Factorising | Revision, Question and Worksheets | MME

# Factorising | Questions, Worksheets and Revision

Level 1-3

## 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.

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

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