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Core Mathematics 2 Algebra 1
Core Mathematics 2
Algebra
Edited by: K V Kumaran
Email: kvkumaran@gmail.com
Core Mathematics 2 Algebra 2
Algebra and functions
Simple algebraic division; use of the Factor Theorem and the
Remainder Theorem.
Only division by (x + a) or (x – a) will be required.
Students should know that if
f(x) = 0 when x = a, then (x – a) is a factor of f(x).
Students may be required to factorise cubic expressions
such as
x3 + 3x2 – 4 and 6x3 + 11x2 – x – 6.
Students should be familiar with the terms ‘quotient’ and
‘remainder’ and be able to determine the remainder when
the polynomial f(x) is divided by (ax + b).
Core Mathematics 2 Algebra 3
Simplifying Algebraic Fractions
Algebraic fractions can be simplified by division.
Example 1: Simplify 3x
3+ 5x
2– 2x
x =
3x3
x+
5x2
x–
2x
x
= 3x2
+ 5x – 2
Example 1: Simplify 3x
3+ 5x
2– 2x
x
= 3x
3
x+
5x2
x–
2x
x
= 3x2
+ 5x – 2
Example 2: Simplify 6x
4– 5x
3+ 2x
2
– 2x =
6x4
– 2x–
5x3
– 2x+
2x2
– 2x = – 3x
3+
5x2
2– x
Example 2: Simplify 6x
4– 5x
3+ 2x
2
– 2x
= 6x
4
– 2x–
5x3
– 2x+
2x2
– 2x
= – 3x3
+5x
2
2– x
Factorising to Cancel
Sometimes you need to factorise before you can simplify
Example 1: Simplify x
2+ 7x + 12
x + 3
= (x + 3)(x + 4)
x + 3
= x + 4
Example 1: Simplify x
2+ 7x + 12
x + 3
= (x + 3)(x + 4)
x + 3
= x + 4
Example 2: Simplify 2x
2+ 5x – 12
2x2
– 7x + 6
= (2x – 3)(x + 4)
(x – 2)(2x – 3) =
x + 4
x – 2
Example 2: Simplify 2x
2+ 5x – 12
2x2
– 7x + 6
= (2x – 3)(x + 4)
(x – 2)(2x – 3)
= x + 4
x – 2
________
Core Mathematics 2 Algebra 4
Dividing a Polynomial by (x±p)
You divide polynomials in the same way as you perform long division.
Example 1: Divide 6x3
+ 28x2
– 7x + 15 by (x + 5) 6x2
– 2x + 3 = (x + 5) 6x3
+ 28x2
– 7x + 15 – 6x3
+ 30x2 – 2x
2– 7x – – 2x
2– 10x 3x + 15 – 3x + 15 0
6x
3+ 28x
2– 7x + 15
x + 5 = 6x
2– 2x + 3
Example 1: Divide 6x3
+ 28x2
– 7x + 15 by (x + 5)
6x2
– 2x + 3
= (x + 5) 6x3
+ 28x2
– 7x + 15
– 6x3
+ 30x2
– 2x2
– 7x
– – 2x2
– 10x
3x + 15
– 3x + 15
0
6x
3+ 28x
2– 7x + 15
x + 5 = 6x
2– 2x + 3
Example 2: Divide – 5x3
– 27x2
+ 23x + 30 by (x + 6) – 5x2 + 3x + 5 = (x + 6) – 5x
3– 27x
2+ 23x + 30 – – 5x
3– 30x
2 3x
2+ 23x – 3x
2+ 18x 5x + 30 – 5x + 30 0
– 5x
3– 27x
2+ 23x + 30
x + 6 = – 5x
2+ 3x + 5
Example 2: Divide – 5x3
– 27x2
+ 23x + 30 by (x + 6)
– 5x2 + 3x + 5
= (x + 6) – 5x3
– 27x2
+ 23x + 30
– – 5x3
– 30x2
3x2
+ 23x
– 3x2
+ 18x
5x + 30
– 5x + 30
0
– 5x
3– 27x
2+ 23x + 30
x + 6 = – 5x
2+ 3x + 5
____________
____________
___________
___________
__________
____________
Core Mathematics 2 Algebra 5
Finding the Remainder
After a division of a polynomial, if there is anything left then it is called the remainder. The answer
is called the quotient. If there is no remainder then what you are dividing by is said to be a factor.
