AQA Level 2 Further Maths · Level 2 Certificate · Chapter 2: Algebra II · Lesson 1

Factorising

About 70 minutesNo calculator62 marks

Hints online, answers free with an account:
mathswariz.uk/worksheets/further-maths-factorising/

Date

Do Now

The first questions are what today's lesson needs. The others bring back earlier topics.

A1
Needed today
Factorise $x^{2} - 5 x - 24$ .
[2 marks]
Answer
A2
Needed today
Factorise fully $6 x^{2} y - 15 x y^{2}$ .
[2 marks]
Answer
A3
Last lesson
A code is four different digits chosen from $1$ to $9$ . How many different codes are possible?
[2 marks]
Answer
A4
From chapter 1
Expand and simplify $(3 + 5) (2 - 5)$ .
[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

Factorise $3 ab - 6 a + 5 b - 10$ .

Hint Take a common factor out of the first two terms, then out of the last two. The brackets should match.

B1
Factorise $2 x y + 6 x - 5 y - 15$ .
[2 marks]
Answer

Example 2

Factorise $6 x^{2} - 7 x - 20$ .

Hint Find two numbers with a product of

6 \times (- 20) = - 120

and a sum of

- 7

. Use them to split

- 7 x

B2
Factorise $10 x^{2} + 11 x - 6$ .
[2 marks]
Answer

Example 3

Factorise $6 x^{2} + x y - 12 y^{2}$ .

Hint The same method: a product of

6 \times (- 12) = - 72

and a sum of

1

. Every term of the split carries

y

as well.

B3
Factorise $5 a^{2} - 13 ab - 6 b^{2}$ .
[2 marks]
Answer

Example 4

Factorise $(2 x + 5)^{2} - (x - 1)^{2}$ .

Hint This is

A^{2} - B^{2} = (A + B) (A - B)

with

A = 2 x + 5

and

B = x - 1

. Take care with the signs in

A - B

B4
Factorise fully $(3 x - 2)^{2} - (x + 4)^{2}$ .
[3 marks]
Answer

Example 5

Factorise fully $12 x^{5} - 75 x^{3} y^{2}$ .

Hint "Fully" means more than one step. Take out the highest common factor first, then look at what is left in the bracket.

B5
Factorise fully $2 a^{7} - 32 a^{3}$ .
[3 marks]
Answer

Practice

Level 2

C1
Factorise $9 x^{2} - 42 x + 49$ .
[2 marks]
Answer
C2
Use factorising to work out $100 6^{2} - 99 4^{2}$ without a calculator.
[2 marks]
Answer

C3
Factorise $2 (y + 1)^{2} + 7 (y + 1) + 3$ . Give your answer in terms of $y$ , with no brackets inside brackets.
[3 marks]
Answer
C4
Factorise $x^{2} - 6 x + 9 - y^{2}$ .
[2 marks]
Answer
C5
Factorise $12 + 5 x - 2 x^{2}$ .
[2 marks]
Answer
C6
Factorise fully $x^{3} - 3 x^{2} - 4 x + 12$ .
[3 marks]
Answer

Exam-style questions

AQA exam style

D1
Jo says, " $12 x^{2} - 27 = (6 x + 9) (2 x - 3)$ , so I have factorised it fully."
Explain why Jo is wrong, and factorise $12 x^{2} - 27$ fully.
[2 marks]
Answer

D2
Each part of this question uses the answer to the part before it.
(a)
Factorise $x^{2} + 5 x - 36$ .
[2 marks]
Answer
(b)
Hence factorise fully $(2 y - 1)^{2} + 5 (2 y - 1) - 36$ .
[2 marks]
Answer
(c)
Hence factorise fully $t^{4} + 5 t^{2} - 36$ .
[2 marks]
Answer
Total for D2: 6 marks
D3
Factorise fully $3 (x - 2)^{7} - 12 (x - 2)^{5}$ .
[3 marks]
Answer
D4
The rectangle has area $(6 x^{2} + 7 x - 20)$ cm². One side is $(2 x + 5)$ cm long.
Not drawn accurately
(a)
Work out an expression for the length of the other side.
[2 marks]
Answer $\dots\dots\dots$ cm
(b)
The perimeter of the rectangle is $52$ cm. Work out the area of the rectangle.
[3 marks]
Answer $\dots\dots\dots$ cm²
Total for D4: 5 marks
D5
Use factorising to show that $2^{16} - 1$ is a multiple of $17$ .
[3 marks]
D6
$E = 3 x^{2} - 10 x y + 3 y^{2}$
(a)
Factorise $E$ .
[2 marks]
Answer
(b)
Hence work out the value of $E$ when $x = 3.4$ and $y = 1.1$ . You must show your working.
[2 marks]
Answer
Total for D6: 4 marks

Extension

Stretch

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

E1
Factorise $x^{4} + 4 y^{4}$ into two quadratic factors. Hence show that $n^{4} + 4$ is not a prime number for any whole number $n > 1$ .
[5 marks]
Hint 1 · What to try
$x^{4} + 4 y^{4}$ is not a difference of two squares as it stands. Add and take away a term so that it becomes one.
Hint 2 · The first line
$x^{4} + 4 y^{4} = (x^{2} + 2 y^{2})^{2} - 4 x^{2} y^{2} = (x^{2} + 2 y^{2})^{2} - (2 x y)^{2}$ .
Hint 3 · The full method
So $x^{4} + 4 y^{4} = (x^{2} + 2 x y + 2 y^{2}) (x^{2} - 2 x y + 2 y^{2})$ . Put $x = n$ and $y = 1$ : $n^{4} + 4 = (n^{2} + 2 n + 2) (n^{2} - 2 n + 2)$ . The smaller factor is $(n - 1)^{2} + 1$ , which is at least $2$ when $n > 1$ , so $n^{4} + 4$ is a product of two whole numbers each bigger than $1$ .

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can factorise four terms by grouping them in pairs.
I can factorise $a x^{2} + b x + c$ , including quadratics in two letters.
I can spot and use a difference of two squares, including with brackets.
I can factorise an expression fully, in more than one step.

Answers and mark scheme

The worked answers and the full mark scheme are free with a Mathswariz account. Teachers and students can both sign up.

Get the answers