## What you need to know

Quadratics are algebraic expressions of the form

$ax^2 + bx + c$

Where $a, b$, and $c$ are all constants (numbers, basically). We are first going to look at factorising quadratics where the number before the $x^2$ term, $a$, is 1, and then later (just for the higher students) we will look at how to factorise quadratics where $a$ is something other than 1.

The aim of factorising is to break something up, numbers or algebra, into factors. We’ve seen factorising a bunch of terms into one bracket by taking all the common factors outside the bracket – if you haven’t seen this, you should head over here (https://mathsmadeeasy.co.uk/gcse-maths-revision/factorising-foundation-gcse-maths-revision-worksheets/). But the trouble here is, if we look at a quadratic, for example

$x^2 - x + 12$,

we can see that the three terms have no common factors, so we aren’t going to be able to factorise it into one bracket. Instead, we factorise it into two brackets. To see how we’re going to do this, let’s expand a pair of brackets (more on bracket expansion here (https://mathsmadeeasy.co.uk/gcse-maths-revision/multiplying-single-double-brackets/)). Consider $(x + 5)(x + 2)$. Expanding this bracket (using whatever your favourite method is), gets

$(x + 5)(x + 2) = x^2 + 2x + 5x + 10$

Collecting the two $x$ terms, this becomes

$x^2 + 7x + 10$

Now, notice that our constant term is 10 because it is the product of the two numbers in the brackets. Furthermore, the coefficient of $x$ is 7 because it is the sum of $2x$ and $5x$, both coefficients corresponding to the numbers in the brackets.

This may all seem a little obvious, but the point is that the constant term of our result is the product of the two numbers in the brackets, and the coefficient of $x$ is the sum of the two numbers in the brackets. This will always be the case, and this is how we will factorise quadratics – by looking for two numbers which times together to make the constant term and add together to make the coefficient of $x$. Let’s take our first quadratic as an example.

Example: Factorise $x^2 - x - 12$.

We are looking for two numbers which multiply to make -12 and add to make -1. Let’s consider some factor pairs of -12.

$-12=(-1)\times 12,\text{ or }(-2)\times 6,\text{ or }(-3)\times 4,\text{ or }(-4)\times 3$

We could keep going, but there’s no need because the last pair, -4 and 3, add to make -1. This pair fills both criteria, (as underlined above) so the factorisation of $x^2 - x - 12$ is

$(x - 4)(x + 3)$

It’s worth noting that it doesn’t make a difference which bracket you put where.

There is a special kind of quadratic, so special that it deserves its own name, which we’ll see now.

Example: Factorise $x^2 - 9$.

Approaching this using the above method, we need to consider that we’re looking for two number which multiply to make -9 and add to make 0 – since there’s no $x$ term, its coefficient must be zero.

For 2 numbers to add to make zero, they must be the same number but with different signs. Furthermore, two of the same numbers being multiplied to make -9 must correspond to a square root. The square root of 9 is 3, and indeed, the pair: 3 and -3 satisfy both criteria of adding to zero and multiplying to make -9. Therefore, the factorisation is

$(x + 3)(x - 3)$

This special case is called the difference of two squares, because it always takes the form of $x^2$ takeaway some other square number.

So, if you see a quadratic with no $x$ term, then you know that it can be factorised quickly – with the square root of the constant term in both brackets – one negative and one positive.

Now, for the higher students only, as mentioned there is another kind of quadratic, one with a number in front of the $x^2$ term. The method above that we now know and love unfortunately does not work anymore, so we have to try something different. This method is much less intuitive and is best explained directly through example.

Example: Factorise $2x^2 + 7x + 3$.

To do this, we must firstly multiply the coefficient of $x^2$ by the constant term, so in this case that is: $2 \times 3 = 6$. Then, we should look for two numbers which multiply to make the result of this, 6, and add to make 7 (this is similar to the earlier method but what we do with the numbers once we’ve found them is different).

$6 = 2 \times 3, \text{ or } 1 \times 6$

1 and 6 add to 7, so 1 are 6 is our relevant pair of numbers. Now we must rewrite the quadratic with the $x$ split into two parts according to the pairing that we found. In other words, because our pair is 1 and 6, we write the $7x$ term as $x + 6x$:

$2x^2 + x + 6x + 3$

Next, we have to treat the quadratic like two separate expressions and factorise the first two terms into their own single bracket, then factorise the last two terms into their own single bracket:

$x(2x + 1) + 3(2x + 1)$

If everything has gone swimmingly, then what’s inside both brackets should be the same thing. This means that we can now take the bracket itself, $(2x + 1)$, out as a factor (this last step is a bit weird, it’s okay if you’re unsure of it right away). We then get

$(2x + 1)(x + 3)$,

which is the full factorisation of the quadratic. Have a couple of reads through this to make sure that you understand how to do each step.

## Example Questions

We are looking for two numbers which add to make 1 and multiply to make -30. Let’s list some factor pairs for -30.

$-30 = (-1) \times 30\text{ or } (-2) \times 15\text{ or } (-5) \times 6$

We can stop here because $-5 + 6 = 1$ as desired, so -5 and 6 is our pair. Therefore, the full factorisation of $a^2 + a - 30$ is

$(a - 5)(a + 6)$

We are looking for two numbers which add to make -5 and multiply to make 6. Let’s list some factor pairs for 6.

$6=1\times 6\text{ or }2\times 3,\text{ or }(-1)\times (-6)\text{ or }(-2)\times (-3)$

These are all the possible factor pairs for 6, and we can see $-2 + (-3) = -5$ as desired, so -2 and -3 is our pair. Therefore, the full factorisation of $k^2 - 5k + 6$ is

$(k - 2)(k - 3)$

Firstly, $4 \times (-6) = -24$. So, we are looking for two numbers which multiply to make -24 and add to make -5. Let’s look at some factor pairs for -24.

$-24 = 1 \times (-24) \text{ or } 2 \times (-12)\text{ or } 3 \times (-8),$

We can stop here because $-8 + 3 = -5$ as desired, so -8 and 3 is our pair. So, now we can rewrite the quadratic, splitting up $-5m$ into $-8m + 3m$:

$4m^2 - 8m + 3m - 6$

Now, we factorise the first two terms into a single bracket and the last two terms into another single bracket to get

$4m(m - 2) + 3(m - 2)$

As per the plan, both of the brackets are the same. This means we can take out $(m - 2)$ as a factor of both terms in the expression. This leaves us with

$(m - 2)(4m + 3)$

Which is the full factorisation of the quadratic.

If you are looking for GCSE Maths materials for your students then the factorising quadratic resources on this page are a great way to cover this topic. Whether you are a private Maths tutor in York or you are a maths teacher in London, the factorising quadratic questions will make a great homework task or could even be used as lesson material.