# How do you factor x^3 + 10x^2 + 33x + 34?

Jul 8, 2016

${x}^{3} + 10 {x}^{2} + 33 x + 34 {=}^{=} \left({x}^{2} + 8 x + 17\right) \left(x + 2\right)$

#### Explanation:

As putting $x = - 2$, we get ${x}^{3} + 10 {x}^{2} + 33 x + 34 = {\left(- 2\right)}^{3} + 10 \times {2}^{2} + 33 \times \left(- 2\right) + 34 = - 8 + 40 - 66 + 34 = 0$. Hence, $- 2$ is a zero of the function and $\left(x + 2\right)$ is a factor of ${x}^{3} + 10 {x}^{2} + 33 x + 34$.

Now dividing the function by $\left(x + 2\right)$, as ${x}^{3} + 10 {x}^{2} + 33 x + 34 {=}^{=} {x}^{2} \left(x + 2\right) + 8 x \left(x + 2\right) + 17 \left(x + 2\right) = \left({x}^{2} + 8 x + 17\right) \left(x + 2\right)$, we get ${x}^{2} + 8 x + 17$.

Now discriminant of ${x}^{2} + 8 x + 17$ is ${8}^{2} - 4 \cdot 1 \cdot 17 = 64 - 68 = - 4$ is negative, as such it cannot be factorized in real factors.

And hence factors of the function are

${x}^{3} + 10 {x}^{2} + 33 x + 34 {=}^{=} \left({x}^{2} + 8 x + 17\right) \left(x + 2\right)$