (0) + 1 = 3(0 + 1)2 + B (0 + 2) (0 + 1) + 1 (0 + 2)
1 = 3 + B (2) (1) + 2
B = −2
We have A = 3, B = − 2 and C = 1.
Therefore: x2 + x + 1x3 + 4x2 + 5x + 2
= 3x + 2 − 2
x + 1 + 1(x + 1)2
Step 7: Substitute partial fractions back into the expression.3x4+12x3+16x2 + 7x + 1
x3 + 4x2 + 5x + 2 = 3x + 3x + 2 − 2
x + 1 + 1(x + 1)2
Q54:
It is clear that we are working with an improper rational function as the degree of the numerator >degree of the denominator. We must therefore use algebraic long division before writing it as thesum of partial fractions.
Step 1: Divide using algebraic long division.
This gives x5 − 2x4 + x3 + x2 − 10x + 5x3 − 2x2 − x + 2
= x2 + 2 + 3x2 − 8x + 1x3 − 2x2 − x + 2
.
Step 2: Now express 3x2 − 8x + 1x3 − 2x2 − x + 2
in partial fractions by first factorising the denominator.
3x2 − 8x + 1x3 − 2x2 − x + 2
= 3x2 − 8x + 1(x + 1)(x − 1)(x − 2)
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Step 3: Now express 3x2 − 8x + 1(x + 1)(x − 1)(x − 2) in partial fractions. In this case it has a distinct linear factors.
3x2 − 8x + 1(x + 1)(x − 1)(x − 2) =
A(x + 1) + B
(x − 1) + C(x − 2)
Step 4: Obtain the fractions with a common denominator.3x2 − 8x + 1
(x + 1)(x − 1)(x − 2) =A(x − 1)(x − 2)
(x + 1)(x − 1)(x − 2) + B(x + 1)(x − 2)(x + 1)(x − 1)(x − 2) + C(x + 1)(x − 1)
(x + 1)(x − 1)(x − 2)
Step 5: Equate numerators since the denominators are equal.
3x2 − 8x + 1 = A (x − 1) (x − 2) + B (x + 1) (x − 2) + C (x + 1) (x − 1) (*)
Step 6: Select values of x and substitute into (*).
Let x = − 1:
3x2 − 8x + 1 = A (x − 1) (x − 2) + B (x + 1) (x − 2) + C (x + 1) (x − 1)
3(−1)2 − 8 (−1) + 1 = A (−1 − 1) (−1 − 2) + B (−1 + 1) (−1 − 2) + C (−1 + 1) (−1 − 1)
3 + 8 + 1 = A (−2) (−3)
12 = 6A
A = 2
Let x = 1:3x2 − 8x + 1 = A (x − 1) (x − 2) + B (x + 1) (x − 2) + C (x + 1) (x − 1)
3(1)2 − 8 (1) + 1 = A (1 − 1) (1 − 2) + B (1 + 1) (1 − 2) + C (1 + 1) (1 − 1)
3 − 8 + 1 = B (2) (−1)
−4 = −2B
B = 2
Let x = 2:3x2 − 8x + 1 = A (x − 1) (x − 2) + B (x + 1) (x − 2) + C (x + 1) (x − 1)
3(2)2 − 8 (2) + 1 = A (2 − 1) (2 − 2) + B (2 + 1) (2 − 2) + C (2 + 1) (2 − 1)
12 − 16 + 1 = C (3) (1)
−3 = 3C
C = −1
Therefore: 3x2 − 8x + 1x3 − 2x2 − x + 2
= 2x + 1 + 2
x − 1 − 1x − 2
Step 7: Substitute partial fractions back into the expression.x5 − 2x4 + x3 + x2 − 10x + 5
x3 − 2x2 − x + 2 = x2 + 2 + 2x + 1 + 2
x − 1 − 1x − 2
Q55:
If we expand the denominator it becomes x3 − 6x2 + 12x − 8. It is then clear that we are workingwith an improper rational function as the degree of the numerator = degree of the denominator. Wemust therefore use algebraic long division before writing it as the sum of partial fractions.
Step 1: Divide using algebraic long division.
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This gives x3 − 5x2 + 9x − 11(x − 2)3
= 1 + x2 − 3x − 3(x − 2)3
.
Step 2: Now express x2 − 3x − 3(x − 2)3
in partial fractions. In this case it has repeated linear factors.
x2 − 3x − 3(x − 2)3
= Ax − 2 + B
(x − 2)2+ C
(x − 2)3
Step 3: Obtain the fractions with a common denominator.
x2 − 3x − 3(x − 2)3
= A(x − 2)2
(x − 2)3+ B(x − 2)
(x − 2)3+ C
(x − 2)3
Step 4: Equate numerators since the denominators are equal.
x2 − 3x − 3 = A(x − 2)2 + B (x − 2) + C (*)
Step 5: Select values of x and equate coefficients.
First we will select a value for x and substitute into (*).
Let x = 2:x2 − 3x − 3 = A(x − 2)2 + B (x − 2) + C
(2)2 − 3 (2) − 3 = A(2 − 2)2 + B (2 − 2) + C
C = −5
Now substitute C = − 5 back into (*), expand the brackets and collect like terms.
x2 − 3x − 3 = A(x − 2)2 + B (x − 2) + C
= A(x2 − 4x − 4
)+ B (x − 2) − 5
= Ax2 − 4Ax − 4A + Bx − 2B − 5
= Ax2 + (−4A + B) x − 4A − 2B − 5
Set the coefficient in front of x2 on the LHS equal to the coefficient in front of x2 on the RHS.
