Cambridge IGCSE Additional Maths SB Sample

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STUDENT’S BOOK

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Additional Maths pl

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Also for Cambridge O Level

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Extensive practice questions with detailed worked examples help students build understanding and knowledge, and include advice on how to tackle difficult content. Problem-solving skills are developed through flagged questions. Additionally, examstyle questions, plus Cambridge past paper questions, are included to help students prepare for their examinations.

Cambridge IGCSE™

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Collins Cambridge IGCSE® Additional Maths Student’s Book provides comprehensive coverage of the syllabus through topic-based chapters. Designed to build on IGCSE® Mathematics, this course contains exercises that test students’ initial understanding and prepare them for Additional Maths content.

Cambridge IGCSE® Additional Maths STUDENT’S BOOK

Cambridge IGCSE® Additional Maths

rigorous quality-assurance process

✓ Developed by subject experts ✓ For Cambridge schools worldwide

David Bird, Su Nicholson, Claire Powis, Peter Ransom, Trevor Senior, Brian Speed, Colin Stobart

4 Factors and Polynomials

C H A P T ER

These topics build on your algebraic skills in factorising and solving quadratic equations by extending to more complex expressions and equations. Before you move on to look in more detail at factorising, review what you know already about factorising quadratics and solving quadratic equations.

Topics

Key words

4.1 The factor theorem

polynomial, variable, cubic, factor theorem, quotient, divisor, long division method, grid method

4.2 The remainder theorem

1 a Factorise x 2 − 2x − 15 = 0.

• Know and use the factor and remainder theorems

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• Find factors of polynomials

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remainder, remainder theorem, dividend

In this chapter you will learn how to

• Solve cubic equations.

Starting point

2 Solve 12x 2 = 28x − 15.

Why this chapter matters Polynomials are used in many real-life situations. For example, you can use them to work out the distance an object has moved in a given time along a track or falling under gravity.

Polynomials are used in economics to model financial situations, for example, to predict patterns in stock markets.

There are lots of different applications of polynomials in engineering and construction. Road builders and roller-coaster designers use polynomials to decide on the size and shape of curves.

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From your Cambridge IGCSE Mathematics course you should be able to

b Identify the roots of the equation x 2 − 2x − 15 = 0.

• Factorise quadratic expressions

o factorisation

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• Solve quadratic equations for real roots, by:

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o the quadratic formula

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o completing the square.

Exploring the topics As you start to explore this topic, discuss possible answers to the following questions in pairs. a Identify some examples of buildings or constructions that have smooth curves in their design. b Identify any words you can think of linked to motion of an object. c Describe how polynomials might be used in either chemistry or in economics.

Chapter 4: Factors and Polynomials

85

4 Factors and Polynomials

C H A P T ER

polynomial, variable, cubic, factor theorem, quotient, divisor, long division method, grid method

4.2 The remainder theorem

remainder, remainder theorem, dividend

In this chapter you will learn how to • Know and use the factor and remainder theorems • Find factors of polynomials • Solve cubic equations.

From your Cambridge IGCSE Mathematics course you should be able to

b Identify the roots of the equation x 2 − 2x − 15 = 0.

2 Solve 12x 2 = 28x − 15.

Why this chapter matters

Polynomials are used in many real-life situations. For example, you can use them to work out the distance an object has moved in a given time along a track or falling under gravity.

Polynomials are used in economics to model financial situations, for example, to predict patterns in stock markets.

There are lots of different applications of polynomials in engineering and construction. Road builders and roller-coaster designers use polynomials to decide on the size and shape of curves.

o completing the square.

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o the quadratic formula

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o factorisation

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• Factorise quadratic expressions • Solve quadratic equations for real roots, by:

ge s

4.1 The factor theorem

1 a Factorise x 2 − 2x − 15 = 0.

Pa

Key words

These topics build on your algebraic skills in factorising and solving quadratic equations by extending to more complex expressions and equations. Before you move on to look in more detail at factorising, review what you know already about factorising quadratics and solving quadratic equations.

e

Topics

Starting point

Exploring the topics

As you start to explore this topic, discuss possible answers to the following questions in pairs. a Identify some examples of buildings or constructions that have smooth curves in their design. b Identify any words you can think of linked to motion of an object. c Describe how polynomials might be used in either chemistry or in economics.

Chapter 4: Factors and Polynomials

85

4.1

Chapter 4 . Topic 1

4.1 The factor theorem

The long division method You can use the long division method to find factors. x +1 x − 3 x 2 − 2x − 3

A polynomial is an algebraic expression with terms that are positive integer powers of the variable and possibly a constant term. Polynomials can be expressions, equations or functions.

− ( x 2 − 3x ) x −3 − ( x − 3) 0

Examples of polynomials are: • x 2 − 2x + 3

This is also called a quadratic expression, as the highest power is 2.

You can factorise

x2

− 2x − 3 to give (x − 3)(x + 1).

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• x 3 + 6x 2 + 8x This is also called a cubic expression, as the highest power is 3.

You can solve the quadratic equation x2 − 2x − 3 = 0 as (x − 3)(x + 1) = 0 so x − 3 = 0 or x + 1 = 0, and the solutions are x = 3 or x = −1.

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You can work backwards from values of x to find quadratic equations.

If the solutions of a quadratic equation are x = −4 and x = 5, then x + 4 = 0 or x − 5 = 0. Combining these, (x + 4)(x − 5) = 0, gives x 2 − x − 20 = 0.

Using the long division method Always look at the highest powers.

x

Dividing the term in the polynomial by the x term in the divisor gives x as the first part of the quotient.

x − 3 x 2 − 2x − 3

Multiplying the divisor (x − 3) by x gives x 2 − 3x.

x − 3 x 2 − 2x − 3

x2

x 2 − 3x Subtracting this from the polynomial gives a remainder of x − 3 (when you ‘bring down’ the next term from the polynomial).