Example 1: Divide 2x3
– 5x2
– 16x + 10 by (x – 4) and state the remainder and quotient 2x2
+ 3x – 4 = (x – 4) 2x3
– 5x2
– 16x + 10 – 2x3
– 8x2 3x
2– 16x – 3x
2– 12x – 4x + 10 – – 4x + 16 – 6 The remainder is – 6 and the quotient is 2x
2+ 3x – 4
Example 1: Divide 2x3
– 5x2
– 16x + 10 by (x – 4) and state the remainder and quotient
2x2
+ 3x – 4
= (x – 4) 2x3
– 5x2
– 16x + 10
– 2x3
– 8x2
3x2
– 16x
– 3x2
– 12x
– 4x + 10
– – 4x + 16
– 6
The remainder is – 6 and the quotient is 2x2
+ 3x – 4
Example 2: Divide 2x3
+ 9x2
+ 25 by (x + 5) and state the remainder and quotient 2x2 – x + 5 = (x + 5) 2x
3+ 9x
2+ 0x + 25 – 2x
3+ 10x
2 – x
2+ 0x – – x
2– 5x 5x + 25 – 5x + 25 0 The remainder is 0 and the quotient is 2x
2– x + 5
Example 2: Divide 2x3
+ 9x2
+ 25 by (x + 5) and state the remainder and quotient
2x2 – x + 5
= (x + 5) 2x3
+ 9x2
+ 0x + 25
– 2x3
+ 10x2
– x2
+ 0x
– – x2
– 5x
5x + 25
– 5x + 25
0
The remainder is 0 and the quotient is 2x2
– x + 5
____________
____________
____________
________
________
________
Remember to use 0x
as there is no x term.
Core Mathematics 2 Algebra 6
Factor Theorem
Notes:
Factor Theorem: (x – a) is a factor of a polynomial f(x) if f(a) = 0.
Remainder Theorem: The remainder when a polynomial f(x) is divided by (x – a) is f(a).
Extended version of the factor theorem:
(ax + b) is a factor of a polynomial f(x) if 0b
fa
.
To find out the factors of a polynomial you can quickly substitute in values of x to see which give
you a value of zero.
Example 1: Given f(x) = x3
+ x2
– 4x – 4. Use the factor theorem to find a factor f(x) = x3
+ x2
– 4x – 4 f(1) = (1)3
+ (1)2
– (4 1) – 4 = – 6 (x – 1) is not a factor f(2) = (2)3
+ (2)2
– (4 2) – 4 = 0 (x – 2) is a factor
Example 1: Given f(x) = x3
+ x2
– 4x – 4. Use the factor theorem to find a factor
f(x) = x3
+ x2
– 4x – 4
f(1) = (1)3
+ (1)2
– (4 1) – 4
= – 6 (x – 1) is not a factor
f(2) = (2)3
+ (2)2
– (4 2) – 4
= 0 (x – 2) is a factor
To go from the x value to the factor simply put it into a bracket and change the sign.
Remember this:-
(x + 3)(x – 2) = 0 x = – 3 and x = 2
(x + 3)(x – 2) = 0
x = – 3 and x = 2
We are now going in reverse by putting the x
value back into the brackets
Core Mathematics 2 Algebra 7
Example 2: Given f(x) = 3x2
+ 8x2 + 3x – 2. Factorise fully the given function f(x) = 3x
3+ 8x
2+ 3x – 2 f(1) = 3 (1)
3+ 8 (1)
2+ (3 1) – 2 = 12 (x – 1) is not a factor f(2) = 3 (2)
3+ 8 (2)
2+ (3 2) – 2 = 60 (x – 2) is not a factor f( – 1) = 3 ( – 1)
3+ 8 ( – 1)
2+ (3 – 1) – 2 = 0 (x + 1) is a factor
Example 2: Given f(x) = 3x2
+ 8x2 + 3x – 2. Factorise fully the given function
f(x) = 3x3
+ 8x2
+ 3x – 2
f(1) = 3 (1)3
+ 8 (1)2
+ (3 1) – 2
= 12 (x – 1) is not a factor
f(2) = 3 (2)3
+ 8 (2)2
+ (3 2) – 2
= 60 (x – 2) is not a factor
f( – 1) = 3 ( – 1)3
+ 8 ( – 1)2
+ (3 – 1) – 2
= 0 (x + 1) is a factor
To fully factorise it we now need to find the quotient. We do this by dividing by the factor we have
just found.
