How do you factor completely  x^2 + 8x+ 12?

May 27, 2016

$x = - 2$ or $x = - 6$

Explanation:

To solve ${x}^{2} + 8 x + 12 = 0$, we should first factorize the polynomial ${x}^{2} + 8 x + 12$

To factorize a quadratic polynomial of type $a {x}^{2} + b x + c$,

one needs to split middle term $b$ in two parts whose product is $a c$. As in ${x}^{2} + 8 x + 12$, the product is $1 \cdot 12 = 12$ and middle term $8$, it can be split in $6$ and $2$.

Hence, ${x}^{2} + 8 x + 12 = 0$ can be written as

${x}^{2} + 6 x + 2 x + 12 = 0$

or $x \left(x + 6\right) + 2 \left(x + 6\right) = 0$

or $\left(x + 2\right) \left(x + 6\right) = 0$

Hence, either $x + 2 = 0$ or $x + 6 = 0$ i.e.

$x = - 2$ or $x = - 6$