# How do you factor  x^3+8x^2+19x+12?

Apr 27, 2016

#### Answer:

${x}^{3} + 8 {x}^{2} + 19 x + 12 = \left(x + 1\right) \left(x + 3\right) \left(x + 4\right)$

#### Explanation:

$f \left(x\right) = {x}^{3} + 8 {x}^{2} + 19 x + 12$

Note that the coefficients of $f \left(- x\right) = - {x}^{3} + 8 {x}^{2} - 19 x + 12$ add up to $0$.

So $f \left(- 1\right) = 0$ and $\left(x + 1\right)$ is a factor:

${x}^{3} + 8 {x}^{2} + 19 x + 12 = \left(x + 1\right) \left({x}^{2} + 7 x + 12\right)$

Then notice that $3 + 4 = 7$ and $3 \times 4$=12#

Hence:

${x}^{2} + 7 x + 12 = \left(x + 3\right) \left(x + 4\right)$

Putting it all together:

${x}^{3} + 8 {x}^{2} + 19 x + 12 = \left(x + 1\right) \left(x + 3\right) \left(x + 4\right)$