aqa level 2 further mathematics · in general, the factor theorem states that: the factor theorem...

AQA Level 2 Further Mathematics

© MEI, 09/08/12 1/1

Further algebra

Section 2: Further equations

Crucial points

1. Be careful with signs when using the elimination method It’s very easy to make mistakes!

2. Think about which method to use If one equation gives, say, y in terms of x, it is usually easier to use the substitution method rather than the elimination method. When one equation is quadratic, you must always use substitution.

3. Remember that for non-linear simultaneous equations there may

be more than one solution When you solve simultaneous equations where one is linear and one is quadratic, you should normally end up with two solutions (unless the quadratic equation you end up with has a repeated root or no roots).

4. Check your answers For simultaneous equations, just substitute your solution into both of the original equations to make sure that it fits. When factorising a cubic, remember that you can check your answer by multiplying out. You can also check the solutions of a cubic equation by substitution.

5. Find a method for factorising that suits you Once you have found a factor of a cubic expression using the factor theorem, there are a number of different ways of dividing so that you can complete the factorisation. Several different methods are shown in the interactive resources on the website. Try some different methods and then stick with the one that you are most comfortable with.

6. Take care with signs Be careful about signs when using the factor theorem:

Wrong f(2) = 0 (x + 2) is a factor

Right f(2) = 0 (x – 2) is a factor


© MEI, 09/08/12 1/2

Further algebra


Section overview Background The first part of this section looks at linear simultaneous equations, and you then go on to look at simultaneous equations for which one equation is linear and the other is quadratic.

Specification content Algebraic and graphical solution of simultaneous equations in two unknowns where the equations could both be linear or one linear and one second order In this section you also look at factorising cubic expressions and solving cubic equations, using the factor theorem. Use of the factor theorem.

Key points Know how to solve simultaneous equations by substituting the linear equation into the quadratic equation. The factor theorem: if f(a) = 0, then (x – a) is a factor of f(x).

Additional resources Flash resources Simultaneous equations shows examples of solving simultaneous equations, one of which is linear and one quadratic. Polynomial division by inspection shows examples of dividing a cubic expression by a linear expression using the inspection method. Polynomial division – the box method shows examples of dividing a cubic expression by a linear expression using a table. Factorising a cubic shows examples of finding one factor using the factor theorem, and then dividing out and factorising completely. Polynomial long division shows examples of dividing a cubic expression by a linear expression using long division. Interactive questions Solving linear simultaneous equations tests you on finding the solution for two linear simultaneous equations. Forming and solving linear simultaneous equations tests you on problems expressed in words. Simultaneous equations involving quadratics tests you on finding the number of solutions for one linear and one quadratic equation.

AQA FM Further algebra Section 2 Overview

© MEI, 09/08/12 2/2

PowerPoint Factorising polynomials takes you through examples of factorising using either inspection or the “box” method. Mathscentre videos The simultaneous equations video looks at solving linear simultaneous equations using elimination or substitution. The Solving cubics video looks at examples of solving cubic equations and sketching graphs, using a given factor and the inspection method of factorising.


© MEI, 09/08/12 1/7

Further algebra


Notes and Examples These notes contain subsections on

Linear simultaneous equations using elimination

Linear simultaneous equations using substitution

One linear and one quadratic equation

The factor theorem

Dividing a cubic expression by a linear factor

Linear simultaneous equations using elimination This work may be revision. You need to make sure that you can solve linear simultaneous equations confidently before you move on to the work on one linear and one quadratic equation, which will probably be new to you. Simultaneous equations involve more than one equation and more than one unknown. To solve them you need the same number of equations as there are unknowns. One method of solving simultaneous equations involves adding or subtracting multiples of the two equations so that one unknown disappears. This method is called elimination, and is shown in the next example. Example 1

Solve the simultaneous equations

3 5

2 4

p q

p q

Solution

3 5

2 4

p q

p q

6 2 10

2 4

7 14

2

p q

p q

p

p

3 2 5

6 5

1

q

q

q

Now substitute this value for p

into one of the original equations – you can use either,

but in this example, is used.

2

Adding:

Adding or subtracting these equations will not eliminate either p or q. However, you can multiply

the first equation by 2, and then add. This will eliminate p.

AQA FM Further algebra Section 2 Notes & Examples

© MEI, 09/08/12 2/7

The solution is p = 2, q = -1.

Notice that, in Example 1, you could have multiplied equation by 3 and then subtracted. This would give the same answer. Sometimes you need to multiply each equation by a different number before you can add or subtract. This is the case in the next example. Example 2

5 pencils and 2 rubbers cost £1.50

8 pencils and 3 rubbers cost £2.35

Find the cost of a pencil and the cost of a rubber.