3x2
+ 5x – 2 = (x + 1) 3x3
+ 8x2
+ 3x – 2 – 3x3
+ 3x2 5x2
+ 3x – 5x2
+ 5x – 2x – 2 – – 2x – 2 0 (x + 1)(3x2
+ 5x – 2) = (x + 1)(3x – 1)(x + 2)
3x2
+ 5x – 2
= (x + 1) 3x3
+ 8x2
+ 3x – 2
– 3x3
+ 3x2
5x2
+ 3x
– 5x2
+ 5x
– 2x – 2
– – 2x – 2
0
(x + 1)(3x2
+ 5x – 2) = (x + 1)(3x – 1)(x + 2)
___________
___________
__________
_________
_________
_________
We have factorised the
quotient further to get the
fully factorised answer
Core Mathematics 2 Algebra 8
Using the Factor Theorem to find the Remainder
If the polynomial f(x) is divided by (ax – b) then the remainder is f
b
a
If the polynomial f(x) is divided by (ax – b) then the remainder is f
b
a
Example 1: Find the remainder when f(x) = 4x4
– 4x2
+ 8x – 1 is divided by (2x – 1)
Using (ax – b) then a = 2, b = 1 so f
1
2
f(x) = 4x4
– 4x2
+ 8x – 1
f
1
2
= 4
1
2
4– 4
1
2
2+
8
1
2
– 1
=4
16 –
4
4 +
8
2 – 1 =
1
4 – 1 + 4 – 1 = 2 1
4 Remainder is 2
1
4 Example 2: Find the remainder if f(x) = x
3– 20x + 3 is divided by (x – 4) Using (ax – b) then a = 1, b = 4 so f(4) f(x) = x
3– 20x + 3 f(4) = (4)
3– (20 4) + 3 = 64 – 80 + 3 = – 13 Remainder is – 13
Example 1: Find the remainder when f(x) = 4x4
– 4x2
+ 8x – 1 is divided by (2x – 1)
Using (ax – b) then a = 2, b = 1 so f
1
2
f(x) = 4x4
– 4x2
+ 8x – 1
f
1
2
= 4
1
2
4– 4
1
2
2+
8
1
2
– 1
=4
16 –
4
4 +
8
2 – 1
=1
4 – 1 + 4 – 1
= 2 1
4
Remainder is 2 1
4
Example 2: Find the remainder if f(x) = x3
– 20x + 3 is divided by (x – 4)
Using (ax – b) then a = 1, b = 4 so f(4)
f(x) = x3
– 20x + 3
f(4) = (4)3
– (20 4) + 3
= 64 – 80 + 3
= – 13
Remainder is – 13
Core Mathematics 2 Algebra 9
(C2-1.1) Name:
Homework Questions 1 – Simplifying Algebraic Fractions
Simplify the following by division
a)
3x3
+ 2x2
+ x
x
3x3
+ 2x2
+ x
x
b)
5x3
+ 4x2
– 2x
– 2x
5x3
+ 4x2
– 2x
– 2x
c)
4x3
+ 3x2
+ 7x
x