Using function notation: f(x) = x 2 − x − 20.

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Since f(−4) = 0 and f(5) = 0, x = −4 and x = 5 are solutions of the quadratic equation x 2 − x − 20 = 0.

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You can extend these ideas to cubic and other polynomials.

Dividing x in the remainder by x in the divisor gives + 1 as the second part of the quotient.

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The factor theorem The factor theorem states:

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If f(x) is a polynomial and a value a can be found so that f(a) = 0, then x = a is a solution of the equation f(x) = 0 and x − a is a factor of the polynomial.  b Note that if ax − b is a factor of the polynomial, then f   = 0.  a  1 For example, if 2x − 1 is a factor of the polynomial, then f   = 0.  2

The quotient When a polynomial is divided by a factor the result is the quotient.

For example: since x 2 − 2x − 3 = (x − 3)(x + 1), then (x 2 − 2x − 3) ÷ (x − 3) = x + 1. where x 2 − 2x − 3 is the polynomial, x − 3 is the factor or divisor and x + 1 is the quotient.

x

This divides exactly so x + 1 is the quotient.

x x − 3 x 2 − 2x − 3 − ( x 2 − 3 x) x −3

Advice and Tips Having identified the multiplier x, to get to x2 from x, you must multiply the whole of the factor by x.

x +1 x − 3 x 2 − 2x − 3 − ( x 2 − 3x ) x −3 − ( x − 3) 0

Similarly, given that x + 4 is a factor of f(x) = x 3 + 2x² − 11x − 12, you can use the long division method to obtain the quotient, which will be a quadratic.

x2 − 2x − 3 x + 4 x 3 + 2 x 2 − 11x − 12 − ( x3 + 4x2) − 2 x 2 − 11x − ( −2 x 2 − 8 x ) − 3 x − 12 − ( −3 x − 12 )

0 So, this gives you x 3 + 2x 2 − 11x − 12 = (x + 4)(x 2 − 2x − 3).

86

4.1 The factor theorem

Chapter 4: Factors and Polynomials

87

4.1

Chapter 4 . Topic 1

4.1 The factor theorem

The long division method You can use the long division method to find factors. x +1 x − 3 x 2 − 2x − 3

A polynomial is an algebraic expression with terms that are positive integer powers of the variable and possibly a constant term. Polynomials can be expressions, equations or functions.

− ( x 2 − 3x ) x −3 − ( x − 3) 0

Examples of polynomials are:

• x 3 + 6x 2 + 8x This is also called a cubic expression, as the highest power is 3.

You can factorise

x2

− 2x − 3 to give (x − 3)(x + 1).

You can solve the quadratic equation x2 − 2x − 3 = 0 as (x − 3)(x + 1) = 0 so x − 3 = 0 or x + 1 = 0, and the solutions are x = 3 or x = −1.

You can work backwards from values of x to find quadratic equations.

If the solutions of a quadratic equation are x = −4 and x = 5, then x + 4 = 0 or x − 5 = 0. Combining these, (x + 4)(x − 5) = 0, gives x 2 − x − 20 = 0.

Using the long division method Always look at the highest powers.

x

Dividing the term in the polynomial by the x term in the divisor gives x as the first part of the quotient.

x − 3 x 2 − 2x − 3

Multiplying the divisor (x − 3) by x gives x 2 − 3x.

x − 3 x 2 − 2x − 3

x2

Subtracting this from the polynomial gives a remainder of x − 3 (when you ‘bring down’ the next term from the polynomial).

The factor theorem states: If f(x) is a polynomial and a value a can be found so that f(a) = 0, then x = a is a solution of the equation f(x) = 0 and x − a is a factor of the polynomial.  b Note that if ax − b is a factor of the polynomial, then f   = 0.  a  1 For example, if 2x − 1 is a factor of the polynomial, then f   = 0.  2

The quotient When a polynomial is divided by a factor the result is the quotient.

For example: since x 2 − 2x − 3 = (x − 3)(x + 1), then (x 2 − 2x − 3) ÷ (x − 3) = x + 1. where x 2 − 2x − 3 is the polynomial, x − 3 is the factor or divisor and x + 1 is the quotient.

x − 3 x 2 − 2x − 3

e

− ( x 2 − 3 x) x −3

This divides exactly so x + 1 is the quotient.

Having identified the multiplier x, to get to x2 from x, you must multiply the whole of the factor by x.

x +1

x − 3 x 2 − 2x − 3

pl

Dividing x in the remainder by x in the divisor gives + 1 as the second part of the quotient.

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The factor theorem

x

Advice and Tips

− ( x 2 − 3x ) x −3 − ( x − 3) 0

Similarly, given that x + 4 is a factor of f(x) = x 3 + 2x² − 11x − 12, you can use the long division method to obtain the quotient, which will be a quadratic.

Sa

You can extend these ideas to cubic and other polynomials.

x

x 2 − 3x

Using function notation: f(x) = x 2 − x − 20.

Since f(−4) = 0 and f(5) = 0, x = −4 and x = 5 are solutions of the quadratic equation x 2 − x − 20 = 0.

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This is also called a quadratic expression, as the highest power is 2.

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• x 2 − 2x + 3

x2 − 2x − 3

x + 4 x 3 + 2 x 2 − 11x − 12 − ( x3 + 4x2) − 2 x 2 − 11x − ( −2 x 2 − 8 x ) − 3 x − 12 − ( −3 x − 12 )

0 So, this gives you x 3 + 2x 2 − 11x − 12 = (x + 4)(x 2 − 2x − 3).

86

4.1 The factor theorem

Chapter 4: Factors and Polynomials

87

4.1 And then factorising

(x 2

This means the term in the bottom middle cell must be −8x.