Solution

5 2 150

8 3 235

p r

p r

15 6 450

16 6 470

20

20

p r

p r

p

p

5 20 2 150

100 2 150

2 50

25

r

r

r

r

A pencil costs 20p and a rubber costs 25p.

Linear simultaneous equations using substitution An alternative method of solving simultaneous equations is called substitution. This can be the easier method to use in cases where one equation gives one of the variables in terms of the other. This is shown in the next example. Example 3


3 2 11

5 2

x y

y x

Let p represent the cost of a pencil and r

represent the cost of a rubber. It is easier to work in pence.

3

2

Subtracting:

Substitute this value of p into equation

The easiest method is to multiply equation by 3 and equation by 2. (You could of course multiply

by 8 and by 5).


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Solution

3 2(5 2 ) 11

3 10 4 11

7 21

3

x x

x x

x

x

5 2 3

5 6

1

y

The solution is x = 3, y = -1

For some practice in examples like the ones above, try the interactive resources Solving linear simultaneous equations and Forming and solving linear simultaneous equations. You can also look at the Simultaneous equations video.

One linear and one quadratic equation When you need to solve a pair of simultaneous equations, one of which is linear and one of which is quadratic, you need to substitute the linear equation into the quadratic equation. Example 4


2 22 6

1

x y

x y

Solution

1x y

2 2

2 2

2

( 1) 2 6

2 1 2 6

3 2 5 0

(3 5)( 1) 0

y y

y y y

y y

y y

53 or 1y y

53

y 5 23 3

1 1x y

1y 1 1 1 2x y

The solutions are 523 3,x y and 2, 1x y

Multiply out the brackets

Substitute the value for x into

the original second equation

Sometimes you will need to use the quadratic formula to solve

the resulting quadratic equation.

Substitute the expression for y given in the second equation, into the first equation.

Start by using the linear equation to write

one variable in terms of the other.

Now substitute this expression for y into the first equation

Multiply out, simplify and factorise

Now substitute each value for y into the linear

equation to find the corresponding values of x


© MEI, 09/08/12 4/7

You can look at some more examples like these using the Flash resource Simultaneous equations.

The factor theorem

You already know that you can solve some quadratics by factorising them. e.g. to solve the quadratic equation x² + 3x – 10 = 0 you factorise: (x + 5)(x – 2) = 0 and deduce the solutions x = -5 and x = 2 Clearly, for f(x) = x² + 3x – 10, f(-5) = 0 and f(2) = 0.

(x + 5) is a factor of f(x) f(-5) = 0

(x – 2) is a factor of f(x) f(2) = 0 This idea can be extended to other polynomials such as cubics. For example, for the cubic function g( ) ( 1)( 2)( 3)x x x x , g(1) = 0, g(-2) = 0

and g(3) = 0.

(x – 1) is a factor of g(x) g(1) = 0

(x + 2) is a factor of g(x) g(-2) = 0

(x – 3) is a factor of g(x) g(3) = 0 In general, the factor theorem states that:

The factor theorem is useful for factorising cubic expressions and for solving cubic equations. As with quadratics, it may or may not be possible to factorise a cubic expression. If it is possible, the first step is to factorise it into a linear factor and a quadratic factor. Then it may be possible to factorise the quadratic factor into two further linear factors.

Example 5

(i) Show that x + 2 is a factor of x³ – 2x² – 5x + 6.

(ii) Show that x³ – 2x² – 5x + 6 = (x + 2)(x² – 4x + 3).

(iii) Hence solve the equation x³ – 2x² – 5x + 6 = 0.

Solution

(i) f(x) = x³ – 2x² – 5x + 6

If x a is a factor of f(x), then f(a) = 0 and x = a is a root of the equation f(x) = 0.

Conversely, if f(a) = 0, then x a is a factor of f(x).


© MEI, 09/08/12 5/7

3 2f ( 2) ( 2) 2( 2) 5 2 6

8 8 10 6

0

f(-2) = 0 so (x + 2) is a factor of f(x).

(ii) 2 2 2

3 2 2

3 2

( 2)( 4 3) ( 4 3) 2( 4 3)

4 3 2 8 6

2 5 6

x x x x x x x x

x x x x x

x x x

(iii) 3 2

2

2 5 6 0

( 2)( 4 3) 0

( 2)( 1)( 3) 0

x x x

x x x

x x x

The roots of the equation are x = -2, x = 1 and x = 3.

In the example above, you were given the factorisation into a linear and a quadratic factor, and you just needed to show that it was correct. More often you will need to find the quadratic factor yourself. So you need a method of dividing a cubic expression by a linear factor to give the quadratic factor. There are several different methods of doing this.

