# How do I factorise a quadratic equation?

*Firstly*, make sure you agree that a quadratic equation is an equation of the form **y = ax ^{2} + 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 = x ^{2} + 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 **ax ^{2} + bx + c = (dx + e)(fx + g)**

*Working with our example,* **y = x ^{2} + x - 6**, we first direct our attention to the constant term c, in this case

**c = -6**.

If we expand the brackets, we get **ax ^{2} + bx + c =(dx+e)(fx + g) = dfx^{2} + (ef +dg)x + eg**.

We should already be able to see that for our example where the coefficient of x^{2} is 1 that d,f = 1 so now we have a simpler equation:

**(x+e)(x+g) = x ^{2} + (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 o**

__f 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) = x ^{2} - 2x + 3x - 6 = x^{2} + x - 6 **

as required!