 Factorising | Revision, Question and Worksheets | MME

# Factorising | Questions, Worksheets and Revision

Level 4-5

## Factorising into Single Brackets

Factorising is putting expressions into brackets, this is the reverse of expanding brackets

Make sure you are happy with the following topics before continuing.

1393
Created on By Matthew Phillpott

Factorising Quiz

Try the MME Factorising quiz.

1 / 4

Factorise fully
$20x^3-88x$

2 / 4

Factorise the following into a single bracket:

$-24x-6$

3 / 4

Factorise the following into a single bracket:
$36+4x$

4 / 4

Factorise the following into a single bracket:

$15x+5$

The average score is 65%

0%
KS3 Level 4-5

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KS3 Level 4-5

## Factorising into single brackets – 3 Key steps

Example:

Fully factorise the following:

$\textcolor{red}{12}\textcolor{limegreen}{x^2} +\textcolor{red}{8}\textcolor{limegreen}{x}$

Step 1 – Take out the largest common factor of both the numbers, and place it in front of the brackets.

Factors of $\textcolor{red}{12}$ are  $1, 2, 3, \textcolor{blue}{4}, 6, 12$

Factors of $\textcolor{red}{8}$ are $1, 2, \textcolor{blue}{4}, 8$

The largest common factor is $\bf{\textcolor{blue}{4}}$

$\textcolor{red}{12} = \textcolor{blue}{4} \times \textcolor{purple}{3}$

$\textcolor{red}{8} = \textcolor{blue}{4} \times \textcolor{purple}{2}$

Step 2 – Take out the highest power of the “Letter” which is a part of every term.

\begin{aligned}\textcolor{red}{12}\textcolor{limegreen}{x^2} = & \textcolor{red}{12} \times \textcolor{limegreen}{x} \times \xcancel{\textcolor{limegreen}{x}}\\ \textcolor{red}{8}\textcolor{limegreen}{x} = & \textcolor{red}{8} \times \xcancel{\textcolor{limegreen}{x}}\end{aligned}

We can take out one $\textcolor{limegreen}{x}$ from each term, placing in front of the brackets.

Step 3 – Place the highest factor and highest power of the letter in front of the bracket, then add the remaining terms inside the bracket

$\textcolor{blue}{4}\textcolor{limegreen}{x}(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)$

$\textcolor{blue}{4}\textcolor{limegreen}{x}(\textcolor{purple}{3}\textcolor{limegreen}{x} +\textcolor{purple}{2})$

To check, you can multiply the bracket back out to see if you have the right answer.

$4x(3x+2) = 12x^2 +8x$

KS3 Level 4-5

## Example 1: Factorising Two Terms

Fully Factorise the following, $\textcolor{red}{3}\textcolor{limegreen}{x}\textcolor{Orange}{y} + \textcolor{red}{6}\textcolor{limegreen}{x^2}$.

[2 marks]

Step 1 – Take out the largest common factor of both the numbers, and place it in front of the brackets.

Factors of $\textcolor{red}{3}$ are  $1, \textcolor{blue}{3}$

Factors of $\textcolor{red}{6}$ are $1, 2, \textcolor{blue}{3}, 6$

The largest common factor is $\bf{\textcolor{blue}{3}}$

$\textcolor{red}{3} = \textcolor{blue}{3} \times \textcolor{purple}{1}$

$\textcolor{red}{6} = \textcolor{blue}{3} \times \textcolor{purple}{2}$

Step 2 – Take out the highest power of the “Letter” which is a part of every term.

\begin{aligned}\textcolor{red}{3}\textcolor{limegreen}{x}\textcolor{Orange}{y} = & \textcolor{red}{3} \times \xcancel{\textcolor{limegreen}{x}} \times \textcolor{orange}{y}\\ \textcolor{red}{6}\textcolor{limegreen}{x^2} = & \textcolor{red}{6} \times \textcolor{limegreen}{x} \times \xcancel{\textcolor{limegreen}{x}}\end{aligned}

We can take out one $\textcolor{limegreen}{x}$ from each term, placing it in front of the brackets.

Step 3 – Place the highest factor and highest letter in front of the bracket, then add the remaining terms inside the bracket

$\textcolor{blue}{3}\textcolor{limegreen}{x}(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)$

$\textcolor{blue}{3}\textcolor{limegreen}{x}(\textcolor{orange}{y} +\textcolor{purple}{2}\textcolor{limegreen}{x})$

To check you can multiply the bracket back out to see if you have the right answer.