4x3
+ 3x2
+ 7x
x
d)
– 8x4
+ 6x3
– 6x2
– 2x2
– 8x4
+ 6x3
– 6x2
– 2x2
e)
5x2
+ 6x – 3
x2
5x2
+ 6x – 3
x2
f)
7x3
– 3x2
+ 5x
x2
7x3
– 3x2
+ 5x
x2
g)
3x5
– 6x3
2x2
3x5
– 6x3
2x2
h)
7x5
+ 3x2
+ 6
3x2
7x5
+ 3x2
+ 6
3x2
3
i)
4x2
+ 6x – 5
2x
4x2
+ 6x – 5
2x
j)
12x3
– 9x2
+ 3x
3x2
12x3
– 9x2
+ 3x
3x2
Core Mathematics 2 Algebra 10
(C2-1.2) Name:
Homework Questions 2 – Factorising to Cancel
Simplify the following by factorizing first
a)
x2
+ 5x + 6
x + 2
x2
+ 5x + 6
x + 2
b)
x2
+ 6x – 7
x – 1
x2
+ 6x – 7
x – 1
c)
x2
+ 2x – 15
x – 3
x2
+ 2x – 15
x – 3
d)
x2
+ 10x + 24
x + 6
x2
+ 10x + 24
x + 6
e)
2x2
– 10x + 12
x – 3
2x2
– 10x + 12
x – 3
f)
x2
– 9x + 20
x – 4
x2
– 9x + 20
x – 4
g)
x2
– 49
x – 7
x2
– 49
x – 7
h)
x2
+ 10x + 9
x + 1
x2
+ 10x + 9
x + 1
i)
4x2
+ 12x + 9
2x + 3
4x2
+ 12x + 9
2x + 3
j)
7x3y – 28xy
3
x – 2y
7x3y – 28xy
3
x – 2y
Core Mathematics 2 Algebra 11
(C2-1.3) Name:
Homework Questions 3 – Dividing a Polynomial
Divide the following to find the quotient
a)
2x3
+ 3x2
– 3x – 2 by (x + 2)
2x3
+ 3x2
– 3x – 2 by (x + 2)
b)
3x3
– 2x2
– 19x – 6 by (x – 3)
3x3
– 2x2
– 19x – 6 by (x – 3)
c)
x3
+ 7x2
– 6x – 72 by (x – 3)
x3
+ 7x2
– 6x – 72 by (x – 3)
e)
4x3
+ 3x2
– 25x – 6 by (x – 2)
4x3
+ 3x2
– 25x – 6 by (x – 2)
Core Mathematics 2 Algebra 12
(C2-1.4) Name:
Homework Questions 4 – Finding the Remainder
Find the remainder by division when the following polynomials are divided by:-
a)
x2
– 4x + 5 by (x – 1)
x2
– 4x + 5 by (x – 1)
b)
2x3
+ 3x2
– 3x + 2 by (x + 2)
2x3
+ 3x2
– 3x + 2 by (x + 2)
c)
2x2
+ 7x – 3 by (2x – 1)
2x2
+ 7x – 3 by (2x – 1)
d)
x3
– 2x2
– 7x – 1 by (x + 1)
x3
– 2x2
– 7x – 1 by (x + 1)
Core Mathematics 2 Algebra 13
(C2-1.5) Name:
Homework Questions 5 – Factor Theorem
Use the factor theorem to show that the following are factors, and hence fully factorise.