Dividing a cubic expression by a linear factor In this course, all the cubic expressions you will see in this course have 1 as the coefficient of x³. This means that the quadratic factor will have 1 as the coefficient of x². So you can express the quadratic factor as x² + px + q, and then multiply out and find the values of p and q by comparing with the original expression.

Example 6

(i) Show that x – 1 is a factor of x³ – 3x² – x + 3.

(ii) Factorise x³ – 3x² – x + 3 completely.

Solution

(i) f(x) = x³ – 3x² – x + 3

f(1) = 1 – 3 – 1 + 3 = 0

so x – 1 is a factor.

(ii) 3 2 2

2 2

3 2 2

3 2

3 3 ( 1)( )

( ) 1( )

( 1) ( )

x x x x x px q

x x px q x px q

x px qx x px q

x p x q p x q

The quadratic factor can be factorised into

two linear factors


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Equating coefficients of x² p – 1 = -3 p = -2

Equating constant terms -q = 3 q = -3

(Check: Coefficients of x = q – p = -3 – (-2) = -1)

3 2 23 3 ( 1)( 2 3)

( 1)( 1)( 3)

x x x x x x

x x x

In the example above, you may have noticed that it is very easy to find the value of q, just by thinking about the constant term. It is possible to do the factorisation ‘in your head’, without writing it out using p and q. It’s also clear that the constant term in the quadratic factor must be -3, to give the constant term 3 using -1 × -3 = 1. So the factorisation can be written like this:

3 2 23 3 ( 1)( 3)x x x x x x

You can then work out the missing coefficient by thinking about either the terms in x² or the terms in x. Multiplying out the linear and quadratic factors will give a term of -x² (from multiplying -1 in the linear factor by x² in the quadratic factor). So we need another -2x² to get a total of -3x². So the x term in the quadratic factor must be -2x, so that we get -2x² by multiplying x in the linear factor by -2x in the quadratic factor.

3 2 23 3 ( 1)( 2 3)x x x x x x

Don’t worry if you find this method too difficult at first. It’s fine to use the method shown in Example 6, or one of the other methods shown in the interactive resources listed below. The methods shown in the Flash resources Polynomial division by inspection, Factorising a cubic and Polynomial division – box method are similar to the ones shown above. The ‘box method’ is just another way of setting out the working which some people prefer. The PowerPoint presentation Factorising polynomials shows the inspection method and the box method. Another method is algebraic long division. You can see examples of this method using the Flash resource Polynomial long division. You can also look at the Solving cubics video. Example 7

f(x) = x³ + px² + 11x – 6 has a factor x – 2.


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Find the value of p and hence factorise f(x) as far as possible.

Solution

x – 2 is a factor of f(x) f(2) = 0

f(2) = 8 + 4p + 22 – 6 = 24 + 4p

24 + 4p = 0 p = -6

f(x) = x³ – 6x² + 11x – 6

3 2 26 11 6 ( 2)( 4 3)

( 2)( 1)( 3)

x x x x x x

x x x


© MEI, 15/10/12 1/1

Further algebra


Exercise

1. Solve the following simultaneous equations:

(i) 2x + 5y = 11 (ii) x + 2y = 6

2x – y = 5 4x + 3y = 4

(iii)3a – 2b = 4 (iv) 2p – 5q = 5

5a + 4b = 3 3p – 2q = –9

(v) 5x + 3y = 9 (vi) 3a + 2b = 1

y = 3x – 4 9a – 4b = 4

2. Solve the following simultaneous equations.

(i) 7x² + y² = 64 (ii) 3x² – 2y² = –5

x + y = 4 y – x = 1

(iii)p² + pq = 2 (iv) 8a² – b² = 2

q – p = 3 2a + b = 1

3. (i) Show that x + 1 is a factor of 3 24 6.x x x

(ii) Hence factorise 3 24 6.x x x completely.

4. x – 2 is a factor of the polynomial 3 2 4 12x ax x .

(i) Find the value of a.

(ii) Factorise the polynomial completely.

5. Solve the equations

(i) 3 22 11 12 0x x x

(ii) 3 24 3 18 0x x x

(iii) 3 19 30 0x x

6. Bob factorises 3 24 7 10x x x and gets ( 1)( 2)( 5)x x x .

Explain how you know that Bob is wrong.

7. A rectangle with length x cm and width y cm has a square

3 cm by 3 cm removed from a corner to leave an L shape.

The area of the L shape is 15 cm2.

The perimeter of the L shape is 20 cm.

Find the values of x and y.


