A-Level Maths (Edexcel 9MA0) · Pure Year 1 · Chapter 7: Algebraic methods · Lesson 2

The factor theorem

71 marks

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Date

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Quick questions to start.

A1
Needed today
Work out $f (- 2)$ when $f (x) = 2 x^{3} - 5 x^{2} + 4 x + 7$ .
[2 marks]
Answer
A2
Last lesson
Divide $x^{3} + 2 x^{2} - 7 x - 2$ by $(x - 2)$ .
[2 marks]
Answer
A3
Needed today
Factorise $12 x^{2} + x - 6$ .
[2 marks]
Answer
A4
From chapter 2
Find the discriminant of $2 x^{2} + 3 x + 7$ and the number of real roots.
[2 marks]
Answer

Example, then your turn

A-Level

Work through each example, then try the one beside it.

Example 1

Show that $(x - 3)$ is a factor of $f (x) = x^{3} - x^{2} - 14 x + 24$ . Hence factorise $f (x)$ completely.

Hint

(x - 3)

is a factor when

f (3) = 0

. Then divide by

(x - 3)

and factorise the quadratic.

B1
Show that $(2 x - 1)$ is a factor of $f (x) = 2 x^{3} + 5 x^{2} - 23 x + 10$ . Hence factorise $f (x)$ completely.
[3 marks]
Answer

Example 2

Factorise $x^{3} + 6 x^{2} - x - 30$ completely. Hence solve $x^{3} + 6 x^{2} - x - 30 = 0$ .

Hint Try

x = \pm 1, \pm 2, \pm 3, \dots

: a root of

x^{3} + \dots - 30

must divide

30

B2
Factorise completely $f (x) = 3 x^{3} - 7 x^{2} - 7 x + 3$ . Hence sketch $y = f (x)$ , showing where it meets the axes.
[5 marks]

Example 3

$(x - 2)$ is a factor of $f (x) = 2 x^{3} + a x^{2} - 7 x - 6$ . Find $a$ . Hence factorise $f (x)$ completely.

Hint

(x - 2)

is a factor, so

f (2) = 0

. That gives an equation in

a

B3
$(x + 3)$ is a factor of $f (x) = 3 x^{3} + k x^{2} - 38 x - 24$ . Find $k$ . Hence factorise $f (x)$ completely.
[4 marks]
Answer $k = \dots\dots$ ; $\dots\dots$

Practice

A-Level

C1
$g (x) = x^{4} + 2 x^{3} - 13 x^{2} - 14 x + 24$ . Show that $(x - 1)$ and $(x + 2)$ are both factors of $g (x)$ . Hence factorise $g (x)$ completely.
[5 marks]
Answer

C2
Given that $(3 x + 2)$ is a factor of $6 x^{3} + 7 x^{2} + k x - 6$ , find the value of $k$ . Hence solve $6 x^{3} + 7 x^{2} + k x - 6 = 0$ .
[5 marks]
Answer $k = \dots$ ; $x = \dots\dots$
C3
Given that $(x + 1)$ and $(x - 3)$ are factors of $p x^{3} + q x^{2} - 5 x + 6$ , find the values of $p$ and $q$ . Hence find the third linear factor.
[5 marks]
Answer $p = \dots$ , $q = \dots$ ; $\dots\dots$

Exam-style questions

Edexcel exam style

D1
For $f (x) = x^{3} + 2 x^{2} - 13 x + 10$ , Sam writes: " $f (2) = 8 + 8 - 26 + 10 = 0$ , so $(x + 2)$ is a factor of $f (x)$ ."
Sam is wrong.
(a)
Explain the mistake Sam has made.
[1 mark]
(b)
Factorise $f (x)$ completely.
[3 marks]
Answer
(c)
Hence solve $(y - 3)^{3} + 2 (y - 3)^{2} - 13 (y - 3) + 10 = 0$ .
[2 marks]
Answer $y = \dots\dots$
Total for D1: 6 marks

D2
$f (x) = 3 x^{3} + a x^{2} + b x + 4$ , where $a$ and $b$ are constants. $(x + 2)$ is a factor of $f (x)$ , and $f (1) = 18$ .
(a)
Find the values of $a$ and $b$ .
[4 marks]
Answer $a = \dots$ , $b = \dots$
(b)
Hence write $f (x)$ in the form $(x + 2) (p x^{2} + q x + r)$ , where $p$ , $q$ and $r$ are integers.
[2 marks]
Answer
(c)
Hence show that the equation $f (x) = 0$ has exactly one real root.
[2 marks]
Total for D2: 8 marks
D3
The curve $C$ has equation $y = x^{3} + x^{2} - 8 x$ . The line $l$ has equation $y = 3 x - 9$ .
(a)
Show that the $x$ -coordinates of the points where $C$ meets $l$ satisfy $x^{3} + x^{2} - 11 x + 9 = 0$ .
[1 mark]
(b)
Use the factor theorem to show that $(x - 1)$ is a factor of $x^{3} + x^{2} - 11 x + 9$ .
[2 marks]
(c)
Hence find the $x$ -coordinates of the other two points where $C$ meets $l$ , giving your answers in the form $p \pm q$ .
[4 marks]
Answer $x = \dots\dots$
Total for D3: 7 marks
D4
An open box is made from a rectangular sheet of card $12$ cm long and $8$ cm wide. A square of side $x$ cm is cut from each corner, and the sides are folded up along the dashed lines.
Not drawn accurately
(a)
Show that the volume of the box, $V$ cm³, is given by $V = 4 x^{3} - 40 x^{2} + 96 x$ .
[2 marks]
(b)
The volume of the box is $64$ cm³. Show that $x^{3} - 10 x^{2} + 24 x - 16 = 0$ .
[1 mark]
(c)
Use the factor theorem to show that $(x - 2)$ is a factor of $x^{3} - 10 x^{2} + 24 x - 16$ . Hence find all the solutions of $x^{3} - 10 x^{2} + 24 x - 16 = 0$ , giving your answers in exact form.
[4 marks]
Answer $x = \dots\dots$
(d)
Explain why only two of these solutions are possible sizes for the squares.
[1 mark]
Total for D4: 8 marks

Extension

Stretch

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

E1
$(x - 3)^{2}$ is a factor of $x^{3} + p x + q$ , where $p$ and $q$ are constants. Find the values of $p$ and $q$ , and find the third factor of $x^{3} + p x + q$ .
[7 marks]
Hint 1 · What to try
$(x - 3)$ is a factor, so putting $x = 3$ gives $27 + 3 p + q = 0$ . That is one equation; you need a second.
Hint 2 · The first line
Divide $x^{3} + 0 x^{2} + p x + q$ by $(x - 3)$ . The quotient is $x^{2} + 3 x + (9 + p)$ , and because $(x - 3)$ is a factor TWICE, $(x - 3)$ is a factor of this quotient too.
Hint 3 · The full method
Put $x = 3$ in the quotient: $9 + 9 + 9 + p = 0$ , so $p = - 27$ and then $q = 54$ . The quotient is $x^{2} + 3 x - 18 = (x - 3) (x + 6)$ .
Answer $p = \dots$ , $q = \dots$ ; $\dots\dots$

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can use the factor theorem to show that a linear expression is a factor of a polynomial.
I can find a factor by trying values, then factorise a cubic completely and sketch it.
I can use given factors to find unknown constants, forming simultaneous equations if needed.
I can use a factorised cubic to solve equations and decide how many real roots there are.

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