Quadratic factorisation worksheet: Factoring Quadratic Equations

Posted on 23 июля, 2023 by alexxlab

2\textcolor{blue}{-3}x+\textcolor{red}{2}

Step 1: First we can write two brackets with an x placed in each bracket.

(x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,) (x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)

Step 2: We can identify that this is a sub-type (b) quadratic, meaning both brackets will contain \large{-}

(x\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,) (x\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,)

Step 3: We have to find two numbers which multiply to make \textcolor{red}{2} and when added together make \textcolor{blue}{-3}.

We know both numbers will be negative.

-2 \times -1 = \textcolor{red}{2}

-2 + -1 = \textcolor{blue}{-3}

Finally add these numbers to the brackets.

(x-2) (x-1)

Level 4-5GCSEAQAEdexcelOCRWJECEdexcel iGCSE

Level 6-7GCSEAQAEdexcelOCRWJECEdexcel iGCSE

Type 2: Factorising quadratics (a> 1)

In this instance the general form of the equation is ax^2+bx+c where a>1. 2+\textcolor{blue}{3x}\textcolor{red}{-1}

Step 1: When a>1 it makes things more complicated. It is not immediately obvious what the coefficient of each x term should be. There are two possible options,

(4x \kern{1 cm} ) (x \kern{1 cm} ) or (2x \kern{1 cm} ) (2x \kern{1 cm} )

Step 2: We can identify that this quadratic is part of sub-type (c) meaning it can contain + and —

This is most important for quadratic pairs which are non-symmetrically creating a third option, all three are shown below.

\begin{aligned}(4x \kern{0.4 cm} +\kern{0.4 cm} )&(x \kern{0.4 cm}-\kern{0.4 cm} )\\ (4x \kern{0.4 cm} -\kern{0.4 cm} )&(x \kern{0.4 cm}+\kern{0.4 cm} )\\(2x \kern{0.4 cm}+\kern{0.4 cm} )&(2x \kern{0.4 cm}-\kern{0.4 cm} )\end{aligned}

Step 3: We need to find two numbers which when multiplied make \textcolor{red}{-1}

\textcolor{red}{-1} has only one factor. 2 — x — 12.

[2 marks]

Step 1: Draw empty brackets

(x \kern{1 cm} ) (x \kern{1 cm} )

Step 2: Identify sub-type (b)

(x \kern{0.4cm} + \kern{0.4cm} ) (x \kern{0.4 cm} — \kern{0.4cm} )

Step 3: We are looking for two numbers which multiply to make \textcolor{red}{-12} and add to make \textcolor{blue}{-1}. Let’s consider some factor pairs of -12.

\begin{aligned}(-1)\times12&=-12 \,\,\text{ and } -1 + 12 = 11\\(-2)\times6&=-12 \,\,\text{ and } -2 + 6 = 4\\ (-6)\times2&=-12 \,\,\text{ and } -6 + 2 = -4\\ (-3)\times4&=-12 \,\,\text{ and } -3 +4 = 1\\ \textcolor{red}{(-4)\times3}&\textcolor{red}{=-12 }\,\,\text{ and } \textcolor{blue}{-4 + 3 = -1}\end{aligned}

We could keep going, but there’s no need because the last pair, -4 and 3, add to make -1. This pair fills both criteria, (as highlighted above) so the factorisation of x^2 — x — 12 is

(x — 4)(x + 3)

Note: You can try expanding the double brackets to check your answer is correct. 2 -5m -6

We can see that last option with +3 and -2 is the correct combination.

This gives the final answer to be:

(4m+3)(m-2)

Factorising Quadratics — GCSE Maths

Introduction

Factorising quadratics video

What is a quadratic expression?

Factorising quadratics worksheets

Common misconceptions

Learning checklist

Next lessons

Still stuck?

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Introduction

Factorising quadratics video

What is a quadratic expression?

Factorising quadratics worksheets

Common misconceptions

Learning checklist

Next lessons

Still stuck?