− 2x − 3) leads to:

x 3 + 2x 2 − 11x − 12 = (x + 4)(x − 3)(x + 1).

The grid method The grid method is another way to divide a polynomial by a linear factor.

Given that f(x) = x 3 + 2x 2 − 11x − 12, and x + 4 is a factor, you can use the grid method to obtain the quotient, which will be quadratic.

x 3 + 2x 2 − 11x − 12 ×

x

x3

x3

x2

x

x3

+4

4x 2

×

x2

−2x

−3

x

x3

−2x 2

−3x

+4

4x 2

−8x

Finally, the bottom right cell will be −3 × 4 = −12. ×

x2

−2x

−3

x

x3

−2x 2

−3x

+4

4x 2

−8x

−12

This method is actually easier than it seems at first glance, and when all of the steps are carried out on the same grid, it is quite compact!

Example 1 It is given that f(x) = x 3 + 6x 2 + 8x.

b Hence factorise f(x) completely and solve the equation f(x) = 0. a f(−4) = (−4)3 + 6(−4)2 + 8(−4)

×

x2

−2x

x

x3

−2x 2

+4

4x 2 4.1 The factor theorem

−8x

a Show that x + 4 is a factor of f(x).

So to make the second term of the polynomial (+2x 2) you need −2x 2 in the top middle cell, which gives the second term of the quotient as −2x.

88

4x 2

x 3 + 2x 2 − 11x − 12 = (x + 4)(x − 3)(x + 1).

This means the term in the bottom left cell must be 4x 2. ×

+4

Factorising the quadratic expression gives:

Sa

+4

−2x 2

or x 3 + 2x 2 − 11x − 12 = (x + 4)(x 2 − 2x − 3).

pl

x

x3

So, (x 3 + 2x 2 − 11x − 12) ÷ (x + 4) = x 2 − 2x − 3

m

x2

x

x 3 + 4x 2 − 2x 2 − 8x − 3x − 12 = x 3 + 2x 2 − 11x − 12

As this is a multiplication grid the first term of the quotient must be x 2, which you can write at the top. ×

−2x

You can then check that the contents of the grid simplify to give the original polynomial.

e

+4

Pa

Write the first term, x 3, of the polynomial in the top left working cell.

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Start by constructing a grid with the linear factor down the left-hand side.

The quotient will appear along the top of the grid.

x2

So to make the third term (−11x) of the polynomial you need −3x in the top right cell which gives the third term of the quotient as −3.

Look at the same example as above.

The contents of the grid will simplify to the polynomial f(x) = x 3 + 2x 2 − 11x − 12.

×

= −64 + 96 − 32 Advice and Tips Note that the sum of the 4x2 and the −2x2 combine to the term in 2x2 in the original polynomial.

=0 So, as f(−4) = 0, x + 4 is a factor of x 3 + 6x 2 + 8x. b There is no constant term, so another factor is x.

So, f(x) is of the form f(x) = x(x + 4)(x + c).

Chapter 4: Factors and Polynomials

89

4.1 This means the term in the bottom middle cell must be −8x.

− 2x − 3) leads to:

x 3 + 2x 2 − 11x − 12 = (x + 4)(x − 3)(x + 1).

The grid method The grid method is another way to divide a polynomial by a linear factor.

Given that f(x) = x 3 + 2x 2 − 11x − 12, and x + 4 is a factor, you can use the grid method to obtain the quotient, which will be quadratic.

Start by constructing a grid with the linear factor down the left-hand side.

−2x

x

x3

−2x 2

+4

4x 2

−8x

×

x2

−2x

−3

x

x3

−2x 2

−3x

+4

4x 2

−8x

Finally, the bottom right cell will be −3 × 4 = −12.

The contents of the grid will simplify to the polynomial f(x) = x 3 + 2x 2 − 11x − 12.

The quotient will appear along the top of the grid.

×

x2

−2x

x

x3

−2x 2

−3x

+4

4x 2

−8x

−12

−3

Pa

Write the first term, x 3, of the polynomial in the top left working cell.

You can then check that the contents of the grid simplify to give the original polynomial.

x 3 + 2x 2 − 11x − 12 ×

x 3 + 4x 2 − 2x 2 − 8x − 3x − 12 = x 3 + 2x 2 − 11x − 12 x3

So, (x 3 + 2x 2 − 11x − 12) ÷ (x + 4) = x 2 − 2x − 3

+4

e

or x 3 + 2x 2 − 11x − 12 = (x + 4)(x 2 − 2x − 3).

Factorising the quadratic expression gives:

x2

x

x3

x 3 + 2x 2 − 11x − 12 = (x + 4)(x − 3)(x + 1).

This method is actually easier than it seems at first glance, and when all of the steps are carried out on the same grid, it is quite compact!

+4

Sa

Example 1

It is given that f(x) = x 3 + 6x 2 + 8x.

This means the term in the bottom left cell must be 4x 2. ×

x2

x

x3

+4

4x 2

a Show that x + 4 is a factor of f(x). b Hence factorise f(x) completely and solve the equation f(x) = 0. a f(−4) = (−4)3 + 6(−4)2 + 8(−4)

So to make the second term of the polynomial (+2x 2) you need −2x 2 in the top middle cell, which gives the second term of the quotient as −2x. ×

x2

−2x

x

x3

−2x 2

+4

4x 2 4.1 The factor theorem

m

×

pl

As this is a multiplication grid the first term of the quotient must be x 2, which you can write at the top.

88

x2

So to make the third term (−11x) of the polynomial you need −3x in the top right cell which gives the third term of the quotient as −3.

