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How do you factorise a quadratic equation?

If you are given a simple quadratic equation, for example x2+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 :
x2+2x+4x+8=0
Simplifying to:
x2+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 xthat are greater than 1. Solving quadratics with negative signs for example: x2-4x-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 x2 is greater than 1. For example: 2x2+5x+2=0. You then have to also consider the factors of the coefficient of x2

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).

3 years ago

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