# Factorising

## Starting with the simple case (explaining with example)

We know that if there's a common factor in every term of an expression then we can factorise it out; $8x + 4 = 4(2x+1)$ Similarly, if that factor is $$x$$ then we get something like $x^2+3x = x(x+3)$ So continuing with this example; \begin{align} x^2+3x&=0 \\[3pt]x(x+3)&=0 \end{align} $x = 0\hspace{6 pt}\text{or }-3$

##### "But why is the root the negative of the value in the bracket?"

We know that if the product of two numbers is $$0$$ then at least one of them is $$0$$. For the equation $$x(x+3)=0$$ the first possibility is obvious, if $$x=0$$ then $0\times(0+3)=0$ We can also see that if $$x=-3$$ then $(-3)(-3+3)=(-3)\times0=0$

## The Usual Case (explaining with example)

More often, there will be a term which you cannot factor $$x$$ from, such as $x^2+5x+6=0$ So we have to try and factorise it into something of the form $(x+a)(x+b)=0$ where $$a$$ and $$b$$ are real numbers and $$x$$ is our variable. Let's take a look at how this would expand \begin{align} (x+a)(x+b) &= x^2 + ax + bx + ab \\[3pt]&= x^2 + (a+b)x + ab \end{align} Looking at the final form of this expansion, we can get an idea of how we're going to factorise our example! The $$x$$ term has a coefficient which is the sum of $$a$$ and $$b$$, while the non-$$x$$ term is the product of $$a$$ and $$b$$.
So we must ask ourselves "What numbers multiply to make $$6$$ and add to make $$5$$?". You may be able to work out in your head what the answer to this example is. However, a more foolproof method is to look at factor pairs of $$6$$, and which of those add to make $$5$$.
Remember negative factors! $\text{Factor pairs of }6\text{: }\{1,6\},\{2,3\},\{-1,-6\},\{-2,-3\}$ $2+3=5$ So our answer is $x^2+5x+6 = (x+2)(x+3) = 0$ $x = -2\hspace{6 pt}\text{or }-3$ We can check by expanding $(x+2)(x+3) = x^2+2x+3x+6 = x^2+5x+6 = 0$

## Full worked example: positive values

$x^2+8x+16=0$ $\text{Factor pairs of }16\text{: }$ $\{1,16\},\{2,8\},\{4,4\},\{-1,-16\},\{-2,-8\},\{-4,-4\}$ $4+4=8$ $(x+4)(x+4)=0$ $x = -4\hspace{10 pt}\text{(repeated root)}$

## Full worked example: negative values

$x^2-6x-7=0$ $\text{Factor pairs of }-7\text{: }\{-1,7\},\{1,-7\}$ $(-1)+7=6$ $(x-1)(x+7)=0$ $x = 1\hspace{6 pt}\text{or }-7$

## Full worked example: mixed values

$x^2-11x+28=0$ $\text{Factor pairs of }28\text{: }$ $\{1,28\},\{2,14\},\{4,7\},\{-1,-28\},\{-2,-14\},\{-4,-7\}$ $(-4)+(-7)=-11$ $(x-4)(x-7)=0$ $x = 4\hspace{6 pt}\text{or }7$