Look at the same example as above.

x

×

ge s

And then factorising

(x 2

= −64 + 96 − 32 Advice and Tips Note that the sum of the 4x2 and the −2x2 combine to the term in 2x2 in the original polynomial.

=0 So, as f(−4) = 0, x + 4 is a factor of x 3 + 6x 2 + 8x. b There is no constant term, so another factor is x.

So, f(x) is of the form f(x) = x(x + 4)(x + c).

Chapter 4: Factors and Polynomials

89

4.1 To make the last term 8x, c = 2, or this suggests the last factor is x + 2.

So, x 3 − 2x 2 − 13x − 10 = (x − 5)(x 2 + 3x + 2)

Checking: f(−2) = (−2)3 + 6(−2)2 + 8(−2)

= (x − 5)(x + 1)(x + 2).

= −8 + 24 − 16

Exercise 4.1

=0

1

So, as f(−2) = 0, x + 2 is a factor of x 3 + 6x 2 + 8x So, f(x) = x(x + 2)(x + 4).

Solving x(x + 2)(x + 4) = 0 gives x = 0 or x + 2 = 0 or x + 4 = 0 So, x = 0, x = −2 and x = −4 are the solutions.

2

Example 2 Find the value of a.

If x − 3 is a factor of x 3 + ax 2 + 11x − 6, then f(3) = 0. f(3) = (3)3 + a(3)2 + 11(3) − 6 = 0

So, 9a = −54 a = −6

Example 3

3

4

e

5

E

m

So, 2x − 5 is a factor of

Sa

Example 4

−

7x 2

f

2x − 3, 2x 3 − 5x 2 − x + 6

b

x3

d

x3

f

6x 3 + 19x 2 + 11x − 6

Factorise each expression. a

x3

c

x3

+

3x 2

− 13x − 15

+

6x 2

+ 11x + 6

Advice and Tips +

5x 2

+ 2x − 8

+

6x 2

− x − 30

Solve each equation.

One way to find integer values of x so that f(x) = 0 is to try different factors of the constant term.

Advice and Tips

a x 3 − 4x 2 − 7x + 10 = 0

b x 3 − 3x 2 + 3x − 1 = 0

c x 3 + 2x 2 − 16x − 32 = 0

d x 3 + 3x 2 − 13x − 15 = 0

e x 3 + 15x 2 + 68x + 84 = 0

f

x 3 − 11x 2 − 10x + 200 = 0

Note that some cubics have only one linear factor and a quadratic factor that does not factorise.

The equation x 3 − 3x 2 + ax + b = 0 has solutions x = 1 and x = 4.

a x + 3 is a factor of x 3 + 2x 2 − 5x + c. Demonstrate that c = −6.

a Given that (x − 3) is a factor of ax 3 − 7x 2 + 2x + 3, use the factor theorem to demonstrate that a = 2.

PS

7

x 3 + 3x 2 + ax + b factorises as (x + c)3. Calculate the values of a, b and c.

E

8

x 3 + 3x 2 − 16x − 48 factorises as (x 2 − a2)(x + b). Identify the three linear factors of x 3 + 3x 2 − 16x − 48.

PS

9

Work out the common factors of x 3 − 5x 2 − 2x + 24 and x 3 − x 2 − 10x − 8.

Using the long division method: E

Problem solving 4.1 1 Work in pairs. Each write down three linear factors, then multiply these factors together to form a cubic expression. Swap your cubic expression with your partner, and factorise each other’s expressions. Provided you make no mistakes, either multiplying the factors together or in factorising the cubic, you should have matching linear factors.

− ( x3 − 5x2)

3 x 2 − 13x − ( 3 x 2 − 15x )

2 x − 10 − ( 2 x − 10 ) 0

90

e x + 7, x 3 − 37x + 84

b Showing all your working, factorise ax 3 − 7x 2 + 2x + 3, and hence solve ax 3 − 7x 2 + 2x + 3 = 0.

(x − 5) is a factor since f(5) = 53 − 2(5)2 − 13(5) − 10 = 125 − 50 − 65 − 10 = 0.

x − 5 x 3 − 2 x 2 − 13x − 10

6

+ x + 10.

Factorise f(x) = x 3 − 2x 2 − 13x − 10.

x2 + 3x + 2

d x + 6, x 3 + 13x 2 + 54x + 72

b Work out the other two factors of the cubic.

125 175 5  5  5  5 2   − 7   +   + 10 = − + + 10 = 0  2  2  2 2 4 4 2x 3

c x − 4, x 3 − 7x 2 + 2x + 40

Work out the third solution to the equation.

pl

2

b x − 3, x 3 + 3x 2 − 13x − 15

g x3 + x2 + x + 1

Show that 2x − 5 is a factor of f(x) = 2x 3 − 7x 2 + x + 10.  5 If 2x − 5 is a factor, then f   = 0.  2 3

a x + 1, x 3 + 6x 2 − 9x − 14

e 2x 3 − 13x 2 + 17x + 12

Pa

27 + 9a + 33 − 6 = 0

ge s

The polynomial f(x) = x 3 + ax 2 + 11x − 6 has a factor of x − 3.

In each part, demonstrate that the expression is a factor of the polynomial.

4.1 The factor theorem

2 A container is a cuboid with lengths 1, x and x 2 − 13x − 128.

Identify the value for x, given that the volume of the container is 240 cubic units. Explain why any other values of x that give V = 240 are not valid.

Chapter 4: Factors and Polynomials

91

4.1 To make the last term 8x, c = 2, or this suggests the last factor is x + 2.

So, x 3 − 2x 2 − 13x − 10 = (x − 5)(x 2 + 3x + 2)

Checking: f(−2) = (−2)3 + 6(−2)2 + 8(−2)

= (x − 5)(x + 1)(x + 2).

= −8 + 24 − 16

Exercise 4.1

So, f(x) = x(x + 2)(x + 4).