Here we will learn about factorising quadratics; we will explore what quadratic expressions are and the steps needed to factorise into double brackets. {2}\color{#00BC89}{+3}x\color{#7C4DFF}{-2}\]

We factorise quadratic expressions of this sort using double brackets. There are different methods we can use depending on whether the coefficient of x² is greater than 1.

What is factorising quadratics?

Factorising, or factoring quadratic equations is the opposite of expanding brackets and is used to solve quadratic equations.

For example, in the form of x² + bx + c requires two brackets (x + d) (x + e).

How to factorise quadratics:

Write out the factor pairs of the last number (c).
Find a pair of factors that + to give the middle number (b) and multiply to give the last number (c).
Write two brackets and put the variable at the start of each one.
Write one factor in the first bracket and the other factor in the second bracket. The order isn’t important, but the signs of the factors are.

What is factorising quadratics?

If you’re looking for a summary of all the different ways you can factorise expressions then you may find it helpful to start with our main factorising lesson or look in detail at the other lessons in this section.

Factorising
Factorising single brackets
Difference of two squares

Quadratic expressions or quadratic equations?

A quadratic equation is a quadratic expression that is equal to something. We can solve quadratic equations by using factorisation (or factoring), the quadratic formula or by completing the square.

Step by step guide: Quadratic equations

Factorising quadratics worksheets

Download two free factorising quadratics worksheets to help your students prepare for GCSEs. Includes reasoning and applied questions.

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Factorising quadratics worksheets

Download two free factorising quadratics worksheets to help your students prepare for GCSEs. Includes reasoning and applied questions.

DOWNLOAD FREE

Factorising quadratics in the form x

² + bx + c

To factorise a quadratic expression in the form x² + bx + c we need double brackets. Factorisation into double brackets is the reverse process of expanding double brackets.

In this case, the coefficient (number in front) of the x² term is 1 (a=1). These are known as monic quadratic.

How to factorise quadratics: x

² + bx + c (double brackets)

In order to factorise a quadratic algebraic expression in the form x² + bx + c into double brackets:

Write out the factor pairs of the last number (c).
Find a pair of factors that + to give the middle number (b) and ✕ to give the last number (c).
Write two brackets and put the variable at the start of each one.
Write one factor in the first bracket and the other factor in the second bracket. The order isn’t important, the signs of the factors are.

Explain how to factorise quadratics: x² + bx + c (double brackets)

Factorising quadratics examples: x

² + bx + c (double brackets)

Example 1: with +x coefficient and a +constant

Fully factorise:

\[x^2 \color{#00BC89}{+ 6x}\color{#7C4DFF} {+ 5}\]

Write out the factor pairs of the last number (5) in order. 2 – 2x – 24\]

Write out the factor pairs of the last number (24) in order

x² – 2x – 24

Factors of 24:

1, 24
2, 12
3, 8
4, 6

We need a pair of factors that + to give the middle number (-2) and ✕ to give the last number (-24).

x² – 2x – 24

Factors of 24:
1, 24
2, 12
3, 8
4, 6

-6 + 4 = -2 ✔

-6 ✕ 4 = -24 ✔

(It’s a good idea to do a quick check that we have the correct numbers)

Remember: to multiply two values together to give a negative answer, the signs must be the different.

Write two brackets and put the variable at the start of each one (x in this case).

(x )(x )

Write one factor in the first bracket and the other factor in the second bracket. The order isn’t important, the signs of the factors are.

(x – 6)(x + 4)

We have now fully factorised the quadratic expression.

We can check the answer by multiplying out the brackets!

(x – 6)(x + 4) = x² – 2x – 24

Example 3: with +x coefficient and a –constant

Fully factorise:

x² + x – 20

Write out the factor pairs of last number (20) in order. 2 – 8x + 15\]

Write out the factor pairs of the last number (15) in order.

Factors of 15:

1, 15

3, 5

We need a pair of factors that + to give the middle number (-8) and ✕ to give the last number (15).

x² – 8x + 15

Factors of 15:
1, 15
3, 5

-3 + -5 = -8 ✔
-3 ✕ -5 = 15 ✔

It’s a good idea to do a quick check that we have the correct numbers.

