# How do you find the factors of f(z) over C if f(z)=2z^3+3z^2-14z-15?

Sep 17, 2016

$\left(z + 1\right) \left(z + 3\right) \left(2 z - 5\right)$.

#### Explanation:

We observe that,

The Sum of the co-effs. of odd powered terms of $z$ is

$2 - 14 = - 12$,

and, that of even powered, $3 - 15 = - 12$

We conclude that $\left(z + 1\right)$ is a factor of $f \left(z\right)$.

$\text{Now, } f \left(z\right) = 2 {z}^{3} + 3 {z}^{2} - 14 z - 15$

$= \underline{2 {z}^{3} + 2 {z}^{2}} + \underline{{z}^{2} + z} - \underline{15 z - 15}$

$= 2 {z}^{2} \left(z + 1\right) + z \left(z + 1\right) - 15 \left(z + 1\right)$

$= \left(z + 1\right) \left(2 {z}^{2} + z - 15\right)$

$= \left(z + 1\right) \left\{\underline{2 {z}^{2} + 6 z} - \underline{5 z - 15}\right\}$

$= \left(z + 1\right) \left\{2 z \left(z + 3\right) - 5 \left(z + 3\right)\right\}$

$= \left(z + 1\right) \left(z + 3\right) \left(2 z - 5\right)$.