Solving x(x + 2)(x + 4) = 0 gives x = 0 or x + 2 = 0 or x + 4 = 0 So, x = 0, x = −2 and x = −4 are the solutions.

2

Example 2 The polynomial f(x) = x 3 + ax 2 + 11x − 6 has a factor of x − 3.

a x + 1, x 3 + 6x 2 − 9x − 14

b x − 3, x 3 + 3x 2 − 13x − 15

c x − 4, x 3 − 7x 2 + 2x + 40

d x + 6, x 3 + 13x 2 + 54x + 72

e x + 7, x 3 − 37x + 84

f

2x − 3, 2x 3 − 5x 2 − x + 6

b

x3

d

x3

f

6x 3 + 19x 2 + 11x − 6

Factorise each expression. a

x3

c

x3

+

3x 2

− 13x − 15

+

6x 2

+ 11x + 6

Find the value of a.

e 2x 3 − 13x 2 + 17x + 12

If x − 3 is a factor of x 3 + ax 2 + 11x − 6, then f(3) = 0.

g x3 + x2 + x + 1

f(3) = (3)3 + a(3)2 + 11(3) − 6 = 0

3

a x 3 − 4x 2 − 7x + 10 = 0

So, 9a = −54

c x 3 + 2x 2 − 16x − 32 = 0

a = −6

e x 3 + 15x 2 + 68x + 84 = 0 4

−

− x − 30

One way to find integer values of x so that f(x) = 0 is to try different factors of the constant term.

Advice and Tips

b x 3 − 3x 2 + 3x − 1 = 0

d x 3 + 3x 2 − 13x − 15 = 0 f

x 3 − 11x 2 − 10x + 200 = 0

Note that some cubics have only one linear factor and a quadratic factor that does not factorise.

The equation x 3 − 3x 2 + ax + b = 0 has solutions x = 1 and x = 4.

e

a x + 3 is a factor of x 3 + 2x 2 − 5x + c. Demonstrate that c = −6.

E

6

+ x + 10.

a Given that (x − 3) is a factor of ax 3 − 7x 2 + 2x + 3, use the factor theorem to demonstrate that a = 2.

m

So, 2x − 5 is a factor of

+

b Work out the other two factors of the cubic.

2

7x 2

+ 2x − 8

6x 2

pl

5

125 175 5  5  5  5 2   − 7   +   + 10 = − + + 10 = 0  2  2  2 2 4 4 2x 3

+

Work out the third solution to the equation.

Show that 2x − 5 is a factor of f(x) = 2x 3 − 7x 2 + x + 10.  5 If 2x − 5 is a factor, then f   = 0.  2 3

Advice and Tips 5x 2

Solve each equation.

27 + 9a + 33 − 6 = 0

Example 3

b Showing all your working, factorise ax 3 − 7x 2 + 2x + 3, and hence solve ax 3 − 7x 2 + 2x + 3 = 0.

PS

Factorise f(x) = x 3 − 2x 2 − 13x − 10.

E

(x − 5) is a factor since f(5) = 53 − 2(5)2 − 13(5) − 10 = 125 − 50 − 65 − 10 = 0.

Using the long division method: E

x2 + 3x + 2 x − 5 x 3 − 2 x 2 − 13x − 10

PS

7

x 3 + 3x 2 + ax + b factorises as (x + c)3. Calculate the values of a, b and c.

8

x 3 + 3x 2 − 16x − 48 factorises as (x 2 − a2)(x + b). Identify the three linear factors of x 3 + 3x 2 − 16x − 48.

9

Work out the common factors of x 3 − 5x 2 − 2x + 24 and x 3 − x 2 − 10x − 8.

Sa

Example 4

Problem solving 4.1 1 Work in pairs. Each write down three linear factors, then multiply these factors together to form a cubic expression. Swap your cubic expression with your partner, and factorise each other’s expressions. Provided you make no mistakes, either multiplying the factors together or in factorising the cubic, you should have matching linear factors.

− ( x3 − 5x2)

3 x 2 − 13x − ( 3 x 2 − 15x )

2 x − 10 − ( 2 x − 10 ) 0

90

In each part, demonstrate that the expression is a factor of the polynomial.

ge s

1

So, as f(−2) = 0, x + 2 is a factor of x 3 + 6x 2 + 8x

Pa

=0

4.1 The factor theorem

2 A container is a cuboid with lengths 1, x and x 2 − 13x − 128.

Identify the value for x, given that the volume of the container is 240 cubic units. Explain why any other values of x that give V = 240 are not valid.

Chapter 4: Factors and Polynomials

91

4.2

Chapter 4 . Topic 2

4.2 The remainder theorem

Example 8 The expression 2x 3 − 9x 2 + ax + b has a factor of x − 5 and leaves a remainder of −33 when divided by x − 2. Find the value of each of the constants a and b.

The remainder theorem states: If you divide a polynomial f(x) by any factor (x − a) the remainder is f(a), where a is a constant.

If there is no remainder, f(a) = 0, which means that (x − a) is a factor of f(x).

You can prove this.

When x = a f(a) = (a − a) g(a) + R, so f(a) = R, where R is the remainder.

A more general form of the remainder theorem states:

You may see the polynomial that you are dividing called the dividend. You should know this term, in case it comes up in a question.

ge s

Let f(x) = (x − a) g(x) + R

Advice and Tips

Pa

b If you divide a polynomial f(x) by any factor (ax − b), the remainder is f   , where a and b are a constants and a is non-zero.

e

For example, if you divide a polynomial f(x) by a factor (2x − 1),  1 the remainder is f   .  2

pl

Example 5

Find the remainder when f(x) = x 3 + 7x 2 − 6x + 2 is divided by (x + 3).