© MEI, 15/10/12 1/5

Further algebra


Solutions to Exercise

1. (i) 2 5 11 (1)

2 5 (2)

x y

x y

Subtracting: 6 6

1

y

y

Substituting into (1): 2 5 1 11

2 6

3

x

x

x

The solution is x = 3, y = 1. Check: 2 5 2 3 5 1 11

2 2 3 1 5

x y

x y

(ii) 2 6 (1) 4

4 3 4 (2)

x y

x y

4 8 24

4 3 4

x y

x y

Subtracting: 5 20

4

y

y

Substituting into (1): 2 4 6

2

x

x

The solution is x = -2, y = 4. Check: 2 2 8 6

4 3 8 12 4

x y

x y

(iii) 3 2 4 (1) 2

5 4 3 (2)

a b

a b

6 4 8

5 4 3

a b

a b

Adding: 11 11

1

a

a

Substituting into (1):

12

3 1 2 4

2 1

b

b

b

The solution is a = 1, b = 12 . Check: 3 2 3 1 4

5 4 5 2 3

a b

a b

(iv) 2 5 5 (1) 3

3 2 9 (2) 2

p q

p q

6 15 15

6 4 18

p q

p q

Subtracting: 11 33

3

q

q

AQA FM Further algebra Section 2 Exercise solutions

© MEI, 15/10/12 2/5

Substituting into (1): 2 5 3 5

2 10

5

p

p

p

The solution is p = -5, q = -3. Check: 2 5 10 15 5

3 2 15 6 9

p q

p q

(v) 5 3 9 (1)

3 4 (2)

x y

y x

Substituting (2) into (1):

32

5 3(3 4) 9

5 9 12 9

14 21

x x

x x

x

x

Substituting into (1): 93 1

2 2 23 4 4y

The solution is 3 12 2,x y . Check: 15 3

2 25 3 9x y

(vi) 3 2 1 (1) 2

9 4 4 (2)

a b

a b

6 4 2

9 4 4

a b

a b

Adding:

25

15 6a

a

Substituting into (1): 25

6 15 5

110

3 2 1

2 1

b

b

b

The solution is a = 25 , b = 1

10 . Check: 6 15 5

18 25 5

3 2 1

9 4 4

a b

a b

2. (i) 2 27 64 (1)

4 (2)

x y

x y

(2) 4y x


2 2

2

2

7 (4 ) 64

7 16 8 64

8 8 48 0

6 0

( 3)( 2) 0

3 or 2

x x

x x x

x x

x x

x x

x x

When x = 3, 4 3 1y

When x = -2, 4 ( 2) 6y


© MEI, 15/10/12 3/5

The solutions are x = 3, y = 1 and x = -2, y = 6

Check: 2 2

2 2

3, 1 7 63 1 64

2, 6 7 28 36 64

x y x y

x y x y

(ii) 2 23 2 5 (1)

1 (2)

x y

y x

(2) 1y x


2 2

2 2

2

3 2(1 ) 5

3 2(1 2 ) 5

3 2 4 2 5

4 3 0

( 1)( 3) 0

1 or 3

x x

x x x

x x x

x x

x x

x x

When x = 1, 1 1 2y

When x = 3, 1 3 4y

The solutions are x = 1, y = 2 and x = 3, y = 4

Check: 2 2

2 2

1, 2 3 2 3 5 5

3, 4 3 2 27 32 5

x y x y

x y x y

(iii) 2 2 (1)

3 (2)

p pq

q p

(2) 3q p

Substituting into (1): 2

2 2

2

12

(3 ) 2

3 2

2 3 2 0

(2 1)( 2) 0

or 2

p p p

p p p

p p

p p

p p

When 71 12 2 2, 3p q

When 2, 3 2 1p q

The solutions are 712 2,p q and p = -2, q = 1.

Check: 7 721 12 2 4 4

2

, 2

2, 1 4 2 2

p q p pq

p q p pq

(iv) 2 28 2 (1)

2 1 (2)

a b

a b

(2) 1 2b a


© MEI, 15/10/12 4/5


2 2

2 2

2

3 12 2

8 (1 2 ) 2

8 (1 4 4 ) 2

8 1 4 4 2

4 4 3 0

(2 3)(2 1) 0

or

a a

a a a

a a a

a a

a a

a a

When 3 32 2, 1 2 1 3 4a b

When 1 12 2, 1 2 1 1 0a b

The solutions are 32 , 4a b and 1

2 , 0a b .

Check: 92 23

2 4

2 21 12 4

, 4 8 8 16 18 16 2

, 0 8 8 0 2

a b a b

a b a b

3. (i) 3 2f( ) 4 6x x x x

3 2f( 1) ( 1) 4( 1) ( 1) 6 1 4 1 6 0

so by the factor theorem, x + 1 is a factor.

