Factorisation

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Factorisation basics

By ‘factoring’ something, we intend to bring out common terms. So, basically, we wish to put an expression back into one which has brackets, as opposed to multiplying out the brackets like we have so far. For this reason, factorisation can be seen as the opposite to multiplying out brackets.

To do this we must find a common term and then bring this out to the front of the brackets. This can then make an expression easier to read and work with.

Example

Factorise the following:

a) $xy+xz$

b) $2a+8b-4c$

c) $3ab+abc+ab^2$

a) Here we have a common term of x so this can be brought out to the front and the other terms put into a bracket without the x. This gives an answer of $x(y+z)$ .

b) For this question we have a common term that is a number and not an algebraic symbol like a letter. Therefore, we can do the same things as we did above but bring out the 2 to give the answer $2(a+4b-2c)$ .

c) Here the common term is a mix of two different letters but exactly the same method is used. We get an answer of $ab(3+c+b)$ .

The thing you must remember when factorising is to bring out the common term(s) and divide the other terms by this before they are put into the bracket. Then when we multiply out the bracket we get the exact same expression that we started with. A common mistake is to factorise something and have the elements in the bracket the same as what we started with. For example: factorising $x^2-4x$ is NOT $x(x^2-4x)$ despite x being the correct common factor. This is because we have not divided by the common factor, and when we multiply out the bracket we get $x^3-4x^2$ which is not equal to the expression that we started with.

We can also factorise when we already have a bracket in the question. For example:

Factorise $a(c-5)-b(c-5)$

Clearly the common term here is $(c-5)$ therefore we can bring this out to the front in exactly the same way and put the remaining terms into brackets, giving an answer of $(c-5)(a-b)$

Here is a step-by-step guide for factorising ANY expression:

1) Identify the common terms that are used in more than one part of the expression.

2) Bring these common terms out to the front and put any remaining terms into brackets, remembering to divide each by the common term.

3) Make sure that the terms in the bracket are not the originals, but have been divided by the common term.

4) Check the answer by multiplying out the bracket and matching to the original expression.

Multiplying out brackets

When factorising out an expression that has a bracketed part as the common term we get an answer that has a bracket multiplied by another bracket, such as in the previous section where we had $(c-5)(a-b)$ . If we wish to check the answer to this we obviously need to know the rules of multiplying two brackets like this.

When we learnt how to multiply a single term by a bracket before, we used the rule of multiplying the term outside by everything in the bracket. When we have two brackets we use a similar rule and multiply everything in the first bracket by everything in the second, doing each term separately.

The arrows around the brackets above show which parts should be multiplied by others to make sure that every term in the first bracket gets multiplied by every term in the second. We can think of doing this as basically going from something that is factorised to something that isn’t. So $(a+b)(c+d)$ is the same as $a(c+d)+b(c+d)$ .

A quick way to check if you have multiplied two brackets correctly is to multiply the number of terms in the first by the number in the second, this is the amount of terms you should have in your answer before any like terms are collected. So, in the example above, we have two terms in the first bracket and two in the second, meaning our answer should have four terms. Which is correct since our answer would be:

$ac+ad+bc+bd$