# How do you factor the expression x^2 + x - 12?

Jan 15, 2016

$\left(x - 3\right) \left(x + 4\right)$

$\textcolor{b l u e}{\text{Look at how I reasoned it out!}}$

#### Explanation:

You look to see if you spot the appropriate factors strait away. If you can not immediately determine them you have to consider all of the potential ones and reason out which ones work.

So for this question:

$\textcolor{b l u e}{\text{Starting point}}$

${x}^{2}$ can only be $x \times x$ so we have a starting point of:

color(blue)((x+?)(x+?))
'~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Consider the constant of } - 12}$

This is a product so one of the factors is negative and the other positive.

$1 \times 12 = 12$
$2 \times 6 = 12$
$3 \times 4 = 12$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Now consider the "+x". This is } 1 \times x}$

So the difference between the factors of 12 must be 1.
The only factors that give us this difference is 3 and 4.

We have $+ x$ so the larger has to be the positive.

$\textcolor{b l u e}{\text{That means we must have "-3" and } + 4}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Lets try it out!}}$

color(blue)((x-3)color(brown)((x+4))

$\textcolor{b l u e}{x \textcolor{b r o w n}{\left(x + 4\right)} - 3 \textcolor{b r o w n}{\left(x + 4\right)}}$

${x}^{2} + 4 x - 3 x - 12$

${x}^{2} + x - 12$

which is where we started from so these are the factors
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{m a \ge n t a}{\text{Factors of } {x}^{2} + x - 12 \to \left(x - 3\right) \left(x + 4\right)}$