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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.

In algebra, this works the same way. If our algebraic expression is $5xy$ , then we can say that $x$ is a factor of this because $x \times 5y = 5xy$ . In fact, spotting the factors of this is maybe even easier than with numbers because it’s immediately broken up into the factors of 5, $x$ , and $y$ .

The aim of factorising is to manipulate an algebraic expression (one with multiple terms) by “taking out” common factors – factors that each term in this expression has. This amounts to writing the common factors on the outside of a pair of brackets and writing the algebraic expression with those factors removed on the inside. Mathematically, this means that everything outside the bracket is being multiplied by everything inside the bracket

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. Having a good understanding of expanding brackets will help with this topic.

Example 1: Factorising Two Terms

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

So, we’re looking for things that both terms have in common. Let’s consider the coefficients (the numbers in front of the algebraic bits) first. One is 3 and the other is 6 – both these numbers 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

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

Now we ask: 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:

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

The terms left inside the bracket have no more common factors, so we are done. Just as a check, if we are to expand the bracket, then that means multiplying $3x$ by $y$ to get $3xy$ , multiplying $3x$ by $2x$ to get $6x^2$ , and finally adding the results together. So indeed, the result of the expansion is

$3xy + 6x^2$

The fact that this is the expression we started with tells us that we probably did the factorising correctly.

Before we go further, it helps to recall the laws of indices (Rules of indices), specifically the multiplication rule:

$a^m \times a^n = a^{m + n}$ .

You’ll probably see why in this next example. The idea is exactly the same as the last one, there’s just going to be a little more going on.

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 (and so removing an $a$ from each term inside 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 $c$ , but rather than taking this out we can actually go one further – they all also have a factor of $c^2$ . Using the multiplication rule of indices, we get

$c^5 = c^2 \times c^3 \text{ and } c^3 = c^2 \times c$ ,

and furthermore,

$2bc^5 = c^2 \times 2bc^3 \text{ and } 4a^{2}c^3 = c^2 \times 4a^{2}c$ .

In general, we want to look for the highest power of a factorthat is shared by every term – here, this is $c^2$ . Finally, the terms inside the brackets have no more common factors, so we are done.

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

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.

Example Questions

1) Factorise fully $10pq + 15pqr$

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.

2) Factorise fully $u^3+3uv^3+2u$

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.

3) Factorise fully $4xy^5 + y^5 + 12y^7$

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.

4) Factorise fully $5xy^2-5x^2y-5x^2y^2$

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.

5) Factorise fully $7abc + 14a^{2}bc + 21ab^{2}c + 49abc^3$

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.