(ii)

3 2 24 6 ( 1)( 5 6)

( 1)( 2)( 3)

x x x x x x

x x x

4. (i) 3 2f( ) 4 12x x ax x

3 2f(2) 2 2 4 2 12

8 4 8 12

4 12

a

a

a

x – 2 is a factor so by the factor theorem f(2) = 0

4 12 0

3

a

a

(ii)

3 2 23 4 12 ( 2)( 6)

( 2)( 2)( 3)

x x x x x x

x x x

5. (i) 3 2f( ) 2 11 12x x x x

f(1) 1 2 11 12 0 so (x – 1) is a factor

3 2

2

2 11 12 0

( 1)( 12) 0

( 1)( 3)( 4) 0

x x x

x x x

x x x

x = 1, x = -3, x = 4


© MEI, 15/10/12 5/5

(ii) 3 2f( ) 4 3 18x x x x

f(1) 1 4 3 18 0

f( 1) 1 4 3 18 0

f(2) 8 16 6 18 0

so (x – 2) is a factor

3 2

2

2

4 3 18 0

( 2)( 6 9) 0

( 2)( 3) 0

x x x

x x x

x x

x = 2 or x = -3

(iii) 3f( ) 19 30x x x

f(1) 1 19 30 0

f(2) 8 38 30 0

f( 2) 8 38 30 0

so x + 2 is a factor

3

2

19 30 0

( 2)( 2 15) 0

( 2)( 3)( 5) 0

x x

x x x

x x x

x = -2 or x = -3 or x = 5

6. If Bob is right, then x = 1, x = 2 and x = -5 would all make 3 24 7 10x x x be zero.

x=1 3 24 7 10 1 4 7 10 0x x x so (x – 1) is a factor.

x=2 3 24 7 10 8 16 14 10 12 0x x x so (x - 2) is not a factor.

7. 9 15xy

2 2 20x y

2

2

10

(10 ) 24

10 24

10 24 0

( 6)( 4) 10

y x

x x

x x

x x

x x

x = 6 and y = 4 (or x = 4 and y = 6)


© MEI, 09/08/12 1/2

Further algebra


Multiple Choice Test Questions 1 and 2 are about the simultaneous equations

3 5

3 5

x y

x y

1) The correct value of x for the solution is

(a) x = 2 (b) x = -1

(c) x = 1 (d) x = -2

(e) I don’t know

2) The correct value of y for the solution is

(a) y = -1 (b) y = 1

(c) y = 2 (d) y = -2

(e) I don’t know

3) For the simultaneous equations

5 7 17

1 3

a b

a b

the correct value of a for the solution is

(a) 32

a (b) 72

a

(c) a = 2 (d) 112

a

(e) I don’t know


2 5 2

6 1 4

x y

y x

the correct value of x for the solution is

(a) 78

x (b) 34

x

(c) 38

x (d) 178

x

(e) I don’t know

AQA FM Further algebra Section 2 MC test

© MEI, 09/08/12 2/2


2 22 6

3 5

s t

s t

the values of t for the solutions are

(a) t = -11 and 16119

t (b) t = 1 and 2919

t

(c) t = -11 and 2919

t (d) t = 1 and 16119

t

(e) I don’t know


2 2 5

2 3 12

x y

x y

the values of y for the solutions are

(a) y = -6 and 629

y (b) y = -2 and 109

y

(c) y = -2 and 629

y (d) y = -6 and 109

y

(e) I don’t know

7) Which of the following is a factor of x³ + x² + 2x + 8?

(a) x + 1 (b) x – 1

(c) x + 2 (d) x – 2

(e) I don’t know

8) x – 2 is a factor of x³ – 5x² + ax + 2.

The value of a is

(a) -5 (b) 5

(c) -13 (d) 13

(e) I don’t know

9) (x – 1) is a factor of x³ + x² – 5x + 3. This expression can be written in the form

(a) (x – 1)(x² – 2x + 3) (b) (x – 1)(x² + x – 2)

(c) (x – 1)(x² + 2x – 3) (d) (x – 1)(x² – x + 2)

(e) I don’t know

10) Factorise x³ – x² – 34x – 56

(a) (x – 2)(x – 4)(x + 7) (b) (x + 2)(x + 4)(x – 7)

(c) (x – 1)(x – 7)(x – 8) (d) (x + 1)(x + 7)(x – 8)

(e) I don’t know

aqa level 2 further mathematics · in general, the factor theorem states that: the factor theorem...

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