AQA Level 2 Further Maths · Level 2 Certificate · Chapter 4: Algebra IV · Lesson 3

The factor theorem

About 85 minutesNo calculator82 marks

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Quick questions to start.

A1
Needed today
Expand and simplify $(x - 3) (x^{2} + 2 x - 4)$ .
[2 marks]
Answer
A2
GCSE Higher
Solve $x^{2} - 4 x - 3 = 0$ . Give your answers in surd form.
[2 marks]
Answer $x = \dots\dots$
A3
Last lesson
Solve the simultaneous equations $y = 2 x - 1$ and $x y = 15$ .
[2 marks]
Answer $(\dots, \dots)$ and $(\dots, \dots)$
A4
From chapter 2
Factorise $6 x^{2} + x - 12$ .
[2 marks]
Answer

Example, then your turn

Level 2

Your teacher works through each example with you. Then try the one beside it.

Example 1

$f (x) = x^{3} - 6 x^{2} + 5 x + 12$ . Work out $f (1)$ , $f (- 1)$ and $f (3)$ . Hence factorise $f (x)$ fully and solve $f (x) = 0$ .

Hint If

f (a) = 0

then

(x - a)

is a factor. Only factors of the constant term

12

are worth trying.

B1
Show that $(x - 2)$ is a factor of $x^{3} - 3 x^{2} - 10 x + 24$ . Hence factorise $x^{3} - 3 x^{2} - 10 x + 24$ fully.
[4 marks]
Answer

Example 2

$(x + 3)$ is a factor of $x^{3} + x^{2} - 4 x + 6$ . Work out the quadratic factor. Show that $(x + 3)$ is the only linear factor.

Hint Write

x^{3} + x^{2} - 4 x + 6 \equiv (x + 3) (x^{2} + b x + 2)

: the

x^{2}

and the

2

must be there to make

x^{3}

and

6

. Then compare the

x^{2}

terms.

B2
$(x - 3)$ is a factor of $x^{3} - 4 x^{2} + 6 x - 9$ . Show that the equation $x^{3} - 4 x^{2} + 6 x - 9 = 0$ has exactly one real root.
[4 marks]

Example 3

Show that $(3 x - 2)$ is a factor of $3 x^{3} + 4 x^{2} - 13 x + 6$ . Hence solve $3 x^{3} + 4 x^{2} - 13 x + 6 = 0$ .

Hint

(3 x - 2)

is zero when

x = \frac{2}{3}

, so work out

f (\frac{2}{3})

. Then

3 x^{3} + 4 x^{2} - 13 x + 6 \equiv (3 x - 2) (x^{2} + b x - 3)

B3
Use the factor theorem to show that $(2 x + 3)$ is a factor of $2 x^{3} + 5 x^{2} - x - 6$ . Hence solve $2 x^{3} + 5 x^{2} - x - 6 = 0$ .
[5 marks]
Answer $x = \dots\dots, \dots\dots, \dots\dots$

Example 4

$f (x) = x^{3} + a x^{2} + b x - 12$ . $(x + 2)$ and $(x - 3)$ are factors of $f (x)$ . Work out $a$ and $b$ , and the third linear factor.

Hint

f (- 2) = 0

and

f (3) = 0

give two equations in

a

and

b

. The constants in the three brackets multiply to give

- 12

B4
$f (x) = x^{3} + a x^{2} + b x + 30$ . $(x - 2)$ and $(x + 3)$ are factors of $f (x)$ . Work out the values of $a$ and $b$ .
[4 marks]
Answer $a = \dots\dots b = \dots\dots$

Example 5

A cuboid measures $x$ cm by $(x + 3)$ cm by $(x - 1)$ cm. Its volume is $84$ cm³. Show that $x^{3} + 2 x^{2} - 3 x - 84 = 0$ . Hence work out the dimensions of the cuboid.

Not drawn accurately

Hint Multiply the three edges and set the product equal to

84

. Try factors of

84

that make

x - 1

positive.

