Cambridge IGCSE Additional Maths SB Sample

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

Factors and Polynomials SamplePages

Starting point

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.

1 a Factorise x 2 2 x 15 = 0.

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

2 Solve 12 x 2 = 28 x 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.

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.

The factor theorem

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.

Examples of polynomials are:

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

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

You can factorise x 2 2 x 3 to give ( x 3)( x + 1).

You can solve the quadratic equation x2 2 x 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 nd 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 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.

You can extend these ideas to cubic and other polynomials.

The factor theorem

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.

Note that if ax b is a factor of the polynomial, then f       b a = 0.

For example, if 2 x 1 is a factor of the polynomial, then f       1 2 = 0.

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The quotient

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

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

The long division method

2 2

+ ( (

x x xx xx x x

) )

You can use the long division method to nd factors. 1 32 3 3 3 3 0

Using the long division method

Always look at the highest powers.

Dividing the x 2 term in the polynomial by the x term in the divisor gives x as the rst part of the quotient.

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

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

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

This divides exactly so x + 1 is the quotient. Similarly, given that x + 4 is a factor of f ( x) = x 3 + 2 x ² 11x 12, you can use the long division method to obtain the quotient, which will be a quadratic.

Advice and Tips

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

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

And then factorising ( x 2 2 x 3) leads to:

x 3 + 2 x 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.

Look at the same example as above.

Given that f ( x) = x 3 + 2 x 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.

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

The quotient will appear along the top of the grid.

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

x 3 + 2 x 2 11x 12

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

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

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

x 2 2 x

x 3 2 x 2

Advice and Tips

Note that the sum of the 4x2 and the 2x2 combine to the term in 2x2 in the original polynomial.

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

× x 2 2 x

x x 3 2 x 2

+ 4 4 x 2 8 x

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.

× x 2 2 x 3

x x 3 2 x 2 3x + 4 4 x 2 8 x

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

× x 2 2 x 3

x x 3 2 x 2 3x + 4 4 x 2 8 x 12

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

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

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

Factorising the quadratic expression gives: x 3 + 2 x 2 11x 12 = ( x + 4)( x 3)( x + 1).

This method is actually easier than it seems at rst 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 + 6 x 2 + 8 x

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) = 64 + 96 32 = 0

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

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).

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

Checking: f ( 2) = ( 2)3 + 6( 2)2 + 8( 2) = 8 + 24 16 = 0

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

Example 2

The polynomial f ( x) = x 3 + ax 2 + 11x 6 has a factor of x 3. 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 27 + 9a + 33 6 = 0 So, 9a = 54 a = 6

Example 3

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

Example 4

So, x 3 2 x 2 13x 10 = ( x 5)( x 2 + 3x + 2) = ( x 5)( x + 1)( x + 2).

1 In each part, demonstrate that the expression is a factor of the polynomial. a x + 1, x 3 + 6 x 2 9x 14 b x 3, x 3 + 3x 2 13x 15 c x 4, x 3 7x 2 + 2 x + 40 d x + 6, x 3 + 13x 2 + 54 x + 72 e x + 7, x 3 37x + 84 f 2 x 3, 2 x 3 5x 2 x + 6

2 Factorise each expression. a x 3 + 3x 2 13x 15 b x 3 + 5x 2 + 2 x 8 c x 3 + 6x 2 + 11x + 6 d x 3 + 6 x 2 x 30 e 2 x 3 13x 2 + 17x + 12 f 6x 3 + 19x 2 + 11x 6 g x 3 + x 2 + x + 1

3 Solve each equation. a x 3 4 x 2 7x + 10 = 0 b x 3 3x 2 + 3x 1 = 0 c x 3 + 2 x 2 16x 32 = 0 d x 3 + 3x 2 13x 15 = 0 e x 3 + 15x 2 + 68 x + 84 = 0 f x 3 11x 2 10x + 200 = 0

4 The equation x 3 3x 2 + ax + b = 0 has solutions x = 1 and x = 4. Work out the third solution to the equation.

Advice and Tips

f ( x) =

Advice and Tips

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

5 a x + 3 is a factor of x 3 + 2 x 2 5x + c. Demonstrate that c = 6. b Work out the other two factors of the cubic. 6 a Given that ( x 3) is a factor of ax 3 7x 2 + 2 x + 3, use the factor theorem to demonstrate that a = 2. b Showing all your working, factorise ax 3 7x 2 + 2 x + 3, and hence solve ax 3 7x 2 + 2 x + 3 = 0. 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

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

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

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.