Remember: to ✕ two values together to give a positive answer, the signs must be the same

Write two brackets and put the variable at the start of each one (x in this case)

(x )(x )

Write one factor in the first bracket and the other factor in the second bracket.

\[(x – 3)(x – 5)\]

We have now fully factorised the quadratic expression.

We can check the answer by multiplying out the brackets!

(x – 3)(x – 5) = x² – 8x + 15

Practice factorising quadratics questions: x

² + bx + c (double brackets)
(x+5)(x+6)

(x+3)(x+2)

(x+1)(x+6)

(x+3)(x+3)

We need numbers that have a product of 6 and a sum of 5 . {2}-10x+24=(x-4)(x-6) .

Factorising quadratics GCSE questions: x

² + bx + c (double brackets)
1. Factorise: x² + 3x – 10

Show answer

(x – 2)(x + 5)

(2 marks)

2. Factorise: y² – 10y + 16

Show answer

(y – 2)(y – 8)

(2 marks)

3. Factorise: x² – 12x + 27

Show answer

(x – 3)(x – 9)

(2 marks)

Factorising quadratics in the form ax

² + bx + c
To factorise a quadratic expression in the form ax² + bx + c we need double brackets. Factorising into double brackets is the reverse process of expanding double brackets.

In this case the coefficient (number in front) of the x² term is greater than 1 (a > 1). These are known as non-monic quadratics.

How to factorise quadratics: ax

² + bx + c (double brackets)
In order to factorise a quadratic algebraic expression in the form ax² + bx + c into double brackets:
1. Multiply the end numbers together (a and c) then write out the factor pairs of this new number in order. 2 + 5x + 3\]
  1. Multiply the end numbers together (2 and 3) then write out the factor pairs of this new number in order.
  2x² + 5x + 3
  
  2 × 3 = 6
  
  Factors of 6:
  1, 6
  2, 3
  
  2We need a pair of factors that + to give the middle number (5) and ✕ to give this new number (6).
  
  2x² + 5x + 3
  
  2 × 3 = 6
  
  Factors of 6:
  1, 6
  2, 3
  
  + 5
  ✕ 6
  
  2 + 3 = 5 ✔
  2 x 3 = 6 ✔
  
  Remember: to x two values together to give a positive answer, the signs must be the same.
  
  3Go back to the original equation and rewrite it this time splitting the middle term into the two factors we found in step 2 – the order of these factors doesn’t matter, the signs do.
```
2x² + 5x + 3
```
```
2x² + 2x + 3x + 3
```
  4Split the equation down the middle into two halves and fully factorise each half – the expressions in the brackets must be the same!
```
2x² + 5x + 3
```
```
2x² + 2x + 3x + 3
```
```
2x(x + 1) + 3(x + 1)
```
  2x(x + 1) + 3(x + 1)
  
  5Now factorise the whole expression by bringing whatever is in the bracket to the front and writing the two other terms in the other bracket. 2 + 3x – 2\]
  
  Multiply the the end numbers together (2 and -2) then write out the factor pairs of this new number in order.
  
  2x² + 3x – 2
  
  2 ✕ -2 = -4
  
  Factors of 4:
  1, 4
  2, 2
  
  We need a pair of factors that + to give the middle number (3) and ✕ to give this new number (-4)
  
  2x² + 3x – 2
  
  2 ✕ -2 = -4
  
  Factors of 4:
  1, 4
  2, 2
  
  ⊕ 3
  ✕ -4
  
  -1 + 4 = 3 ✔
  -1 ✕ 4 = -4 ✔
  
  Remember: to x two values together to give a negative answer, the signs must be different
  
  Go back to the original equation and rewrite it this time splitting the middle term into the two factors we found in step 2 – the order of these factors doesn’t matter, the signs do.
```
2x² + 3x - 2
```
```
2x² - x + 4x - 2
```
  Split the equation down the middle into two halves and fully factorise each half – the expressions in the brackets must be the same!
```
2x² + 3x - 2
2x² - x + 4x - 2
x(2x + 1) + 2(2x - 1)
```
  Now factorise the whole expression by bringing whatever is in the bracket to the front and writing the two other terms in the other bracket. 2 – 2x – 8\]
  
  Multiply the the end numbers together (3 and -8) then write out the factor pairs of this new number in order.
  