$3x(y+2x) = 3xy+6x^2$

KS3 Level 4-5

## Example 2: Factorising with three terms

Fully factorise the following,  $\textcolor{red}{8}\textcolor{limegreen}{x}\textcolor{Orange}{y} + \textcolor{red}{12}\textcolor{limegreen}{x^2}\textcolor{Orange}{y} - \textcolor{red}{4}\textcolor{limegreen}{x^2}\textcolor{Orange}{y^2}$.

[3 marks]

Step 1 – Take out the largest common factor of all of the numbers, and place it in front of the brackets.

Factors of $\textcolor{red}{8}$ are  $1, 2, \textcolor{blue}{4}, 8$

Factors of $\textcolor{red}{12}$ are $1, 2, 3, \textcolor{blue}{4}, 6, 12$

Factors of $\textcolor{red}{4}$ are $1, 2, \textcolor{blue}{4}$

The highest common factor of all three is $\textcolor{blue}{4}$

$\textcolor{red}{8} = \textcolor{blue}{4} \times \textcolor{purple}{2}$

$\textcolor{red}{12} = \textcolor{blue}{4} \times \textcolor{purple}{3}$

$\textcolor{red}{4}= \textcolor{blue}{4} \times \textcolor{purple}{1}$

Step 2 – Take out the highest power of the “Letter” which is a part of every term.

\begin{aligned}\textcolor{red}{8}\textcolor{limegreen}{x}\textcolor{Orange}{y} = & \textcolor{red}{8} \times \xcancel{\textcolor{limegreen}{x}} \times \xcancel{\textcolor{Orange}{y}}\\ \textcolor{red}{12}\textcolor{limegreen}{x^2}\textcolor{Orange}{y} = & \textcolor{red}{12} \times \xcancel{\textcolor{limegreen}{x}} \times\textcolor{limegreen}{x} \times \xcancel{\textcolor{Orange}{y}} \\ \textcolor{red}{4}\textcolor{limegreen}{x^2}\textcolor{Orange}{y^2} = & \textcolor{red}{4} \times \xcancel{\textcolor{limegreen}{x}} \times \textcolor{limegreen}{x} \times \xcancel{\textcolor{Orange}{y}} \times \textcolor{Orange}{y} \end{aligned}

We can take out one $\textcolor{limegreen}{x}$ and one $\textcolor{Orange}{y}$ from each term.

Step 3 – Place the highest factor and highest letter in front of the bracket, then add the remaining terms inside the bracket.

$\textcolor{blue}{4}\textcolor{limegreen}{x}\textcolor{Orange}{y}(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)$

$\textcolor{blue}{4}\textcolor{limegreen}{x}\textcolor{Orange}{y}(\textcolor{purple}{2} + \textcolor{purple}{3}\textcolor{limegreen}{x} - \textcolor{limegreen}{x}\textcolor{Orange}{y})$

To check you can multiply the bracket back out to see if you have the right answer.

$4xy(2 + 3x-xy) = 8xy + 12x^2y-4x^2y^2$

Level 4-5

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### Example Questions

Take out a factor of 5 from both terms to get

$5(2pq + 3pqr)$

There is both a $p$ and a $q$ in the two terms inside the bracket. Taking out both $p$ and $q$, we get

$5pq(2 + 3r)$

The two numbers in the bracket have nothing more in common so we are done.

Take out a factor of $u$ from both terms to get,

$u(u^2+3v^3+2)$

The terms inside the bracket have no more common factors, so we are done.

The first and last term have a factor of 4 in common, but the middle term doesn’t, so we can’t take any numbers out as factors.

All 3 terms have a factor of $y$ in them. Specifically, the highest power of $y$ that all 3 terms have in common is $y^5$. Taking $y^5$ out as a factor, we get,

$y^5(4x + 1 + 12y^2)$

The terms in the bracket have no more common factors, so we are done.

Take out a factor of 5 from every term to get

$5(xy^2-x^2y-x^2y^2)$

Now, clearly each term has a factor of $x$ and $y$, so we just need to determine what the highest power of each factor we can take out is,

$5xy(y-x-xy)$

The terms inside the bracket have no more common factors, so we are done.

Take out a factor of 7 from every term to get

$7(abc + 2a^{2}bc + 3ab^{2}c + 7abc^3)$

Now, clearly each term has a factor of $a, b$, and $c$, so we just need to determine what the highest power of each factor we can take out is.

The first term only has the three factors $a, b$, and $c$ to the power of 1 (note that we don’t write the power of 1 since $x^1 = x$), which means that this is the highest power of each factor we can take out is 1. Taking out a factor of $a, b$, and $c$, we get

$7abc(1 + 2a + 3b + 7c^2)$

The terms inside the bracket have no more common factors, so we are done.

### Worksheets and Exam Questions

#### (NEW) Factorising (Foundation) Exam Style Questions - MME

Level 4-5 New Official MME