Quadratic Factorisation

Whether you are taking HL, SL or Studies, you can be certain that there will be at least one quadratic expression to factorise in your exams – probably several. Here is my quick guide.

If I multiply out these brackets:


I will get:


The result is a quadratic expression: a quadratic must have a term in x^2, and can also have a term in x and a constant (a number). If you reverse the process, you will be factorising the quadratic. So, here are the quadratic expressions you might be faced with:

Quadratic with no constant term

This is easy because all you have to do is find the common factor. For example:



Quadratic with no term in x

Generally these quadratics cannot be factorised, unless in the form of the difference of two squares, ie x^2-a^2. This factorises to (x+a)(x-a). For example:



It is important to remember that expressions of the form x^2+a^2 do not factorise. For example, x^2+16\ne (x+4)^2 since


Remember that when factorising any expression you should look for a common factor first:


Quadratic with all three terms

If the coefficient of x2 is 1:

Look at the expansion at the top: the 2x came from 3x – and -3 from 3\times -1. So in reverse it’a quite simple: the two numbers in the brackets will add to give the term in x and multiply to give the constant term. You just have to be really careful with negative numbers. So, some examples:

Example 1:   Factorise x^2+10x+21.

So, what two numbers add to give 10 and multiply to give 21? Answer: 3 and 7.

x^2+10x+21 = (x+3)(x+7)

Example 2:   Factorise x^2-2x-15.

What two numbers add to give -2 and multiply to give -15? Answer: -5 and 3.

x^2-2x-15 = (x-5)(x+3)

Example 3:   Factorise x^2-6x+5

Since the numbers multiply to give a positive, but add to give a negative, they must both be negative. They are therefore -5 and -1. (It would be very easy to come up, incorrectly, with -2 and -3 – it just shows how important it is that you check your answer). 


If the coefficient of x2 is greater than 1:

There is a method for doing these, but it’s quite complicated to remember. The questions are rarely too challenging, so trial and error should get you there – but don’t forget you can always use the quadratic formula if you need to. Also, you must first check whether there is a common factor, which will simplify things considerably:

Example 4:   Factorise 3x^2-x-10

The brackets will begin with 3x and x, so that means the number in the second bracket will be multiplied by 3. The two numbers are probably 5 and 2, with one of them being negative. Trying all the possibilities quickly leads to the correct answer:


Example 5:   Factorise 5x^2-30x+25

5 is a common factor:


Solving quadratic equations by factorisation

So far we have been factorising quadratic expressions. Quadratic equations can be solved by factorisation, but only if the right hand  side is zero. This is because when a\times b = 0, either a or b must be zero. So if:


then either (x – 2) = 0, giving x = 2, or (x + 5) = 0, giving x = -5.

Example 6:   Solve x^2+2x-10=x+2

First rearrange to make the right hand side equal zero:


Now factorise:


And solve:

x=-4 or x=3

Example 7:   Solve 3x^2-12x=0

There is no constant term, but there is a common factor:



x=0 or x=4

I hope you found this useful. Do give me feedback if there are other topics you would like me to cover.

  • Nathalie Jimenez
    August 20, 2016


    I’m a math studies student and these posts really help me out!

    Would it be possible for you to write a post covering the “oh so confusing” topic of functions (more specifically exponential functions)?

    Thank you once again for these informative posts. Keep up the good work!

    • Ian Lucas
      August 21, 2016

      Hi Nathalie

      Glad you find the posts about mathematics topics useful – I intend to do lots of these! There will definitely be one about exponential functions.


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