  3x² – 2x – 8
  
  3 ✕ -8 = -24
  
  Factors of 24:
  1, 24
  2, 12
  3, 8
  4, 6
  
  We need a pair of factors that + to give the middle number (-2) and to ✕ give this new number (-24)
  
  3x² – 2x – 8
  
  3 ✕ -8 = -24
  
  Factors of 24:
  1, 24
  2, 12
  3, 8
  4, 6
  
  ⊕ -2
  ✕ -24
  
  -6 + 4 = -2 ✔
  -6 ✕ 4 = -24 ✔
  
  Remember: to ✕ two values together to give a negative answer, the signs must be different
  
  Go back to the original equation and rewrite it this time splitting the middle term into the two factors we found in step 2 – the order of these factors doesn’t matter, the signs do.
```
3x² - 2x - 8
```
```
3x² - 6x + 4x - 8
```
  Split the equation down the middle into two halves and fully factorise each half – the expressions in the brackets must be the same!
```
3x² - 2x - 8
```
```
3x² - 6x + 4x - 8
```
```
3x(x - 2) + 4(x - 2)
```
  Now factorise the whole expression by bringing whatever is in the bracket to the front and writing the two other terms in the other bracket. 2 – 7x + 2 \]
  
  Multiply the the end numbers together (6 and 2) then write out the factor pairs of this new number in order.
  
  6x² – 7x + 2
  
  6 ✕ 2 = 12
  
  Factors of 12:
  1, 12
  2, 6
  3, 4
  
  We need a pair of factors that + to give the middle number (-7) and ✕ to give this new number (12)
  
  6x² – 7x + 2
  
  6 ✕ 2 = 12
  
  Factors of 12:
  1, 12
  2, 6
  3, 4
  
  + -7
  ✕ -24
  
  -3 + -4 = -7 ✔
  -3 ✕ -4 = 12 ✔
  
  Remember: to ✕ two values together to give a positive answer, the signs must be the same
  
  Go back to the original equation and rewrite it this time splitting the middle term into the two factors we found in step 2 – the order of these factors doesn’t matter, the signs do.
```
6x² - 7x + 2
```
```
6x² - 3x - 4x + 2
```
  Split the equation down the middle into two halves and fully factorise each half – the expressions in the brackets must be the same!
```
6x² - 7x + 2
```
```
6x² - 3x - 4x + 2
```
```
3x(2x - 1) - 2(2x - 1)
```
  Now factorise the whole expression by bringing whatever is in the bracket to the front and writing the two other terms in the other bracket. {2}-8x-x+4] or 2[2x(x-4)-(x-4)] so that the fully factorised expression is 2(2x-1)(x-4) .
  
  Factorising quadratics GCSE questions: ax
  
  ² + bx + c (double brackets)
  1. Factorise: 2x² + 9x + 4
  
  Show answer
  
  (2x + 1)(x + 4)
  
  (2 marks)
  
  2. Factorise: 2y² – y – 3
  
  Show answer
  
  (2y – 3)(y + 1)
  
  (2 marks)
  
  3. Factorise: 2x² – x – 10
  
  Show answer
  
  (2x – 5)(x + 2)
  
  (2 marks)
  
  Common misconceptions
  - The order of the brackets
    When we multiply two values the order doesn’t matter. This is true for the brackets when factorising quadratics
    
    e.g. 2 ✕3 = 3 ✕2
    
    It is exactly the same here.
    
    (x – 6)(x + 4) means (x – 6)(x + 4)
    
    So,
    
    (x – 6)(x + 4)=(x + 4)(x – 6)
  + ✕ + = +
  
  e.g 2 ✕ 3 = 6
  
  4 ✕ 5=20
  
  – ✕ – = +
  
  e.g -2 ✕ -3 = 6
  
  -4 ✕ -5= 20
  
  + ✕ – = –
  
  e.g 2 ✕ -3= -6
  
  4 ✕ -5= -20
  
  – ✕ + = –
  
  e. {2} + bx + c (H)