Example 6

The remainder theorem tells you that f(2) = −33. f(5) = 2(5)3 − 9(5)2 + a(5) + b = 0 250 − 225 + 5a + b = 0 25 + 5a + b = 0 f(2) = 2(2)3 − 9(2)2 + a(2) + b = −33 16 − 36 + 2a + b = −33 13 + 2a + b = 0 Since b = −25 − 5a and b = −13 − 2a −25 − 5a = −13 − 2a −3a = 12 a = −4 b = −25 − 5(−4) b = −25 + 20 b = −5

Exercise 4.2 1

Sa

Find the remainder when f(x) = x 3 − 4x 2 − 2x + 1 is divided by (2x − 1).

3 2 1 7  1  1  1  1 The remainder is f   =   − 4   − 2  + 1 = − 1 − 1 + 1 = −  2  2  2  2 8 8

c Find the remainder when f(x) = 2x 3 − 4x 2 − 2x − 3 is divided by (x − 2). d Find the remainder when f(x) = 3x 3 + x 2 − x − 8 is divided by (x + 3).

2

b When f(x) = 2x 3 − x 2 + ax + 6 is divided by (x + 2), the remainder is −4. c When f(x) = x 3 − ax 2 + x + 2 is divided by (x − 3), the remainder is 8.

When f(x) = 3x 3 + ax 2 + x + 4 is divided by (x − 1), the remainder is 6. f(1) = 6

d When f(x) = ax 3 + 2x 2 − 4x + 3 is divided by (x + 2), the remainder is 8.

3

The expression 2x 3 − x 2 + ax + b has a factor x + 2 and leaves a remainder of 10 when divided by x + 1. Find the value of each of the constants a and b.

4

The expression ax 3 + bx 2 − 26x + 8 has a factor x + 4 and leaves a remainder of −10 when divided by x − 1. Find the value of each of the constants a and b.

3(1)3 + a(1)2 + (1) + 4 = 6 3+a+1+4=6 a+8=6

In each part, work out the value of a. a When f(x) = x 3 − 3x 2 + x + a is divided by (x − 1), the remainder is 5.

Example 7

Work out the value of a.

a Find the remainder when f(x) = x 3 + 2x 2 − 3x + 2 is divided by (x − 1). b Find the remainder when f(x) = x 3 − 5x 2 + x + 1 is divided by (x + 1).

m

The remainder is f(−3) = (−3)3 + 7(−3)2 − 6(−3) + 2 = −27 + 63 + 18 + 2 = 56

The factor theorem tells you that f(5) = 0.

a = −2

92

4.2 The remainder theorem

Chapter 4: Factors and Polynomials

93

4.2

Chapter 4 . Topic 2

4.2 The remainder theorem

Example 8 The expression 2x 3 − 9x 2 + ax + b has a factor of x − 5 and leaves a remainder of −33 when divided by x − 2. Find the value of each of the constants a and b.

If you divide a polynomial f(x) by any factor (x − a) the remainder is f(a), where a is a constant.

If there is no remainder, f(a) = 0, which means that (x − a) is a factor of f(x).

You can prove this.

Let f(x) = (x − a) g(x) + R

Advice and Tips You may see the polynomial that you are dividing called the dividend. You should know this term, in case it comes up in a question.

The factor theorem tells you that f(5) = 0.

The remainder theorem tells you that f(2) = −33. f(5) = 2(5)3 − 9(5)2 + a(5) + b = 0 250 − 225 + 5a + b = 0 25 + 5a + b = 0 f(2) = 2(2)3 − 9(2)2 + a(2) + b = −33

When x = a

16 − 36 + 2a + b = −33

f(a) = (a − a) g(a) + R,

13 + 2a + b = 0

so f(a) = R, where R is the remainder.

Since b = −25 − 5a and b = −13 − 2a

A more general form of the remainder theorem states:

−25 − 5a = −13 − 2a

Example 5 Find the remainder when f(x) = x 3 + 7x 2 − 6x + 2 is divided by (x + 3).

The remainder is f(−3) = (−3)3 + 7(−3)2 − 6(−3) + 2 = −27 + 63 + 18 + 2 = 56

Pa

a = −4 b = −25 − 5(−4) b = −25 + 20 b = −5

Exercise 4.2 1

e

For example, if you divide a polynomial f(x) by a factor (2x − 1),  1 the remainder is f   .  2

−3a = 12

pl

b If you divide a polynomial f(x) by any factor (ax − b), the remainder is f   , where a and b are a constants and a is non-zero.

ge s

The remainder theorem states:

a Find the remainder when f(x) = x 3 + 2x 2 − 3x + 2 is divided by (x − 1).

m

b Find the remainder when f(x) = x 3 − 5x 2 + x + 1 is divided by (x + 1). c Find the remainder when f(x) = 2x 3 − 4x 2 − 2x − 3 is divided by (x − 2).

Find the remainder when f(x) = x 3 − 4x 2 − 2x + 1 is divided by (2x − 1).

3 2 1 7  1  1  1  1 The remainder is f   =   − 4   − 2  + 1 = − 1 − 1 + 1 = −  2  2  2  2 8 8

d Find the remainder when f(x) = 3x 3 + x 2 − x − 8 is divided by (x + 3).

Example 7

Sa

Example 6 2

a When f(x) = x 3 − 3x 2 + x + a is divided by (x − 1), the remainder is 5. b When f(x) = 2x 3 − x 2 + ax + 6 is divided by (x + 2), the remainder is −4.

c When f(x) = x 3 − ax 2 + x + 2 is divided by (x − 3), the remainder is 8.

When f(x) = 3x 3 + ax 2 + x + 4 is divided by (x − 1), the remainder is 6.

Work out the value of a. f(1) = 6

d When f(x) = ax 3 + 2x 2 − 4x + 3 is divided by (x + 2), the remainder is 8.