1 = A
Set the coefficient in front of x on the LHS equal to the coefficient in front of x on the RHS.
−3 = − 4A + B so B = 1 since we know A = 1.
We have A = 1, B = 1 and C = − 5.
Therefore: x2 − 3x − 3(x − 2)3
= 1x − 2 + 1
(x − 2)2− 5
(x − 2)3
Step 6: Substitute partial fractions back into the expression.x3 − 5x2 + 9x − 11
(x − 2)3= 1 + 1
x − 2 + 1(x − 2)2
− 5(x − 2)3
Q56:
It is clear that we are working with an improper rational function as the degree of the numerator >degree of the denominator. We must therefore use algebraic long division before writing it as thesum of partial fractions.
Step 1: Divide using algebraic long division.
This gives 2x4 + 9x3 + 22x2 + 27x + 152x3 + 7x2 + 5x + 6 = x + 1 + 10x2 + 16x + 9
2x3 + 7x2 + 5x + 6 .
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76 ANSWERS: UNIT 1 TOPIC 1
Step 2: Now express 10x2 + 16x + 92x3 + 7x2 + 5x + 6
in partial fractions by first factorising the denominator.
10x2 + 16x + 92x3 + 7x2 + 5x + 6
= 10x2 + 16x + 9(x + 3)(2x2 + x + 2)
Step 3: Now express 10x2 + 16x + 9(x + 3)(2x2 + x + 2)
in partial fractions. In this case it has a distinct linear factorand an irreducible quadratic factor.
10x2 + 16x + 9(x + 3)(2x2 + x + 2)
= A(x + 3) + Bx + C
(2x2 + x + 2)
Step 4: Obtain the fractions with a common denominator.
10x2 + 16x + 9(x + 3)(2x2 + x + 2) =
A(2x2 + x + 2)(x + 3)(2x2 + x + 2) + (Bx + C)(x + 3)
(x + 3)(2x2 + x + 2)
Step 5: Equate numerators since the denominators are equal.
10x2 + 16x + 9 = A(2x2 + x + 2
)+ (Bx + C) (x + 3) (*)
Step 6: Select values of x and substitute into (*).
Let x = − 3:
10x2 + 16x + 9 = A(2x2 + x + 2
)+ (Bx + C) (x + 3)
10(−3)2 + 16 (−3) + 9 = A(2(−3)2 + (−3) + 2
)+ (B (−3) + C) (−3 + 3)
90 − 48 + 9 = A (18 − 3 + 2)
51 = 17A
A = 3
Now substitute A = 3 back into (*) and select another value of x to simplify the calculation.
Let x = 0:10x2 + 16x + 9 = A
(2x2 + x + 2
)+ (Bx + C) (x + 3)
10(0)2 + 16 (0) + 9 = 3(2(0)2 + (0) + 2
)+ (B (0) + C) (0 + 3)
9 = 6 + 3C
C = 1
Now substitute A = 3 and C = 1 back into (*) and select another value of x to simplify thecalculation.
Let x = 1:10x2 + 16x + 9 = A
(2x2 + x + 2
)+ (Bx + C) (x + 3)
10(1)2 + 16 (1) + 9 = 3(2(1)2 + (1) + 2
)+ (B (1) + 1) (1 + 3)
35 = 19 + 4B
16 = 4B
B = 4
We have A = 3, B = 4 and C = 1.
Therefore: 10x2 + 16x + 92x3 + 7x2 + 5x + 6
= 3x + 3 + 4x + 1
2x2 + x + 2
Step 7: Substitute partial fractions back into the expression.2x4 + 9x3 + 22x2 + 27x + 15
2x3 + 7x2 + 5x + 6 = x + 1 + 3x + 3 + 4x + 1
2x2 + x + 2
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End of topic 1 test (page 37)
Q57: 3x − 1 + 2
x − 2
Q58: − 1x + 3 − 5
x + 1
Q59: − 3x − 2 − 1
x + 4
Q60: − 1x + 1 + 3
x + 2 + 2x − 3
Q61: 9x − 1 + 13
x − 2 + 1x + 1
Q62: 2x − 2 − 3
(x − 2)2
Q63: 2x + 3 + 2
(x + 3)2
Q64: 3x − 1 − 2
x + 2 − 3(x + 2)2
Q65: 2x − 2 + 3
(x − 2)2+ 4
(x − 2)3
Q66: 3x + 1 + x + 2
x2 + 4x + 7
Q67: − 72(x − 1) + 7x + 9
2(x2 − 2x + 3)
Q68: x − 3 + 7x + 6x2 + 3x + 2
Q69: x − 3 − 1x + 1 + 8
x + 2
Q70: 2x2 + 4x + 6 − 5x − 8x2 − 3x + 2
Q71: 2x2 + 4x + 6 − 3x − 1 − 2
x − 2
Q72: 4x + 4x − 11x2 − 4x + 4
Q73: 4x + 4x − 2 − 3
(x − 2)2
Q74: −3x+ 5− x2−x+7x3−2x2+2x−5
Q75: −3x + 5 − 1x − 1 − 2
x2 − 3x + 5
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