a)
(x – 3) is a factor of x3
– 3x2
– 4x + 12
(x – 3) is a factor of x3
– 3x2
– 4x + 12
b)
(x + 2) is a factor of x3
– 3x2
– 6x + 8
(x + 2) is a factor of x3
– 3x2
– 6x + 8
c)
(x + 1) is a factor of x3
+ 8x2
+ 19x + 12
(x + 1) is a factor of x3
+ 8x2
+ 19x + 12
d)
(x – 2) is a factor of x3
– 4x2
– 11x + 30
(x – 2) is a factor of x3
– 4x2
– 11x + 30
e)
(x + 3) is a factor of x3
+ 2x2
– 5x – 6
(x + 3) is a factor of x3
+ 2x2
– 5x – 6
Core Mathematics 2 Algebra 14
(C2-1.6) Name:
Homework Questions 6 – Finding Remainder Using Factor Theorem
Find the remainder when the following polynomial is divided by:-
a)
3x3
– 2x2
+ 5x – 12 by (x – 1)
3x3
– 2x2
+ 5x – 12 by (x – 1)
b)
7x3
– 5x2
+ 12x + 16 by (x + 2)
7x3
– 5x2
+ 12x + 16 by (x + 2)
c)
6x3
+ 4x2
– 9x + 10 by (x + 1)
6x3
+ 4x2
– 9x + 10 by (x + 1)
d)
5x5
– 4x3
+ 2x2
– 8x + 11 by (x + 3)
5x5
– 4x3
+ 2x2
– 8x + 11 by (x + 3)
e)
7x4
+ 7x3
– 6x2
+ 9x + 2 by (x – 2)
7x4
+ 7x3
– 6x2
+ 9x + 2 by (x – 2)
f)
5x3
– 5x2
+ 9x + 12 by (x + 1)
5x3
– 5x2
+ 9x + 12 by (x + 1)
g)
7x4
– 4x3
+ 5x2
– 6x + 15 by (x + 2)
7x4
– 4x3
+ 5x2
– 6x + 15 by (x + 2)
h)
4x3
+ 6x2
+ 3x + 4 by (x – 3)
4x3
+ 6x2
+ 3x + 4 by (x – 3)
i)
7x4
– 9x3
– 8x2
– 7x + 5 by (x – 1)
7x4
– 9x3
– 8x2
– 7x + 5 by (x – 1)
j)
6x5
+ 4x3
+ 2x2
– x – 4 by (x – 3)
6x5
+ 4x3
+ 2x2
– x – 4 by (x – 3)
Core Mathematics 2 Algebra 15
Past examination questions.
1. f(x) = x3 – 2x2 + ax + b, where a and b are constants.
When f(x) is divided by (x – 2), the remainder is 1.
When f(x) is divided by (x + 1), the remainder is 28.
(a) Find the value of a and the value of b. (6)
(b) Show that (x – 3) is a factor of f(x). (2)
(C2, Jan 2005 Q5)
2. (a) Use the factor theorem to show that (x + 4) is a factor of 2x3 + x2 – 25x + 12.
(2)
(b) Factorise 2x3 + x2 – 25x + 12 completely.
(4)
(C2, June 2005 Q3)
3. f(x) = 2x3 + x2 – 5x + c, where c is a constant.
Given that f(1) = 0,
(a) find the value of c,
(2)
(b) factorise f(x) completely,
(4)
(c) find the remainder when f(x) is divided by (2x – 3).
(2)
(C2, Jan 2006 Q1)
4. f(x) = 2x3 + 3x2 – 29x – 60.
(a) Find the remainder when f(x) is divided by (x + 2).
(2)
(b) Use the factor theorem to show that (x + 3) is a factor of f(x).
(2)
(c) Factorise f(x) completely.
(4)
(C2, May 2006 Q4)
5. f(x) = x3 + 4x2 + x – 6.
(a) Use the factor theorem to show that (x + 2) is a factor of f(x). (2)
(b) Factorise f(x) completely.
Core Mathematics 2 Algebra 16
(4)
(c) Write down all the solutions to the equation
x3 + 4x2 + x – 6 = 0. (1)
(C2, Jan 2007 Q5)
6. f(x) = 3x3 – 5x2 – 16x + 12.
(a) Find the remainder when f(x) is divided by (x – 2).
(2)
Given that (x + 2) is a factor of f(x),
(b) factorise f(x) completely.
(4)
(C2, May 2007 Q2)
7. (a) Find the remainder when
x3 – 2x2 – 4x + 8
is divided by
(i) x – 3,
(ii) x + 2.
(3)
(b) Hence, or otherwise, find all the solutions to the equation
x3 – 2x2 – 4x + 8 = 0.
(4)
(C2, Jan 2008 Q1)
8. f(x) = 2x3 – 3x2 – 39x + 20
(a) Use the factor theorem to show that (x + 4) is a factor of f (x).
(2)
(b) Factorise f (x) completely.
(4)
(C2, June 2008 Q1)
9. f (x) = x4 + 5x3 + ax + b,
where a and b are constants.
The remainder when f(x) is divided by (x – 2) is equal to the remainder when f(x) is divided by (x +
1).