3

The expression 2x 3 − x 2 + ax + b has a factor x + 2 and leaves a remainder of 10 when divided by x + 1. Find the value of each of the constants a and b.

4

The expression ax 3 + bx 2 − 26x + 8 has a factor x + 4 and leaves a remainder of −10 when divided by x − 1. Find the value of each of the constants a and b.

3(1)3 + a(1)2 + (1) + 4 = 6 3+a+1+4=6 a+8=6

In each part, work out the value of a.

a = −2

92

4.2 The remainder theorem

Chapter 4: Factors and Polynomials

93

4.2 5 PS

E

E

6

7

8

Advice and Tips

b When f(x) is divided by x − a, the remainder is 60. Find the values of a.

This polynomial can be factorised in the same way as the quadratic y2 − 25y + 144, where y = x2.

The polynomials f(x) = x3 − 4x2 + x + 6 and g(x) = x3 + ax2 − 9x − 14 have two common factors. When g(x) is divided by x − 3, the remainder is 40. Work out the factors of g(x).

Chapter review 1

2

ax 2

Problem solving 4.2

Pa

You can often tackle problem solving activities by means of a graph. For example, you can work out problems based on quadratics, such as how high a dirt-bike course should be in the O2 Arena, based on the graph of the parabolic function of the path of the bike and the height of the ramps.

Bearing that principle in mind, try these questions.

1 Work in small groups to solve this problem.

(2 marks)

b Factorise f(x).

(3 marks)

a Show that x − 2 is a factor of x 3 − 6x 2 + 12x − 8.

5

Solve these equations. a x 2 − 3x − 4 = 0

(3 marks)

b x 3 − 3x 2 − 4x = 0

(4 marks)

c E

6 7

8

− 4x + 12 = 0

(5 marks)

e

x − 35 = 3x .

11 − x The remainder when f(x) = 3x 3 − 4x 2 − 6x + a is divided by (x − 2) is 11.

Solve

(5 marks)

Show that the value of a is 15.

(3 marks)

a The cubic equation x 3 + ax 2 + bx − 40 = 0 has three positive integer roots. Two of the roots are 2 and 4. Find the other root and the value of each of the integers a and b.

(4 marks) Solve the equation x 3 − 5x 2 − 46x − 40 = 0 given that it has three integer roots, only one of which is positive.

pl

m

−

3x 2

b Do not use a calculator in this question.

(4 marks)

Cambridge IGCSE Additional Mathematics 0606, Paper 22, Q7 October/November 2019 9

The polynomial f(x) = ax 3 − 15x 2 + bx − 2 has a factor of 2x − 1 and a remainder of 5 when divided by x − 1. i

Sa

4.2 The remainder theorem

x3

3

Show that b = 8 and find the value of a.

(4 marks)

ii Using the values of a and b from part (i), express f(x) in the form (2x − 1)g(x), where g(x) is a quadratic factor to be found.

(4 marks)

Cambridge IGCSE Additional Mathematics 0606, Paper 11, Q6 June 2015 The polynomial p(x) = mx 3 − 29x 2 +39x + n, where m and n are constants, has a factor 3x − 1, and remainder 6 when divided by x − 1. Show that x − 2 is a factor of p(x).

(6 marks) Cambridge IGCSE Additional Mathematics 0606, Paper 22, Q4 May/June 2021

I can use the factor theorem to identify linear factors of polynomials.

94

(3 marks)

(5 marks)

Progression checklist

I can use the remainder theorem to work out unknown coefficients in polynomials.

+ 12x − 8.

Solve the equation x 3 + 6x 2 + 11x + 6 = 0.

10

I can solve cubic equations.

(2 marks)

4

2 Create a roller-coaster ride that represents a polynomial function and that will last a maximum of 30 seconds. Your ride must intersect with ground level at least once and return to ground level at the end of the ride. State the domain and range. Use a graphing program or a graphical calculator to explore how to design the ride. Then draw your graph on graph paper, using metres for the vertical scale and seconds for the horizontal scale. Use your graph to describe your ride.

I can factorise polynomials.

−

6x 2

(4 marks)

• The remainder when f(x) is divided by x − 1 is 9

I can find factors of polynomials.

x3

Factorise x 3 − 4x 2 − 25x + 100.

• f(x) has only one linear factor: a + c = 0 4 • A root of the equation f(x) = 0 is x = 5 • The remainder when f(x) is divided by x is −12

• a is a prime number.

(1 mark)

3

The function f(x) = ax 3 + bx 2 + cx + d, where a, b, c and d are integers.

Use the following information to work out the values of a, b, c and d and find the factors of f(x):

a Show that x + 4 is a factor of f(x).

b Factorise

The function f(x) = + + x + 5, where a is a constant. Given that the remainder when f(x) is divided by x − 1 is twice the remainder when f(x) is divided by x + 1, work out the value of a. x3

f(x) = x 3 − x 2 − 14x + 24

c Solve the equation f(x) = 0.

The function f(x) = ax 3 − 3x 2 + bx + 12, where a and b are constants, is such that 2x − 1 is a factor. Given that the remainder when f(x) is divided by x − 1 is −12, work out the other factors of f(x).

ge s

PS

a Factorise f(x) = x4 − 25x 2 + 144.

11

(x − 3) is a factor of f(x) = ax 3 + bx 2 − 9x − 9, where a and b are constants. a Show that 3a + b = 4.

(2 marks)

b When f(x) is divided by x − 2, the remainder is −15.

Show that 2a + b = 3. c Find the value of a and the value of b.

(3 marks) (3 marks) Chapter 4: Factors and Polynomials

95

4.2

E

E

6

7

8

b When f(x) is divided by x − a, the remainder is 60. Find the values of a.

