AS PURE MATHEMATCICS WORKED SOLUTIONS: THE FACTOR THEOREM 1

Exercise 5A

Question 1:

a.

2

3 2

3 2

2

2

6 223 3 4 1

3 6 4 6 18 22 1 22 66 67

x xx x x x

x xx xx x

xx

− ++ − + −

+− +− −

−+−

3 2 2 673 4 1 6 223

x x x x xx

∴ − + − = − + −+

The rest can be done similarly.

b. 22 2x x+ +

c. 3 2 5 3 22 2 1

x xx

+ + −−

d. 3 2 6759 42 1694

x x xx

− + − ++

e. 4 3 2 233 5 112

x x x xx

+ + + + +−

f. 25 23 165 559 644 16 64 4x x

x+ + +

+

g. 2 84 52

x xx

− + − ++

h. 5 4 3 22 13 110 523 2696 11779 58166 7293 9 27 81 243 729 3 5x x x x x

x+ + + + + +

−

i. 4 3 2 2312 2 9 26 793

x x x xx

− − − − − −−

j. 5 4 3 2 62117411 95 852 7669 690209

x x x x xx

+ + + + + +−

k. 23 5 3 5 82 4 8 1 2x x

x− − + +

−

l. 21 13 55 27 644 16 64 3 4x x

x− + + +

−

AS PURE MATHEMATCICS WORKED SOLUTIONS: THE FACTOR THEOREM 2

Question 2:

2

3 2

3 2

2

2

5 11 4 6 2

5 6 5 5 2 1 1

x xx x x x

x xx xx x

xx

+ −− + − +

−−−− +− +

3 224 6 2 15 1

1 1x x x x x

x x+ − +∴ = + − +

− −

AS PURE MATHEMATCICS WORKED SOLUTIONS: THE FACTOR THEOREM 3

Question 3:

2

3 2

3 2

2

2

3 6 112 3 0 1

3 6 6 6 12 11 1 11 22 23

x xx x x x

x xx xx x

xx

− ++ + − −

+− −− −

−+−

( ) ( )3 2 231 2 3 6 112

x x x x xx

∴ − − ÷ + = − + −+

AS PURE MATHEMATCICS WORKED SOLUTIONS: THE FACTOR THEOREM 4

Question 4:

a.

2

3 2

3 2

2

2

8 203 5 4 1

3 8 4 8 24 20 1 22 60 61

x xx x x x

x xx xx x

xx

+ +− + − +

−−−

+−

So the remainder is 61 .

b. The remainder is 2− .

c. The remainder is 116

−

d. The remainder is 18 .

AS PURE MATHEMATCICS WORKED SOLUTIONS: THE FACTOR THEOREM 5

Exercise 5B

Question 1:

a. Let 3 2( ) 3 2 5 1f x x x x= + − + . Then

( ) ( ) ( )3 2( 2) 3 2 2 2 5 2 15

f − = − + − − − += −

So by the remainder theorem, the remainder is 5− .

b. Let 4 3( ) 2 2 5 1f x x x x= − − + + .Then

( ) ( ) ( )4 3(3) 2 3 2 3 5 3 1200

f = − − + += −

So by the remainder theorem, the remainder is 200−

c. Let 5 4 3( ) 8 4 6 4 4f x x x x x= + + − + .Then

5 4 31 1 1 1 18 4 6 4 42 2 2 2 2

134

f ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

=

So by the remainder theorem, the remainder is 134

.

The rest can be done similarly.

d. The remainder is 6297

e. The remainder is 53627

f. The remainder is 338

g. The remainder is 0

h. The remainder is 534

.

AS PURE MATHEMATCICS WORKED SOLUTIONS: THE FACTOR THEOREM 6

Exercise 5C

Question 1:

a. Let 3 2( ) 2 3 11 6f x x x x= + − − .

Then ( ) ( ) ( )3 2(2) 2 2 3 2 11 2 60

f = + − −=

So by the factor theorem, 2x − is a factor of 3 22 3 11 6x x x+ − − .

b.

2

3 2

3 2

2

2

2 7 32 2 3 11 6

2 4 7 11 7 14 3 6 3 6 0

x xx x x x

x xx xx x

xx

+ +− + − −

−−−

−−

( )( )

( )( )( )

3 2 22 3 11 6 2 2 7 3

2 2 1 3

x x x x x x

x x x

∴ + − − = − + +

= − + +

AS PURE MATHEMATCICS WORKED SOLUTIONS: THE FACTOR THEOREM 7

Question 2:

a. Let 3 2( ) 2 9 6 5f x x x x= − − + . Then

( ) ( ) ( )3 2(5) 2 5 9 5 6 5 50

f = − − +=

So 5x = is a solution to ( ) 0f x = .

b. 2

3 2

3 2

2

2

2 15 2 9 6 5

2 10 6 5 5 5 0

x xx x x x

x xx xx x

xx

+ −− − − +

−−−− +− +

So we can factorise

( )( )( )( )( )

3 2 22 9 6 5 5 2 1

5 2 1 1

x x x x x x

x x x

− − + = − + −

= − − +

11, ,52

x∴ = −

AS PURE MATHEMATCICS WORKED SOLUTIONS: THE FACTOR THEOREM 8

Question 3:

a. If 2x = then ( ) ( ) ( )3 22 2 6 2 8 2 24 0y = + − − = so 2x − is a factor. Then

2

3 2

3 2

2

2

2 10 122 2 6 8 24

2 4 10 8 10 20 12 24 12 24 0

x xx x x x

x xx xx x

xx

+ +− + − −

−−−

−−

So

( )( )( )( )( )

22 2 10 12

2 2 2 3

y x x x

x x x

= − + −

= − + +

b.