(a) Find the value of a. (5)
Core Mathematics 2 Algebra 17
Given that (x + 3) is a factor of f(x),
(b) find the value of b. (3)
(C2, Jan 2009 Q6)
10. f(x) = (3x − 2)(x − k) − 8
where k is a constant.
(a) Write down the value of f (k).
(1)
When f(x) is divided by (x − 2) the remainder is 4.
(b) Find the value of k.
(2)
(c) Factorise f (x) completely.
(3)
(C2, June 2009 Q3)
11. f(x) = 2x3 + ax2 + bx – 6,
where a and b are constants.
When f(x) is divided by (2x – 1) the remainder is –5.
When f(x) is divided by (x + 2) there is no remainder.
(a) Find the value of a and the value of b.
(6)
(b) Factorise f(x) completely.
(3)
(C2, Jan 2010 Q3)
12. f(x) = 3x3 − 5x2 − 58x + 40.
(a) Find the remainder when f (x) is divided by (x − 3) .
(2)
Given that (x − 5) is a factor of f(x) ,
(b) find all the solutions of f(x) = 0 .
(5)
(C2, June 2010 Q2)
Core Mathematics 2 Algebra 18
13. f(x) = x4 + x3 + 2x2 + ax + b,
where a and b are constants.
When f(x) is divided by (x – 1), the remainder is 7.
(a) Show that a + b = 3.
(2)
When f(x) is divided by (x + 2), the remainder is 8.
(b) Find the value of a and the value of b.
(5)
(C2, Jan 2011 Q1)
14. f(x) = 2x3 − 7x2 − 5x + 4
(a) Find the remainder when f(x) is divided by (x – 1).
(2)
(b) Use the factor theorem to show that (x + 1) is a factor of f(x).
(2)
(c) Factorise f(x) completely.
(4)
(C2, May 2011 Q1)
15. f(x) = x3 + ax2 + bx + 3, where a and b are constants.
Given that when f (x) is divided by (x + 2) the remainder is 7,
(a) show that 2a − b = 6.
(2)
Given also that when f(x) is divided by (x −1) the remainder is 4,
(b) find the value of a and the value of b.
(4)
(C2, Jan 2012 Q5)
16. f(x) = 2x3 – 7x2 – 10x + 24.
(a) Use the factor theorem to show that (x + 2) is a factor of f(x).
(2)
(b) Factorise f(x) completely.
(4)
(C2, May 2012 Q4)
Core Mathematics 2 Algebra 19
17. f(x) = ax3 + bx2 − 4x − 3, where a and b are constants.
Given that (x – 1) is a factor of f(x),
(a) show that a + b = 7.
(2)
Given also that, when f(x) is divided by (x + 2), the remainder is 9,
(b) find the value of a and the value of b, showing each step in your working.
(4)
(C2, Jan 2013 Q2)
18. f(x) = 2x3 – 5x2 + ax + 18
where a is a constant.
Given that (x – 3) is a factor of f(x),
(a) show that a = –9,
(2)
(b) factorise f(x) completely.
(4)
Given that
g(y) = 2(33y) – 5(32y) – 9(3y) + 18,
(c) find the values of y that satisfy g(y) = 0, giving your answers to 2 decimal places where
appropriate.
(3)
(C2, May 2013 Q3)
19. f(x) = –4x3 + ax2 + 9x – 18, where a is a constant.
Given that (x – 2) is a factor of f(x),
(a) find the value of a,
(2)
(b) factorise f(x) completely,
(3)
(c) find the remainder when f(x) is divided by (2x – 1).
(2)
(C2, May 2014 Q4)
Core Mathematics 2 Algebra 20
20. f(x) = 2x3 – 7x2 + 4x + 4.
(a) Use the factor theorem to show that (x – 2) is a factor of f(x).
(2)
(b) Factorise f(x) completely.
(4)
(C2, May 2014 Q2)
21. f(x) = 6x3 + 3x2 + Ax + B, where A and B are constants.
Given that when f(x) is divided by (x + 1) the remainder is 45,
(a) show that B – A = 48.
(2)
Given also that (2x + 1) is a factor of f(x),
(b) find the value of A and the value of B.
(4)
(c) Factorise f(x) fully.
(3)
(C2, May 2015 Q3)
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