# How do you factorise a quadratic equation?

If you are given a simple quadratic equation, for example x^{2}+6x+8=0, then in order to factorise this you must find two numbers that __add together to make the coefficient of x__ in the equation (in this case the coefficient is 6) and __multiply together to find the constant__ (in this case 8).

In order to do this you must find the __pairs of factors__ that multiply together to make the constant, so for this example the factors of +8 are 1&8, 2&4, (-1)&(-8) and (-2)&(-4), then using these factors you have to find a pair that will add together to make the coefficient of x, which we know is 6. Therefore the only pair of factors that will add up to 6 are 2 & 4.

So we place these into brackets like so:

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

To check your answer you can simply expand the brackets again, which would give you :

x^{2}+2x+4x+8=0

Simplifying to:

x^{2}+6x+8=0, which is what we originally started with therefore showing that we have factorised correctly.

Factorising quadratics can become more complicated when there are negatives or coefficients of x^{2 }that are greater than 1. Solving quadratics with negative signs for example: x^{2}-4x-12=0 is done the same way as before. The factors of -12 are: 1&(-12), (-1)&12, 2&(-6), (-2)&6, 3&(-4) and (-3)&4. Then added together the only factors that make -4 are (-6)&2. Therefore the answer would be (x-6)(x+2)=0

They become even more tricky when the coefficient of x^{2} is greater than 1. For example: 2x^{2}+5x+2=0. You then have to also consider the factors of the coefficient of x^{2}

The only factors of the constant 2 are 1&2 and (-2)&(-1). However one of the factors in a pair will be multiplied by 2, so these factors become: 2&2, 1&4, (-4)&(-1) and (-2)&(-2). We then need to find out which of these factors add up to 5, which we can see is 1&4.

So our final answer is (2x+1)(x+4).