Factorising Quadratics (AQA GCSE Further Maths)

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Written by: Jamie Wood

Reviewed by: Dan Finlay

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Factorising Simple Quadratics

What is a quadratic expression?

A quadratic expression is in the form:
- ax² + bx + c (as long as a ≠ 0)
If there are any higher powers of x (like x³ say) then it is not a quadratic
If a = 1 e.g. $x^{2} - 2 x - 8$ , it can be called a “monic” quadratic expression
If a ≠ 1 e.g. $2 x^{2} - 2 x - 8$ , it can be called a “non-monic” quadratic expression

Method 1: Factorising "by inspection"

This is shown easiest through an example; factorising $x^{2} - 2 x - 8$

We need a pair of numbers that for $x^{2} + b x + c$
- multiply to c
  - which in this case is -8
- and add to b
  - which in this case is -2
- -4 and +2 satisfy these conditions
- Write these numbers in a pair of brackets like this:
  - $(x + 2) (x - 4)$

Method 2: Factorising "by grouping"

This is shown easiest through an example; factorising $x^{2} - 2 x - 8$

We need a pair of numbers that for $x^{2} + b x + c$
- multiply to c
  - which in this case is -8
- and add to b
  - which in this case is -2
- 2 and -4 satisfy these conditions
- Rewrite the middle term by using 2x and -4x
  - $x^{2} + 2 x - 4 x - 8$
- Group and factorise the first two terms, using x as the highest common factor, and group and factorise the second two terms, using -4 as the factor
  - $x (x + 2) - 4 (x + 2)$
- Note that these now have a common factor of (x + 2) so this whole bracket can be factorised out
  - $(x + 2) (x - 4)$

Method 3: Factorising "by using a grid"

This is shown easiest through an example; factorising $x^{2} - 2 x - 8$

We need a pair of numbers that for $x^{2} + b x + c$
- multiply to c
  - which in this case is -8
- and add to b
  - which in this case is -2
- -4 and +2 satisfy these conditions
- Write the quadratic equation in a grid (as if you had used a grid to expand the brackets), splitting the middle term as -4x and 2x
- The grid works by multiplying the row and column headings, to give a product in the boxes in the middle


	x²	-4x
	+2x	-8

Write a heading for the first row, using x as the highest common factor of x² and -4x


x	x²	-4x
	+2x	-8

You can then use this to find the headings for the columns, e.g. “What does x need to be multiplied by to give x²?”

	x	-4
x	x²	-4x
	+2x	-8

We can then fill in the remaining row heading using the same idea, e.g. “What does x need to be multiplied by to give +2x?”

	x	-4
x	x²	-4x
+2	+2x	-8

We can now read-off the factors from the column and row headings
- $(x + 2) (x - 4)$

Which method should I use for factorising simple quadratics?

The first method, by inspection, is by far the quickest so is recommended in an exam for simple quadratics (where a = 1)
However the other two methods (grouping, or using a grid) can be used for harder quadratic equations where a ≠ 1 so you should learn at least one of them too

Examiner Tips and Tricks

As a check, expand your answer and make sure you get the same expression as the one you were trying to factorise.

Worked Example

(a) Factorise $x^{2} - 4 x - 21$ .

We will factorise by inspection.

We need two numbers that:

multiply to -21, and sum to -4

-7, and +3 satisfy this

Write down the brackets.

(x + 3)(x - 7)

(b) Factorise $x^{2} - 5 x + 6$ .

We will factorise by splitting the middle term and grouping.

We need two numbers that:

multiply to 6, and sum to -5

-3, and -2 satisfy this

Split the middle term.

x² - 2x - 3x + 6

Factorise x out of the first two terms.

x(x - 2) - 3x +6

Factorise -3 out of the last two terms.

x(x - 2) - 3(x - 2)

These have a common factor of (x - 2) which can be factored out.

(x - 2)(x - 3)

We will factorise by using a grid.

