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

Expanding brackets and factorising

77 marks

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Date

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Quick questions to start.

A1
Needed today
Factorise $x^{2} - 2 x - 15$ .
[2 marks]
Answer
A2
Needed today
Expand and simplify $(3 x + 2) (x - 5)$ .
[2 marks]
Answer
A3
Last lesson
Evaluate $1 6^{- \frac{3}{4}}$ .
[2 marks]
Answer
A4
Last lesson
Write $\frac{4 x ^{2} + 3}{x}$ in the form $a x^{p} + b x^{q}$ .
[2 marks]
Answer

Example, then your turn

A-Level

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

Example 1

Expand and simplify $(x + 2) (2 x - y + 3)$ .

Hint

2 \times 3 = 6

products, then collect like terms.

B1
Expand and simplify $(3 x - y) (x + 4 y - 2)$ .
[3 marks]
Answer

Example 2

Expand and simplify $(x - 3) (2 x + 5) (x + 1)$ .

Hint Multiply two of the brackets first, then multiply the quadratic by the third bracket.

B2
Expand and simplify $(2 x + 3) (x - 2) (3 x - 1)$ .
[4 marks]
Answer

Example 3

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

Hint Find two numbers that multiply to

a c = 6 \times (- 20) = - 120

and add to

7

. Split the

7 x

with them, then factorise in pairs.

B3
Factorise (a) $10 x^{2} - 11 x - 6$ (b) $25 a^{2} - 81 b^{2}$
[3 marks]
Answer(a) $\dots\dots$ (b) $\dots\dots$

Example 4

Factorise completely (a) $3 x^{3} - 12 x^{2} - 36 x$ (b) $18 x - 8 x^{3}$

Hint Take out the highest common factor first. Then factorise what is left in the bracket, if it will go.

B4
Factorise completely (a) $4 x^{3} + 2 x^{2} - 30 x$ (b) $75 x - 3 x^{3}$
[4 marks]
Answer(a) $\dots\dots$ (b) $\dots\dots$

Example 5

Write $x^{4} - 34 x^{2} + 225$ as the product of four linear factors.

Hint Think of it as a quadratic in

x^{2}

y^{2} - 34 y + 225

with

y = x^{2}

. Factorise that, then each bracket is a difference of two squares.

B5
Factorise completely $9 x^{4} - 37 x^{2} + 4$ .
[4 marks]
Answer

Practice

A-Level

C1
A square of side $(x - 1)$ cm is cut from the corner of a rectangle that is $(2 x + y + 3)$ cm long and $(x + 4)$ cm wide. Find an expanded and simplified expression for the area that is left, shaded in the diagram.
[3 marks]
Not drawn accurately
Answer

C2
Factorise $12 + 5 x - 2 x^{2}$ .
[2 marks]
Answer
C3
Given that $(a x + 1) (x + b) (x - 2) \equiv 3 x^{3} + p x^{2} + q x + 8$ , where $a$ , $b$ , $p$ and $q$ are constants, find the values of $a$ , $b$ , $p$ and $q$ .
[4 marks]
Answer $a = \dots$ $b = \dots$ $p = \dots$ $q = \dots$
C4
Factorise completely $2 x^{5} - 32 x$ .
[3 marks]
Answer
C5
Simplify fully $\frac{10 x ^{2} + 13 x - 3}{25 x ^{2} - 1}$ .
[3 marks]
Answer

Exam-style questions

Edexcel exam style

D1
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
(a)
Factorise $10 x^{2} + 9 x - 7$ .
[2 marks]
Answer
(b)
Hence, without a calculator, write $9263$ as the product of two prime numbers.
[4 marks]
Answer $9263 = \dots\dots \times \dots\dots$
Total for D1: 6 marks

D2
A cuboid has edges of length $(x + 3)$ cm, $(2 x - 1)$ cm and $(x + 2)$ cm, where $x > \frac{1}{2}$ .
Not drawn accurately
(a)
Show that the volume of the cuboid is $(2 x^{3} + 9 x^{2} + 7 x - 6)$ cm³.
[3 marks]
(b)
The total surface area of the cuboid is $(10 x^{2} + 26 x + 2)$ cm², and it is equal to $94$ cm². Find the value of $x$ and hence the volume of the cuboid.
[4 marks]
Answer $x = \dots\dots$ volume $= \dots\dots$ cm³
Total for D2: 7 marks
D3
Priya is told that $2 x^{3} + x^{2} - 25 x + 12 \equiv (x + 4) (2 x^{2} + b x + c)$ . She writes: "The $x^{2}$ terms give $b = 1$ . The constant terms give $4 c = 12$ , so $c = 3$ ."
Priya has made a mistake.
(a)
Explain the mistake Priya has made.
[1 mark]
(b)
Find the values of $b$ and $c$ , and use the $x$ terms to check them.
[3 marks]
Answer $b = \dots\dots$ $c = \dots\dots$
(c)
Hence factorise $2 x^{3} + x^{2} - 25 x + 12$ completely.
[2 marks]
Answer
(d)
Hence solve $2 x^{3} + x^{2} - 25 x + 12 = 0$ .
[2 marks]
Answer
Total for D3: 8 marks
D4
$f (x) = 4 x^{4} - 17 x^{2} + 4$
(a)
Write $f (x)$ as the product of four linear factors.
[4 marks]
Answer
(b)
Hence solve $f (x) = 0$ .
[2 marks]
Answer
(c)
Hence solve $4^{2 y + 1} - 17 \times 4^{y} + 4 = 0$ .
[3 marks]
Answer $y = \dots\dots$ or $y = \dots\dots$
Total for D4: 9 marks

Extension

Stretch

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

E1
Find every positive integer $n$ for which $n^{4} + 4$ is a prime number, and prove that there are no others.
[6 marks]
Hint 1 · What to try
$n^{4} + 4$ has no obvious factors, but it is close to a perfect square. Add and subtract a term to make a difference of two squares.
Hint 2 · The first line
$n^{4} + 4 = (n^{2} + 2)^{2} - 4 n^{2} = (n^{2} + 2)^{2} - (2 n)^{2}$ .
Hint 3 · The full method
So $n^{4} + 4 = (n^{2} + 2 n + 2) (n^{2} - 2 n + 2)$ . The second factor is $(n - 1)^{2} + 1$ , which is $1$ only when $n = 1$ and at least $2$ otherwise. So for $n \geq 2$ both factors are at least $2$ and $n^{4} + 4$ is not prime. For $n = 1$ , $n^{4} + 4 = 5$ , which is prime.
Answer

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can expand two or three brackets and collect like terms.
I can factorise a quadratic when the x squared term is not just x squared.
I can factorise completely, using common factors and the difference of two squares.
I can find unknown constants by comparing coefficients.

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