This polynomial can be factorised in the same way as the quadratic y2 − 25y + 144, where y = x2.

The polynomials f(x) = x3 − 4x2 + x + 6 and g(x) = x3 + ax2 − 9x − 14 have two common factors. When g(x) is divided by x − 3, the remainder is 40. Work out the factors of g(x).

Chapter review 1

a Show that x + 4 is a factor of f(x).

(2 marks)

b Factorise f(x).

(3 marks)

c Solve the equation f(x) = 0.

2

The function f(x) = ax 3 − 3x 2 + bx + 12, where a and b are constants, is such that 2x − 1 is a factor. Given that the remainder when f(x) is divided by x − 1 is −12, work out the other factors of f(x). ax 2

−

6x 2

4

Solve the equation x 3 + 6x 2 + 11x + 6 = 0.

(5 marks)

5

Solve these equations. b x 3 − 3x 2 − 4x = 0 c

6 7

8

x − 35 = 3x .

11 − x The remainder when f(x) = 3x 3 − 4x 2 − 6x + a is divided by (x − 2) is 11.

I can use the remainder theorem to work out unknown coefficients in polynomials.

4.2 The remainder theorem

(4 marks) (5 marks) (5 marks)

(3 marks)

e

a The cubic equation x 3 + ax 2 + bx − 40 = 0 has three positive integer roots. Two of the roots are 2 and 4. Find the other root and the value of each of the integers a and b.

(4 marks)

pl

Solve the equation x 3 − 5x 2 − 46x − 40 = 0 given that it has three integer roots, only one of which is positive.

(4 marks)

Cambridge IGCSE Additional Mathematics 0606, Paper 22, Q7 October/November 2019 9

The polynomial f(x) = ax 3 − 15x 2 + bx − 2 has a factor of 2x − 1 and a remainder of 5 when divided by x − 1.

Show that b = 8 and find the value of a.

(4 marks)

Sa

i

ii Using the values of a and b from part (i), express f(x) in the form (2x − 1)g(x), where g(x) is a quadratic factor to be found.

(4 marks)

Cambridge IGCSE Additional Mathematics 0606, Paper 11, Q6 June 2015

The polynomial p(x) = mx 3 − 29x 2 +39x + n, where m and n are constants, has a factor 3x − 1, and remainder 6 when divided by x − 1. Show that x − 2 is a factor of p(x).

(6 marks) Cambridge IGCSE Additional Mathematics 0606, Paper 22, Q4 May/June 2021

I can use the factor theorem to identify linear factors of polynomials.

I can solve cubic equations.

(3 marks)

b Do not use a calculator in this question.

Progression checklist

I can factorise polynomials.

− 4x + 12 = 0

Show that the value of a is 15.

10

I can find factors of polynomials.

−

3x 2

Solve

• The remainder when f(x) is divided by x − 1 is 9 2 Create a roller-coaster ride that represents a polynomial function and that will last a maximum of 30 seconds. Your ride must intersect with ground level at least once and return to ground level at the end of the ride. State the domain and range. Use a graphing program or a graphical calculator to explore how to design the ride. Then draw your graph on graph paper, using metres for the vertical scale and seconds for the horizontal scale. Use your graph to describe your ride.

x3

3

• f(x) has only one linear factor: a + c = 0 4 • A root of the equation f(x) = 0 is x = 5 • The remainder when f(x) is divided by x is −12

• a is a prime number.

(3 marks) (4 marks)

You can often tackle problem solving activities by means of a graph. For example, you can work out problems based on quadratics, such as how high a dirt-bike course should be in the O2 Arena, based on the graph of the parabolic function of the path of the bike and the height of the ramps.

Bearing that principle in mind, try these questions.

Use the following information to work out the values of a, b, c and d and find the factors of f(x):

+ 12x − 8.

Factorise x 3 − 4x 2 − 25x + 100.

a x 2 − 3x − 4 = 0

E

(2 marks)

3

The function f(x) = ax 3 + bx 2 + cx + d, where a, b, c and d are integers.

94

x3

Problem solving 4.2

1 Work in small groups to solve this problem.

(1 mark)

a Show that x − 2 is a factor of x 3 − 6x 2 + 12x − 8. b Factorise

The function f(x) = + + x + 5, where a is a constant. Given that the remainder when f(x) is divided by x − 1 is twice the remainder when f(x) is divided by x + 1, work out the value of a. x3

f(x) = x 3 − x 2 − 14x + 24

m

PS

Advice and Tips

ge s

PS

a Factorise f(x) = x4 − 25x 2 + 144.

Pa

5

11

(x − 3) is a factor of f(x) = ax 3 + bx 2 − 9x − 9, where a and b are constants. a Show that 3a + b = 4.

(2 marks)

b When f(x) is divided by x − 2, the remainder is −15.

Show that 2a + b = 3. c Find the value of a and the value of b.

(3 marks) (3 marks) Chapter 4: Factors and Polynomials

95

Chapter 4 . Topic 1 12

E

When f(x) = 2x 3 + ax 2 + bx − 12 is divided by x − 2, the remainder is 20.

When g(x) = 2x 3 + bx 2 + ax − 12 is divided by x − 2, the remainder is 18.

Work out the values of a and b.

13

(5 marks)

The polynomial p(x) = 2x 3 − 3x 2 + ax + b, where a and b are integers, has a factor of x + 2. a Given that p(2) = 2p(1), find the values of a and b.

(4 marks)

b Using your values of a and b, i

find the remainder when p(x) is divided by x − 3

(2 marks)

Sa

m

pl

e

Pa

ge s

ii factorise p(x).

(2 marks)

Past paper questions reproduced by permission of Cambridge Assessment International Education.

Other exam-style questions have been written by the authors.

96

Chapter review