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.

One way to nd integer values of x so that

0 is to try different factors of the constant term.

4.2 The remainder theorem

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. Let f ( x) = ( x a) g( x) + R 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:

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.

If you divide a polynomial f ( x) by any factor (ax b), the remainder is f b a

, where a and b are constants and a is non-zero. For example, if you divide a polynomial f ( x) by a factor (2 x 1), the remainder is 

1 2

Example 5

Find the remainder when f ( x) = x 3 + 7x 2 6 x + 2 is divided by ( x + 3). The remainder is f ( 3) = ( 3)3 + 7( 3)2 6( 3) + 2 = 27 + 63 + 18 + 2 = 56

Example 6

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

Example 7

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

Example 8

The expression 2 x 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 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 + 5 a + b = 0

25 + 5 a + 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 5 a and b = 13 2a

25 5 a = 13 2a

3a = 12 a = 4 b = 25 5( 4) b = 25 + 20 b = 5

Exercise 4.2

1 a Find the remainder when f ( x) = x 3 + 2 x 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).

c Find the remainder when f ( x) = 2 x 3 4 x 2 2 x 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 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.

b When f ( x) = 2 x 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.

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

3

The expression 2 x 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

5

a Factorise f ( x) = x4 25x 2 + 144.

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

6 The polynomials f( x) = x 3 4x 2 + x + 6 and g( x) = x 3 + ax 2 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).

7

Advice and Tips

This polynomial can be factorised in the same way as the quadratic y 2 25y + 144, where y = x2

The function f ( x) = ax 3 3x 2 + bx + 12, where a and b are constants, is such that 2 x 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).

8 The function f ( x) = x 3 + ax 2 + 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

Problem solving 4.2

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.

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 nd the factors of f ( x):

• f ( x) has only one linear factor: a + c = 0

•A root of the equation f ( x) = 0 is x = 4 5

•The remainder when f ( x) is divided by x is 12

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

• a is a prime number.

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.

Progression checklist

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

I can nd factors of polynomials.

I can factorise polynomials.

I can solve cubic equations.

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

1 f ( x) = x 3 x 2 14 x + 24

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. (1 mark)

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

3 Factorise x 3 4 x 2 25x + 100. (4 marks)

4 Solve the equation x 3 + 6x 2 + 11x + 6 = 0. (5 marks)

5 Solve these equations. a x 2 3x 4 = 0 (3 marks) b x 3 3x 2 4 x = 0 (4 marks) c x 3 3x 2 4 x + 12 = 0 (5 marks)

6 Solve = x x x 35 11 3 3 . (5 marks)

7 The remainder when f ( x) = 3x 3 4 x 2 6 x + a is divided by ( x 2) is 11. Show that the value of a is 15. (3 marks)

8 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)

b Do not use a calculator in this question. 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 2 x 1 and a remainder of 5 when divided by x 1.

i Show that b = 8 and nd the value of a (4 marks)

ii Using the values of a and b from part (i), express f ( x) in the form (2 x 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

10 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

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. (3 marks)

c Find the value of a and the value of b. (3 marks)

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

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

Work out the values of a and b (5 marks)

13 The polynomial p( x) = 2 x 3 3x 2 + ax + b, where a and b are integers, has a factor of x + 2.

a Given that p(2) = 2p(1), nd the values of a and b. (4 marks)

b Using your values of a and b, i nd the remainder when p( x) is divided by x 3 (2 marks) ii factorise p( x). (2 marks)

SamplePages

Past paper questions reproduced by permission of Cambridge Assessment International Education. Other exam-style questions have been written by the authors.