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Unit 10: Factor Theorem
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1
MATHEMATICS
Learner’s Study and
Revision Guide for
Grade 12
FACTOR THEOREM
Revision Notes, Exercises and Solution Hints by
Roseinnes Phahle
Examination Questions by the Department of Basic Education
Preparation for the Mathematics examination brought to you by Kagiso Trust
Contents
Unit 10
Remainder Theorem 3
Factor Theorem 3
Division 4
Sketching cubic functions 5
Exercise 10 6
Answers 7
This Unit is in preparation for Unit 11 which deals with applications of differentiation to finding the coordinates of the maximum and minimum turning points of the graphs of cubic functions. Units 10
and 11 must therefore be studied together.
How to use this revision and study guide
1. Study the revision notes given at the beginning. The notes are interactive in that in some parts you are required to make a response based on your prior learning of the topic in class or from a textbook.
2. “Warm‐up” exercises follow the notes. Some exercises carry solution HINTS in the answer section. Do not read the answer or hints until you have tried to work out a question and are having difficulty.
3. The notes and exercises are followed by questions from past examination papers.
4. The examination questions are followed by blank spaces or boxes inside a table. Do the working out of the question inside these spaces or boxes.
5. Alongside the blank boxes are HINTS in case you have difficulty solving a part of the question. Do not read the hints until you have tried to work out the question and are having difficulty.
6. What follows next are more questions taken from past examination papers.
7. Answers to the extra past examination questions appear at the end. Some answers carry HINTS and notes to enrich your knowledge.
8. Finally, don’t be a loner. Work through this guide in a team with your classmates.
Factor theorem
REVISION UNIT 10: THE REMAINDER AND FACTOR THEOREMS
THE REMAINDER THEORM
Look at the division of numbers, say 17 by 5.
We say that the number of times 5 divides into 17 is 3 plus a remainder of 2. We could write this as 17 = 5X3+2. In Exercise 1.4 you brushed up on your knowledge of algebraic division.
Example: If ( ) 3532 23 ++−= xxxxf , verify that the “number of times” 1+x divides into ( )xf is
1052 2 +− xx plus remainder ‐7. You work out this example.
In the same way as we did with the division of a number by a number, we can write that
( )( ) ( )7105213532 223 −++−+=++− xxxxxx
or ( ) ( )( ) 710521 2 −+−+= xxxxf
or ( ) ( ) ( ) RQ1 +×+= xxxf
NOTE: If ( )xf is a cubic and we are dividing by a linear function, then ( )xQ must be a quadratic. Why?
Is R a function or a constant in this case?
If we now replace x by ‐1, we get:
( ) ( ) ( ) R1Q111 +−×+−=−f
which is ( ) ( ) R1Q01 +−×=−f
or simply ( ) R1 =−f , that is when ( )xf is divided by ( )1+x the remainder is ( )1−f .
In general, when ( )xf is divided by ( )α−x then the remainder is ( )αf .
THE FACTOR THEOREM
What if the remainder is zero? In other words, what if ( )α−x divides exactly into ( )xf without leaving
a remainder?
Then because R = 0, we will have
( ) ( ) ( )xxxf Q⋅−= α
which shows that ( )α−x is then a factor of ( )xf .
Preparation for the Mathematics examination brought to you by Kagiso Trust
DIVISION
What we noted above is that if ( )xf is a cubic and ( )α−x is its factor, we can write
( ) ( ) ( )cba 2 ++⋅−= xxxxf α
The values of a, b and c can then be determined by one of two methods:
Either by
1. dividing ( )xf by ax − ; or
2. equating coefficients as demonstrated below.
Example: Factorise 15162 23 +−− xxx .
First put ( )xf = 15162 23 +−− xxx
We want a value of x that makes ( ) 0=xf .
The coefficient of 3x and the constant term in ( )xf give a clue as to what values to try.
The coefficient of 3x is 2 and its factors suggest we try values such as 1± or 2± .
The constant term is 15 and its factors suggest we try values such as 1± or 3± or 5± or 15± .
Try 1−=x : ( ) ( ) ( ) ( ) 0281516121511611.21 23 ≠=++−−=+−−−−−=−f
So 1+x is not a factor of ( )xf .
Try 1=x : ( ) 01717151612151.1611.21 23 =−=+−−=+−−=f
Therefore, 1−x is a factor of ( )xf .
Method of division (it is left as an exercise for you to carry out the division below):
151621 23 +−−− xxxx
Factor theorem
Having done the division, can you complete the following statement:
( )( ) 115162 23 −=+−− xxxx
Method of equating coefficients:
We can write
15162 23 +−− xxx = ( )( )cba1 2 ++− xxx
You now multiply out the RHS and then compare coefficients of like terms on either side. For example
the coefficient of the 3x term on the LHS must equal the coefficient of the 3x term on the RHS; and so
on with the 2x terms, the x terms and the constants.
