# How do you factor  a^2+a-12?

Jun 25, 2018

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

#### Explanation:

${a}_{1 , 2} = - \frac{1}{2} \pm \sqrt{\frac{1}{4} + \frac{48}{4}}$
${a}_{1 , 2} = - \frac{1}{2} \pm \frac{7}{2}$

we get

${a}_{1} = 3$

${a}_{2} = - 4$

so our factorization is given by

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

Jun 25, 2018

$\left(a + 4\right) \left(a - 3\right)$
If we factor it into linear brackets $\left(a - b\right) \left(a - c\right)$, then ${a}^{2} + a - 12 = {a}^{2} - \left(b + c\right) a + b c$. So $b c = - 12$ and $- \left(b + c\right) = 1$. So we're looking for values that are of opposite sign, multiply to -12, and whose absolute values are 1 apart. 4 and 3 is where we are looking, and specifically, -4 and +3 are the answers, which we find by a small amount of trial and error.
So ${a}^{2} + a - 12 = \left(a + 4\right) \left(a - 3\right)$.