We need two numbers that:

multiply to -24, and sum to -2

+4, and -6 satisfy this

Use these to split the -2x term and write in a grid.


	x²	+4x
	-6x	-24

Write a heading using a common factor for the first row:


x	x²	+4x
	-6x	-24

Work out the headings for the rows, e.g. “What does x need to be multiplied by to make x²?”

	x	+4
x	x²	+4x
	-6x	-24

Repeat for the heading for the remaining row, e.g. “What does x need to be multiplied by to make -6x?”

	x	+4
x	x²	+4x
-6	-6x	-24

Read-off the factors from the column and row headings.

(x + 4)(x - 6)

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Factorising Harder Quadratics

How do I factorise a harder quadratic expression?

Factorising a ≠ 1 "by grouping"

This is shown easiest through an example; factorising $4 x^{2} - 25 x - 21$

We need a pair of numbers that for $a x^{2} + b x + c$
- multiply to ac
  - which in this case is 4 × -21 = -84
- and add to b
  - which in this case is -25
- -28 and +3 satisfy these conditions
- Rewrite the middle term using -28x and +3x
  - $4 x^{2} - 28 x + 3 x - 21$
- Group and factorise the first two terms, using 4x as the highest common factor, and group and factorise the second two terms, using 3 as the factor
  - $4 x (x - 7) + 3 (x - 7)$
- Note that these terms now have a common factor of (x - 7) so this whole bracket can be factorised out, leaving 4x + 3 in its own bracket
  - $(x - 7) (4 x + 3)$

Factorising a ≠ 1 "by using a grid"

This is shown easiest through an example; factorising $4 x^{2} - 25 x - 21$

We need a pair of numbers that for $a x^{2} + b x + c$
- multiply to ac
  - which in this case is 4 × -21 = -84
- and add to b
  - which in this case is -25
- -28 and +3 satisfy these conditions
- Write the quadratic equation in a grid (as if you had used a grid to expand the brackets), splitting the middle term as -28x and +3x
- The grid works by multiplying the row and column headings, to give a product in the boxes in the middle


	4x²	-28x
	+3x	-21

Write a heading for the first row, using 4x as the highest common factor of 4x² and -28x


4x	4x²	-28x
	+3x	-21

You can then use this to find the headings for the columns, e.g. “What does 4x need to be multiplied by to give 4x²?”

	x	-7
4x	4x²	-28x
	+3x	-21

We can then fill in the remaining row heading using the same idea, e.g. “What does x need to be multiplied by to give +3x?”

	x	-7
4x	4x²	-28x
+3	+3x	-21

We can now read-off the factors from the column and row headings
- $(x - 7) (4 x + 3)$

How do I factorise a quadratic with two variables?

To factorise 3x² + 13xy - 10y²
- Factorise the easier quadratic 3x² + 13x - 10
  - (3x - 2)(x + 5)
- Insert y's on the last terms in the brackets
  - (3x - 2y)(x + 5y)
Check by expanding (3x - 2y)(x + 5y)
- 3x² + 15xy - 2yx - 10y²
  - 3x² + 13xy - 10y² ✓

Examiner Tips and Tricks

As a check, expand your answer and make sure you get the same expression as the one you were trying to factorise.

Worked Example

(a) Factorise $6 x^{2} - 7 x - 3$ .

We will factorise by splitting the middle term and grouping.

We need two numbers that:

multiply to 6 × -3 = -18, and sum to -7

-9, and +2 satisfy this

Split the middle term.

6x² + 2x - 9x - 3

Factorise 2x out of the first two terms.

2x(3x + 1) - 9x - 3

Factorise -3 of out the last two terms.

2x(3x + 1) - 3(3x + 1)

These have a common factor of (3x + 1) which can be factored out.

(3x + 1)(2x - 3)

(b) Factorise $10 x^{2} + 9 x - 7$ .

We will factorise by using a grid.

We need two numbers that:

multiply to 10 × -7 = -70, and sum to +9

-5, and +14 satisfy this

Use these to split the 9x term and write in a grid.


