KT Classroom Unit 10: Factor Theorem

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 1

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 f ( x ) = 2 x 3 − 3 x 2 + 5 x + 3 , verify that the “number of times” x + 1 divides into f ( x ) is

2 x 2 − 5 x + 10 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

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2 x 3 − 3 x 2 + 5 x + 3 = ( x + 1) 2 x 2 − 5 x + 10 + (− 7 )

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or

f ( x ) = ( x + 1) 2 x 2 − 5 x + 10 − 7

or

f ( x ) = ( x + 1) × Q( x ) + R

NOTE: If f ( x ) is a cubic and we are dividing by a linear function, then Q( x ) must be a quadratic. Why? Is R a function or a constant in this case?

If we now replace x by ‐1, we get:

f (− 1) = (− 1 + 1) × Q(− 1) + R

which is

f (− 1) = 0 × Q(− 1) + R

or simply

f (− 1) = R , that is when f ( x ) is divided by (x + 1) the remainder is f (− 1) .

In general, when f ( x ) is divided by (x − α ) then the remainder is f (α ) .

THE FACTOR THEOREM What if the remainder is zero? In other words, what if (x − α ) divides exactly into f ( x ) without leaving a remainder?

Then because R = 0, we will have

f ( x ) = (x − α ) ⋅ Q( x )

which shows that (x − α ) is then a factor of f ( x ) .

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DIVISION What we noted above is that if f ( x ) is a cubic and (x − α ) is its factor, we can write

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f ( x ) = ( x − α ) ⋅ ax 2 + bx + c

The values of a, b and c can then be determined by one of two methods: Either by 1. dividing f ( x ) by x − a ; or 2. equating coefficients as demonstrated below.

Example: Factorise 2 x 3 − x 2 − 16 x + 15 .

First put f ( x ) = 2 x 3 − x 2 − 16 x + 15 We want a value of x that makes f ( x ) = 0 .

The coefficient of x 3 and the constant term in f ( x ) give a clue as to what values to try.

The coefficient of x 3 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 x = −1 :

f (− 1) = 2.(− 1) − (− 1) − 16(− 1) + 15 = −2 − 1 + 16 + 15 = 28 ≠ 0

So x + 1 is not a factor of f ( x ) .

Try x = 1 :

3

2

f (1) = 2.13 − 12 − 16.1 + 15 = 2 − 1 − 16 + 15 = 17 − 17 = 0 Therefore, x − 1 is a factor of f ( x ) .

Method of division (it is left as an exercise for you to carry out the division below):

x − 1 2 x 3 − x 2 − 16 x + 15

Factor theorem

Having done the division, can you complete the following statement:

2 x 3 − x 2 − 16 x + 15 = ( x − 1)(

)

Method of equating coefficients: We can write

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2 x 3 − x 2 − 16 x + 15 = ( x − 1) ax 2 + bx + c

You now multiply out the RHS and then compare coefficients of like terms on either side. For example the coefficient of the x 3 term on the LHS must equal the coefficient of the x 3 term on the RHS; and so on with the x 2 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 x 2 or x .

Coefficient of x on LHS: ‐16 Coefficient of x on RHS: ‐b+c=‐b‐15 Therefore –b‐15=‐16 or b=1

Coefficient of x 2 on LHS: ‐1 Coefficient of x 2 on RHS: ‐a+b = ‐2+b Therefore ‐2+b=‐1 or b=1

Hence, 2 x 3 − x 2 − 16 x + 15 = ( x − 1) 2 x 2 + x - 15 = ( x − 1)( x + 3)(2 x − 5)

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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:

x − 1 = 0 that is x = 1

x + 3 = 0 that is x = −3

2 x − 5 = 0 that is x = 2,5

The y ‐intercept is always very easy to determine because it is given by x = 0 , that is f (0) = 15 .

The x ‐intercepts are thus at (1: 0), (‐3; 0) and (2,5; 0).

And the y ‐intercept is at (0; 15).

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Illustration

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Below is the graph of y = 2 x 3 − x 2 − 16 x + 15 = ( x − 1) 2 x 2 + x - 15 = ( x − 1)( x + 3)(2 x − 5) y

33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 -3

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-1

x 1

-1 -2 -3 -4 -5 -6 -7

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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

3 2 10.1.1 x − 2 x − 5 x + 6

10.1.2 x 3 − 2 x 2 − 9 x + 18

10.2 The remainder obtained when 2 x 3 + ax 2 − 6 x + 1 is divided by ( x + 2) is twice the remainder when the same expression is divided by ( x − 1) . Find the value of a .

10.3 Find the coordinates of the points where the curve y = x 3 + 6 x 2 + 11x + 6 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 x 3 + 6 x 2 + 11x + 6 ≤ 0 .

10.4 Find the coordinates of the points where the curve y = x 3 − 5 x 2 + 2 x + 8 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 x 3 − 5 x 2 > −2 x − 8 .

Factor theorem

ANSWERS EXERCISE 10

Recall that A cubic = (a linear expression) × (ax 2 + bx + c) 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 f ( x ) = ( x − 1)( x + 2)( x − 3)

10.1.2 f ( x ) = ( x − 2)( x − 3)(x + 3)

Question 10.2 10.2 All you need is the following:

Determine f (− 2) Determine f (1)

Put f (− 2) = 2 ⋅ f (1)

and solve this equation for a .

Answer: 3 a=− 2

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Question 10.3 The y ‐intercept is (0;6) The x ‐intercepts are (‐1;0). (‐2;0) and (‐3;0) Where is f ( x ) 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: x ≤ −3 and − 2 ≤ x ≤ −1 4

2

0 -10

-5

0

5

10

-2

-4

Question 10.4 The y ‐intercept is (0;8) The x ‐intercepts are (‐1;0). (2;0) and (4;0) x 3 − 5 x 2 > −2 x − 8 is really x 3 − 5 x 2 + 2 x + 8 > 0 .

So you must inspect the graph for f ( x ) > 0 . That is for what values of x is y or f ( x ) positive (implying above the x ‐axis)?

Answer: − 1 < x < 2 and x > 4 8

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4

2

0 -4

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0

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