Factorising Maths | Question and Worksheets | MME

# Factorising Maths | Questions, Worksheets and Revision

Level 1 Level 2 Level 3

## What you need to know

### Factorising

A factor of some value is a number that can be multiplied by another number to get that value. For example, 2 is a factor of 10 because $2 \times 5 = 10$, and for this reason 5 is also a factor of 10.

The aim of factorising is to manipulate an algebraic expression (one with multiple terms) by “taking out” common factors. We wrote the common factors on the outside of a pair of brackets and the algebraic expression with those factors removed on the inside.

Having a good understanding of expanding brackets and the rules of indices, will help with this topic.

### Take Note:

We are always looking for ways to reduce the amount of errors we make when completing any mathematical procedure.

If you the expand brackets you have factorised, you should end up back where you started – you can use this to check your answer is correct.

### Example 1: Factorising Two Terms

Factorise fully $3xy + 6x^2$.

Both numbers in the question have a common factor of 3, so that means we can take a 3 out to the front of our brackets.

If we remove the factor of 3 from the first term, all that’s left is $xy$, but if we remove a factor of 3 from the second term, there is still a factor of 2 there (as well as the algebra part), meaning the result is $2x^2$. So, we get

$3xy + 6x^2 = 3 \times xy + 3 \times 2x^2 = 3(xy + 2x^2)$

Are there any other common factors? Yes, both terms in the bracket on the right-hand side have a factor of $x$ So, taking the $x$ out as a factor we get:

$3x(y + 2x)$

as our factorised expression.

### Example 2: Factorising With Indices

Factorise fully $4abc^5 + 2ac^2 + 8a^{3}c^3$.

We look at the numbers first. They are all multiples of 2, so we can take a factor of 2 out. This leaves us with

$4abc^5 + 2ac^2 + 8a^{3}c^3 = 2(2abc^5 + ac^2 + 4a^{3}c^3)$

We can see that they all have a factor of $a$ in common. Taking this outside the bracket, we get

$2a(2bc^5 + c^2 + 4a^{2}c^3)$

Now, the first term is the only one with a $b$, so we can’t take that out as a factor. They all have a factor of $c^2$. Taking this out leaves us with

$2ac^2(2bc^3 + 1 + 4a^{2}c)$

Note:removing the $c^2$ as a factor from the middle term does not mean the middle term disappears, it means that there is a 1 left in its place. If you aren’t sure about why this is the case, try expanding the bracket with/without the 1 there. When factorising we want to look for the highest power of a factorthat is shared by every term.

### 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.

Level 4-5

Level 4-5

GCSE MATHS

GCSE MATHS

GCSE MATHS

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