AS PURE MATHEMATCICS WORKED SOLUTIONS: THE FACTOR THEOREM 9

Question 4:

Let 3 2( ) 4 5 2f x ax x x= − − + . Since 1x + is a factor of ( )f x we have ( )1 0f − = . Then

( ) ( ) ( )3 21 4 1 5 1 2 0 3a a− − − − − + = ⇒ = . 2

3 2

3 2

2

2

3 7 21 3 4 5 2

3 3 7 5 7 7 2 2 2 2 0

x xx x x x

x xx xx x

xx

− ++ − − +

+− −− −

++

( )( )( )( )( )

3 2 23 4 5 2 1 3 7 2

1 3 1 2

x x x x x x

x x x

∴ − − + = + − +

= + − −

AS PURE MATHEMATCICS WORKED SOLUTIONS: THE FACTOR THEOREM 10

Exercise 5C

Question 1:

2

3 2

3 2

2

2

2 93 2 5 6 1

2 6 6 3 9 1 9 27 26

x xx x x x

x xx xx x

xx

+ +− − + −

++−

−−

3 222 5 6 1 262 9

3 3x x x x x

x x− + −∴ = + + +

− −

AS PURE MATHEMATCICS WORKED SOLUTIONS: THE FACTOR THEOREM 11

Question 2:

a. Let 3 2( ) 4 8 7 5f x x x x= − − +

i. ( ) ( ) ( )3 2( 3) 4 3 8 3 7 3 5 154f − = − − − − − + = − is the remainder

ii. ( ) ( ) ( )3 2( 1) 4 1 8 1 7 1 5 0f − = − − − − − + = is the remainder.

b. Since 1x + is a factor and 2

3 2

3 2

2

2

4 12 51 4 8 7 5

4 4 12 7 12 12 5 5 5 5 0

x xx x x x

x xx xx x

xx

− ++ − − +

+− −− −

++

then we can factorise ( )( )( )( )( )

3 2 24 8 7 5 1 4 12 5

1 2 1 2 5

x x x x x x

x x x

− − + = + − +

= + − −

1 51, ,2 2

x∴ = −

AS PURE MATHEMATCICS WORKED SOLUTIONS: THE FACTOR THEOREM 12

Question 3:

a. Let 3 2( ) 5 2 8f x x x x= − + + then since ( 1) 1 5 2 8 0f − = − − − + = we know that 1x + is a

factor. Then 2

3 2

3 2

2

2

6 81 5 2 8

6 2 6 6 8 8 8 8 0

x xx x x x

x xx xx x

xx

− ++ − + +

+− +− −

++

( )( )

( )( )( )

3 2 25 2 8 1 6 8

1 2 4

x x x x x x

x x x

∴ − + + = + − +

= + − −

b. Let y x= so the equation is 3 22 5 8 0y y y+ − + = whose solutions are 1,2,4y = − from

above. Then 4,16x = since the square root is non-negative.

AS PURE MATHEMATCICS WORKED SOLUTIONS: THE FACTOR THEOREM 13

Question 4:

a.

2

3 2

3 2

2

2

6 101 5 4 10

6 4 6 6 10 10 10 10 0

x xx x x x

x xx xx x

xx

− ++ − + +

+− +− −

++

So ( )( )3 2 25 4 10 1 6 10x x x x x x− + + = + − + , but the discriminant of the quadratic factor is

( )( )26 4 1 10 4 0Δ = − = − < so it has no real roots. Hence 1x = − is the only real root.

b.

AS PURE MATHEMATCICS WORKED SOLUTIONS: THE FACTOR THEOREM 14

Question 5:

a. 2x − is a factor of ( )f x b. From the remainder theorem, we have ( 1) 18f − = and (2) 0f = , that is

1 10 18 98 4 20 0 4 12

a b a ba b a b

− + + + = ⇒ + =+ − + = ⇒ + =

Subtracting the first equation from the second gives 3 3 1a a= ⇒ = and so b = 8 .

c. You can show that

x + 4( ) x − 2( ) x −1( )

AS PURE MATHEMATCICS WORKED SOLUTIONS: THE FACTOR THEOREM 15

Question 6:

Let 3 2( ) 16f x x x ax= − + + .Sinceboth x b+ and x b− arefactorsthenwehave ( ) 0f b = and

( ) 0f b− = .Thatis

3 2

3 2

16 016 0

b b abb b ab− + + =

− − − + =

Addingthetwoequationsgives 2 22 32 0 16b b− + = ⇒ = .

4b∴ = sinceb ispositive.