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Factorising Quadratics (CIE IGCSE Maths: Extended)

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Jamie W

Last updated

21 November 2023

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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

How to Factorise Quadratics

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 Tip

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
  - $(x - 7) (4 x + 3)$

Examiner Tip

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

Examiner Tip

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

(a)

Factorise

9 x^{2} - 16

Recognise that $9 x^{2}$ and $16$ are both squared terms and the second term is subtracted from the first term - you can factorise using the difference of two squares.

$9 x^{2} - 16 = {(3 x)}^{2} - {(4)}^{2}$

Rewrite the expression with the square root of each term added together in the first bracket and subtracted from each other in the second bracket.

$(3 x + 4) (3 x - 4)$

(b)

Factorise

4 x^{2} - 25

Recognise that $4 x^{2}$ and $25$ are both squared terms and the second term is subtracted from the first term - you can factorise using the difference of two squares.

$4 x^{2} - 25 = {(2 x)}^{2} - {(5)}^{2}$

Rewrite the expression with the square root of each term added together in the first bracket and subtracted from each other in the second bracket.

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

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Quadratics Factorising Methods

How do I know if it factorises?

Method 1: Use a calculator to solve the quadratic expression equal to 0
- If the solutions are integers or fractions (without square roots), then the quadratic expression factorises
Method 2: Find the value under the square root in the quadratic formula, b² – 4ac (called the discriminant)
- If this number is a perfect square number, then the quadratic expression factorises

Which factorisation method should I use for a quadratic expression?

Does it have 2 terms only?
- Yes, like $x^{2} - 7 x$
  - Use "basic factorisation" to take out the highest common factor
  - $x (x - 7)$
- Yes, like $x^{2} - 9$
  - Use the "difference of two squares" to factorise
  - $(x + 3) (x - 3)$
Does it have 3 terms?
- Yes, starting with x² like $x^{2} - 3 x - 10$
  - Use "factorising simple quadratics" by finding two numbers that add to -3 and multiply to -10
  - $(x + 2) (x - 5)$
- Yes, starting with ax² like $3 x^{2} + 15 x + 18$
  - Check to see if the 3 in front of x² is a common factor for all three terms (which it is in this case), then use "basic factorisation" to factorise it out first
  - $3 (x^{2} + 5 x + 6)$
  - The quadratic expression inside the brackets is now x² +... , which factorises more easily
  - $3 (x + 2) (x + 3)$
- Yes, starting with ax² like $3 x^{2} - 5 x - 2$
  - The 3 in front of x² is not a common factor for all three term
  - Use "factorising harder quadratics", for example factorising by grouping or factorising using a grid
  - $(3 x + 1) (x - 2)$

Worked example

Factorise $- 8 x^{2} + 100 x - 48$ .

Spot the common factor of -4 and put outside a set of brackets, work out the terms inside the brackets by dividing the terms in the original expression by -4.

$- 8 x^{2} + 100 x - 48 = - 4 (2 x^{2} - 25 x + 12)$

Check the discriminant for the expression inside the brackets, $(b^{2} - 4 a c)$ , to see if it will factorise.

$\begin{array}{rcl} {(- 25)}^{2} - 4 \times 2 \times 12 \\ = & 625 - 96 \\ = & 529 \end{array}$

$529 = 13^{2}$ , it is a perfect square so the expression will factorise.

Proceed with factorising $2 x^{2} - 25 x + 12$ as you would for a harder quadratic, where $a \neq 1$ .
"+12" means the signs will be the same.
"-25" means that both signs will be negative.

$a \times c = 2 \times 12 = 24$

The only numbers which multiply to give 24 and follow the rules for the signs above are:
$(- 1) \times (- 24)$ and $(- 2) \times (- 12)$ and $(- 3) \times (- 8)$ and $(- 4) \times (- 6)$
but only the first pair add to give $- 25$ .

Split the $- 25 x$ term into $- 24 x - x$ .

$\begin{array}{rcl} 2 x^{2} - 24 x - x + 12 \end{array}$

Group and factorise the first two terms, using $2 x$ as the highest common factor and group and factorise the last two terms using $1$ as the highest common factor.

$\begin{array}{rcl} 2 x (x - 12) + 1 (x - 12) \end{array}$

These factorised terms now have a common term of $(x - 12)$ , so this can now be factorised out.

$(2 x + 1) (x - 12)$

Put it all together.

$- 8 x^{2} + 100 x - 48 = - 4 (2 x^{2} - 25 x + 12) = - 4 (2 x - 1) (x - 12)$

$- 4 (2 x - 1) (x - 12)$

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Factorising Quadratics (CIE IGCSE Maths: Extended)

Revision Note

What is a quadratic expression?

How to Factorise Quadratics

Which method should I use for factorising simple quadratics?

How do I factorise a harder quadratic expression?

What is the difference of two squares?

How do I factorise the difference of two squares?

How do I know if it factorises?

Which factorisation method should I use for a quadratic expression?

You've read 0 of your 10 free revision notes

Unlock more, it's free!

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

Author: Jamie W