However, it may be quicker to determine these values by inspection.
On inspection, it is obvious that a = 2 and c = ‐15. Do you agree?
Now find the value of b by equating coefficients of either 2x or x .
Coefficient of 2x on LHS: ‐1
Coefficient of 2x on RHS: ‐a+b = ‐2+b Therefore ‐2+b=‐1 or b=1
Coefficient of x on LHS: ‐16 Coefficient of x on RHS: ‐b+c=‐b‐15 Therefore –b‐15=‐16 or b=1
Hence, 15162 23 +−− xxx = ( )( ) ( )( )( )523115-21 2 −+−=+− xxxxxx
SKETCHING CUBIC FUNCTIONS
Just like in the case of quadratic functions as we saw in Unit 5, the factors of a cubic indicate the x ‐intercepts of the graph. In the above example, the intercepts will be given by:
01=−x that is 1=x
03 =+x that is 3−=x
052 =−x that is 5,2=x
The y ‐intercept is always very easy to determine because it is given by 0=x , that is ( ) 150 =f .
The x ‐intercepts are thus at (1: 0), (‐3; 0) and (2,5; 0).
And the y ‐intercept is at (0; 15).
Preparation for the Mathematics examination brought to you by Kagiso Trust
Illustration
Below is the graph of 15162 23 +−−= xxxy = ( )( ) ( )( )( )523115-21 2 −+−=+− xxxxxx
-3 -2 -1 1 2 3
-7-6-5-4-3-2-1
123456789
101112131415161718192021222324252627282930313233
x
y
Compare the x ‐intercepts with the factors of the cubic function.
We see that the graph of the cubic function also has turning points. But unlike in the case of the turning points of a quadratic function (as shown in Unit 5), there is no algebraic method for working out the coordinates of the turning points of a cubic. We will need to apply differentiation to work out the coordinates of the turning points of a cubic (and we can also apply the method to quadratics).
Exercise 10
10.1. Factorise the following expression
10.1.1 652 23 +−− xxx
10.1.2 1892 23 +−− xxx 10.2 The remainder obtained when
162 23 +−+ xaxx is divided by )2( +x is twice the remainder when the same expression is divided by )1( −x . Find the value of a .
10.3 Find the coordinates of the points where the
curve 6116 23 +++= xxxy cuts (a) the y ‐axis, (b) the x ‐axis. Hence make a sketch of the curve and state the range of values of x can take for
06116 23 ≤+++ xxx . 10.4 Find the coordinates of the points where the
curve 825 23 ++−= xxxy cuts (a) the y ‐axis, (b) the x ‐axis. Hence make a sketch of the curve and state the range of values of x can take for
825 23 −−>− xxx .
Factor theorem
ANSWERS
EXERCISE 10
Recall that A cubic = (a linear expression) )( 2 cbxax ++× The linear expression you find by using the remainder theorem. The quadratic expression you find either by long division or by synthetic division. Practise both of these methods to arrive at the following answers: Question 10.1 10.1.1 ( ) ( )( )( )321 −+−= xxxxf 10.1.2 ( ) ( )( )( )332 +−−= xxxxf Question 10.2 10.2 All you need is the following: Determine ( )2−f
Determine ( )1f Put ( ) ( )122 ff ⋅=− and solve this equation for a . Answer:
23
−=a
Preparation for the Mathematics examination brought to you by Kagiso Trust
Question 10.3 The y ‐intercept is (0;6) The x ‐intercepts are (‐1;0). (‐2;0) and (‐3;0) Where is ( )xf negative? Determine the answer by inspecting the graph and finding for what values of x the graph is negative (which implies being below the x ‐axis or y taking negative values). Answer: 3−≤x and 12 −≤≤− x
Question 10.4 The y ‐intercept is (0;8) The x ‐intercepts are (‐1;0). (2;0) and (4;0)
825 23 −−>− xxx is really 0825 23 >++− xxx . So you must inspect the graph for ( ) 0>xf . That is for what values of x is y or ( )xf positive (implying above the x ‐axis)? Answer: 21 <<− x and 4>x
1050-5-10
4
2
0
-2
-4
1050-5-10
4
2
0
-2
-4
1050-5-10
4
2
0
-2
-4
1050-5-10
4
2
0
-2
-4
420-2-4
8
6
4
2
0
-2
-4
420-2-4
8
6
4
2
0
-2
-4
420-2-4
8
6
4
2
0
-2
-4
420-2-4
8
6
4
2
0
-2
-4