Factorising quadratics

What is a quadratic?

You need to be able to recognise these. One of the tell-tale signs is that your teacher will ask 'what kind of expression/equation is this?'. The other is that it has an x² in it* - the word 'quadratic' comes from the Latin for square.

What do you do with them?

Most often, you have to factorise them (put into brackets) or to solve them (figure out what x has to be). In this post, we're not going to talk about solving them using the quadratic formula, which you use when they ask for answers to a number of significant figures or decimal places. Instead, we'll look at putting them into brackets.

What's the method?

1. Take the number that doesn't have any xs attached to it, including any minus sign


2. List the factor pairs of this number (remember to think about positive and negative options), and what the numbers add up to


3. Find a pair that adds up to the middle number


4. Rewrite the x term of the expression with the x split up into the numbers you've just found (see below for an example)


5. You should now have an expression with four parts to it. Find a common factor for the first two and a common factor for the second two - bracket them off


6. Tidy up the brackets


7. If it's an equation, figure out what x has to be to make the brackets equal 0

Let's start with an example.

Say we have to solve x² -5x + 4 = 0**.

Step 1: We want the number on its own, which is +4
Step 2: What are the factor pairs of +4? We could have:

Factors	Times to make	Add to make
+1 and +4	+4	+5
+2 and +2	+4	+4
-1 and -4	+4	-5
-2 and -2	+4	+4

Step 3: We want them to add to the middle number - in this equation, it's -5. So the pair we want is -1 and -4.
Step 4: We can rewrite -5x as (-1x + -4x): that makes our equation x² -1x -4x + 4
Step 5: The first two terms have an x in common, so we can write them as x ( x-1 ). The second two terms have a -4*** we can take out, so we can write them as -4 (x-1). Hey, look! The bracket is the same for both of them!
Step 6: That means we can write the whole equation as (x-4)(x-1) = 0.
Step 7: It's an equation (there's an equals sign), so we think about what could make each bracket 0. If x=4, the first bracket is 0 and 0 multiplied by anything is 0 so the equation works. If x=1, the second bracket is 0, so the equation also works. The two answers are x=4 and x=1.

What if there's a number in front of the x²?

Excellent question, I was hoping you'd ask that. The method is almost identical, except that instead of just taking the number on its own, you multiply it by the number of x²s there are. For instance, if you had 4x² + 5x -6=0, instead of taking -6 as your number, you'd take 4 times -6 = -24.

Let's do that.

-24 has a lot of factors.

Factors	Times to make	Add to make
+1 and -24	-24	-23
+2 and -12	-24	-10
+3 and -8	-24	-5
+4 and -6	-24	-2
-1 and +24	-24	+23
-2 and +12	-24	+10
-3 and +8	-24	+5
-4 and +6	-24	+2

We want -3 and +8 because they add up to +5: the middle number. We can split up the 5x into -3x + 8x to get:

4x² - 3x + 8x - 6 = 0

If we take an x out of the first pair and a 2 out of the second, we get:

x(4x - 3) + 2(4x-3) = (x+2)(4x-3) = 0

And we're done! x = -2 or x = 3/4.

* Strictly, the x² has to be the highest power - if there's an x³ in the equation, it's a cubic, x⁴ is a quartic - you don't need to know those, you just need to know that they're not quadratics.

** CHORUS: "We have to solve x² - 5x + 4 = 0!"

*** We could take out a 4, sure. However, we really want the bracket afterwards to be the same - generally, we want to take out the minus sign if there is one).