Sometimes you might be asked to factorise expressions. Factorising is the opposite of multiplying out brackets – i.e. putting an expression into two brackets. A quadratic is just an express that contains a squared term, like x².

So basically ‘factorising quadratics’ just means putting an equation in the form ax² + bx + c into two brackets. Sounds easy right? Well sometimes it is and sometimes it’s pretty difficult! Let’s look at the easy cases first.

### How to factorise quadratics when a=1

When you only have one x² then you do the following:

- Rearrange equation into the format ax² + bx + c = 0
- Write out the brackets with x in – (x )(x ) = 0
- Find two numbers that multiply to give c, but add or subtract to form b and add to the brackets
- Check brackets multiply out to form original equation
- Solve by setting each bracket equal to 0

Phew! OK that might be a bit hard to follow, but have a look at the below example which will make it much easier.

**Example:** Solve x² – 3x = 10

1. Rearrange equation into format ax² + bx + c = 0

If you take away 10 from each side you get:

x² – 3x – 10 = 0

2. Write out the brackets with x in – (x )(x ) = 0

3. Find numbers that multiply to -10 and add up to -3

So first thing to do here is to look at those numbers that multiply to get 10. So basically you have:

1 & 10

2 & 5

Now you need to play around with + and – signs to figure out who to get any of these pairs to multiply to -10 and add/subtract to -3. Here you can see the only pair that would do this are +5 and -2.

So go ahead and add these to the brackets – (x + 5)(x – 2) = 0

4. Check brackets multiply out to form original equation

This is fairly standard but if you need help have a look at our guide to multiplying out brackets.

5. Solve by setting each bracket equal to 0

To solve you need to find the numbers that make each bracket zero

So if you wanted (x – 5) = 0 then you would need x = 5

If you wanted (x + 2) = 0 you would need x = -2

And so those are your answers! x = 5 and x = -2

### How to factorise quadratics when a isn’t 1

This get’s a bit tricky, so buckle up.

The method is basically the same – except steps 1 and 3. You still find numbers that multiply to give c, but then you need to try these out in the brackets to see if you can get b after multiplying out.

**Example:** Solve 2x² + 7x + 3 = 0

The equation here is already in the right format so go right ahead and write out the brackets – (2x )(x ) = 0

You now need to play around with the numbers that multiply to give 3. Luckily in this example there is only one set of numbers that do this i.e. 1 & 3. However, when a isn’t 1, you need to be careful the order you put the numbers in the brackets.

If you put the 1 in the second brackets you get:

(2x 3)(x 1) = 0

However when the brackets are multiplied out give 2x and 3x which can’t add up to 7x! So try the numbers in a different position.

(2x 1)(x 3) = 0

Now you get 1x and 6x which can add up to 7x! Boom! S0 now you just need to add in your signs, which here will both be +. You can then solve the equation by setting each bracket to equal 0:

(2x + 1) = 0 therefore x = -0.5

(x + 3) = 0 therefore x = -3

This type of question is pretty though so you might want to read back through. This is pretty much one of the hardest things at GCSE though, so if you have managed to figure this out you are doing pretty well.

### Watch Out For Tricky Exam Questions Using Quadratics

**Area/Volume Questions**

Examiners LOVE to throw in a quadratic in unexpected places – so always be on the look out. A good example is area or volume questions.

Example: The area of a rectangle is 560 square inches. The length is 3 inches more than twice the width. Find the length and the width.

Doesn’t look like a standard quadratic question does it…but it tells you the length is twice the width add 3 and if you look closely they are really looking for you to write down this equation.

(2W + 3) (W) = 560 – where W equals the width.

This one is tough…so you might want to use the quadratic formula on the next page.

Lastly remember on questions involving area to ignore any answer that is negative…as that can’t happen in real life!!

**Quadratics in Fractions**

Sometimes you will see an x as part of the bottom half of a fraction – so you’ll start thinking about multiplying by x….but then somewhere you might end up with an x squared.

As soon as you see an x squared check whether you can arrange the equation into the ax² + bx + c format and think about using the quadratic formula.**Here’s a bunch of example questions. To reveal the answer click on the “answer” tab.**

**(x + 4)(x + 5)**– This is a fairly straightforward one if you follow the method set out above. Remember to multiply out the brackets when you get your answer to check it’s correct.

**(x + 4)(x − 1)**– following the above method you just write out the bracket and play around with the numbers until you get the numbers that multiply to -4 but can add or subtract to form 3. Here those numbers are -1 and 4

**(m – 5)(m – 3)**– Don’t get worried if they change the letter from an “x” to an “m” or any other letter in the exam, you still just follow the same method as above.

**(2x + 3)(2x + 5)**– when the coefficient of the x² term (i.e. the number before the x² term) is not one, you have to try out more answers.

Given a 2x² term the solution must be of the form (2x + a)(x + b). You then have to play around expanding out possible solutions to get the above answer.

**(3x − 2)(2x + 6)**– when you get a 6x² in the question then the solution could possibly be be in the form (6x+a)(x+b) or (3x+a)(2x+b).

Also, since a×b=−12 we have six possible pairs of values to test:

−12 and 1

12 and −1

−2 and 6

2 and −6

−3 and 4

3 and −4

When you plug these in you can see the solution is that given above.