Firstly, make sure you agree that a quadratic equation is an equation of the form y = ax² + bx + c where a,b,c are (real) constants (a is not 0), and x is the variable. (Note: the equation has an "x squared" term).

An example would be: y = x² + x - 6.

We want to factorise this equation so we can find the values of x for which y = 0 (the points where the curve crosses the x-axis).

Secondly, we need to be aware that factorising a quadratic mens expressing the equations as a product of two brackets which each contain an x term.

So, in the form ax² + bx + c = (dx + e)(fx + g)

Working with our example, y = x² + x - 6, we first direct our attention to the constant term c, in this case c = -6.

If we expand the brackets, we get ax² + bx + c =(dx+e)(fx + g) = dfx² + (ef +dg)x + eg.

We should already be able to see that for our example where the coefficient of x² is 1 that d,f = 1 so now we have a simpler equation:

(x+e)(x+g) = x² + (e + g)x + eg. (1.1)

We can then use eg = c = -6, our constant term ie. the constant terms in the brackets multiple to make our origanal constant term.

To find e,h, we think of pairs of numbers which multiply to give -6 :

-1 x 6 // -2 x 3 // -3 x 2 // -6 x 1

So how do we decide which pair of number will give the correct equation?

Well we could ~~test each pair and multiply out the brackets until we get the right equation~~, but this could take some time if we have more than four options, so instead we'll take a shortcut:

See which pairs add to give the coefficient of x

From (1.1), we can equate e + g = b.

-1 + 6 = 5 // -2 + 3 = 1 // -3 + 2 = -1 // -6 + 1 = -5

In our example, b = 1, so we can tell that our constants, e,h are -2,+3 and our answer is: y = (x - 2)(x + 3).

Check! (x - 2)(x + 3) = x² - 2x + 3x - 6 = x² + x - 6

as required!

Solve the simultaneous equations: 2x-3y = 16 x+2y= -6

Related Maths GCSE answers

Represent x = 0.0154 recurring as a fraction.

2x^2+21x=-40

How do I factorise a quadratic equation?

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