	10x²	-5x
	+14x	-7

Write a heading using a common factor for the first row:


5x	10x²	-5x
	+14x	-7

Work out the headings for the rows, e.g. “What does 5x need to be multiplied by to make 10x²?”

	2x	-1
5x	10x²	-5x
	+14x	-7

Repeat for the heading for the remaining row, e.g. “What does 2x need to be multiplied by to make +14x?”

	2x	-1
5x	10x²	-5x
+7	+14x	-7

Read-off the factors from the column and row headings.

(2x - 1)(5x + 7)

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Difference Of Two Squares

What is the difference of two squares?

When a "squared" quantity is subtracted from another "squared" quantity, you get the difference of two squares
- for example,
  - a² - b²
  - 9² - 5²
  - (x + 1)² - (x - 4)²
  - 4m² - 25n², which is (2m)² - (5n)²

How do I factorise the difference of two squares?

Expand the brackets (a + b)(a - b)
- = a² - ab + ba - b²
- ab is the same quantity as ba, so -ab and +ba cancel out
- = a² - b²
From the working above, the difference of two squares, a² - b², factorises to

$(a + b) (a - b)$

It is fine to write the second bracket first, (a - b)(a + b)
- but the a and the b cannot swap positions
  - a² - b² must have the a's first in the brackets and the b's second in the brackets
It might not be obvious that you can use the difference of two squares
- Try factoring out any common factors first
- $18 x^{2} - 50 y^{2} = 2 (9 x^{2} - 25 y^{2}) = 2 (3 x + 5 y) (3 x - 5 y)$

Examiner Tips and Tricks

The difference of two squares is a very important rule to learn as it often appears in harder questions involving factorisation, e.g. in algebraic fractions
The word difference in maths means a subtraction, it should remind you that you are subtracting one squared term from another
You should be able to recognise factorised difference of two squares expressions

Worked Example

(i) Factorise fully $20 x^{3} - 45 x$ .

(ii) Factorise ${(7 x + 3)}^{2} - {(3 x - 2)}^{2}$ .

(i)

The highest common factor of $20 x^{3}$ and $- 45 x$ is $5 x$ , so take this out as a factor

$20 x^{3} - 45 x = 5 x (4 x^{2} - 9)$

$(4 x^{2} - 9)$ is a difference of two squares, as $4 x^{2} = {(2 x)}^{2}$ and $9 = 3^{2}$
We can factorise the bracket into two further brackets using the difference of two squares

$5 x (4 x^{2} - 9) = 5 x (2 x - 3) (2 x + 3)$

$5 x (2 x - 3) (2 x + 3)$

(ii)

${(7 x + 3)}^{2} - {(3 x - 2)}^{2}$ is a difference of two squares, as both brackets are squared (and one is being subtracted from the other)

Use the pattern $a^{2} - b^{2} = (a - b) (a + b)$ to help you
Here, $a = (7 x + 3)$ and $b = (3 x - 2)$

${(7 x + 3)}^{2} - {(3 x - 2)}^{2} = ((7 x + 3) - (3 x - 2)) \times ((7 x + 3) + (3 x - 2))$

Simplifying

$((7 x + 3) - (3 x - 2)) \times ((7 x + 3) + (3 x - 2)) = (4 x + 5) (10 x + 1)$

$(4 x + 5) (10 x + 1)$

Last updated: 7 October 2024

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Factorising Quadratics (AQA GCSE Further Maths)

Revision Note

Factorising Simple Quadratics

What is a quadratic expression?

Which method should I use for factorising simple quadratics?

Examiner Tips and Tricks

Worked Example

Factorising Harder Quadratics

How do I factorise a harder quadratic expression?

How do I factorise a quadratic with two variables?

Examiner Tips and Tricks

Worked Example

Difference Of Two Squares

What is the difference of two squares?

How do I factorise the difference of two squares?

Examiner Tips and Tricks

Worked Example

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Join the 100,000+ Students that ❤️ Save My Exams

Author: Jamie Wood

Author: Dan Finlay