B5
A cuboid measures $x$ cm by $(x + 1)$ cm by $(x - 3)$ cm. Its volume is $126$ cm³. Show that $x^{3} - 2 x^{2} - 3 x - 126 = 0$ . Hence work out the dimensions of the cuboid.
[5 marks]
Not drawn accurately
Answer $\dots\dots$ cm by $\dots\dots$ cm by $\dots\dots$ cm

Practice

Level 2

C1
$f (x) = x^{3} + 4 x^{2} - 7 x - 10$ . For each of $(x - 2)$ , $(x + 1)$ , $(x + 5)$ and $(x - 5)$ , decide whether it is a factor of $f (x)$ . You must show your working.
[4 marks]
C2
Solve $2 x^{3} - 5 x^{2} - 12 x = 0$ .
[3 marks]
Answer $x = \dots\dots, \dots\dots, \dots\dots$

C3
Factorise fully $x^{5} - 5 x^{3} + 4 x$ .
[3 marks]
Answer
C4
$x = - 4$ is a root of the equation $x^{3} + k x^{2} - 7 x - 12 = 0$ . Work out the value of $k$ and the other two roots. Give the roots in surd form.
[5 marks]
Answer $k = \dots\dots x = \dots\dots$
C5
The sketch shows the graph of $y = x^{3} + a x^{2} + b x + c$ . It crosses the $x$ -axis at $x = - 2$ , $x = 1$ and $x = 4$ . Work out the values of $a$ , $b$ and $c$ .
[3 marks]
Not drawn accurately
Answer $a = \dots\dots b = \dots\dots c = \dots\dots$
C6
Show that $(x - a)$ is a factor of $x^{3} - 2 a x^{2} + 2 a^{2} x - a^{3}$ . Hence write $x^{3} - 2 a x^{2} + 2 a^{2} x - a^{3}$ as a product of a linear factor and a quadratic factor.
[3 marks]
Answer

Exam-style questions

AQA exam style

D1
$f (x) = 6 x^{3} + 11 x^{2} - 19 x + 6$
(a)
Use the factor theorem to show that $(2 x - 1)$ is a factor of $f (x)$ .
[2 marks]
(b)
Hence solve $f (x) = 0$
[3 marks]
Answer $x = \dots\dots, \dots\dots, \dots\dots$
Total for D1: 5 marks

D2
$f (x) = x^{3} - 4 x^{2} - 3 x + 18$ . Jo works out $f (3) = 0$ and says, "So $(x + 3)$ is a factor of $f (x)$ ."
Jo is wrong.
(a)
Show that $(x + 3)$ is not a factor of $f (x)$ , and say which factor $f (3) = 0$ does show.
[1 mark]
(b)
Factorise $f (x)$ fully.
[3 marks]
Answer
Total for D2: 4 marks
D3
$(x + 2)$ and $(x - 5)$ are factors of $x^{3} + a x^{2} + b x - 40$
(a)
Work out the third linear factor.
[2 marks]
Answer $(\dots\dots\dots\dots)$
(b)
Work out the values of $a$ and $b$ .
[3 marks]
Answer $a = \dots\dots b = \dots\dots$
Total for D3: 5 marks
D4
Which of these is a factor of $x^{3} + 27$ ?
Circle your answer. $(x - 3) (x + 3) (x - 27) (x + 27)$
[1 mark]
D5
The curve $y = x^{3} + 2 x^{2} - 4 x$ and the line $y = 7 x + 12$ meet at the points $A$ , $B$ and $C$ .
Not drawn accurately
(a)
Show that the $x$ -coordinates of $A$ , $B$ and $C$ satisfy $x^{3} + 2 x^{2} - 11 x - 12 = 0$ .
[1 mark]
(b)
Hence work out the coordinates of $A$ , $B$ and $C$ . You must show your working.
[5 marks]
Answer $(\dots, \dots)$ , $(\dots, \dots)$ and $(\dots, \dots)$
Total for D5: 6 marks
D6
$(3 x - 1)$ is a factor of $3 x^{3} + k x^{2} + 8 x - 4$
(a)
Work out the value of $k$ .
[3 marks]
Answer $k = \dots\dots$
(b)
Hence factorise $3 x^{3} + k x^{2} + 8 x - 4$ fully.
[2 marks]
Answer
Total for D6: 5 marks

Extension

Stretch

No route is given. The hints are at the end of the sheet. Use one at a time.

E1
Write $x^{6} - 64$ as a product of two linear factors and two quadratic factors.
[5 marks]
Hint 1 · What to try
$x^{6}$ and $64$ are both squares. Start with the difference of two squares.
Hint 2 · The first line
$x^{6} - 64 = (x^{3} - 8) (x^{3} + 8)$ . Now use the factor theorem on each cubic: try $x = 2$ and $x = - 2$ .
Hint 3 · The full method
$x^{3} - 8 = (x - 2) (x^{2} + 2 x + 4)$ and $x^{3} + 8 = (x + 2) (x^{2} - 2 x + 4)$ . Neither quadratic has real roots.
Answer

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can use f(a) = 0 to find a factor.
I can factorise a cubic and solve it.
I can find unknown coefficients from factors.
I can show a cubic has only one real root.

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