# How do you find the rest of the zeros given one of the zero c=-1 and the function f(x)=x^5+2x^4-12x^3-38x^2-37x-12?

Aug 1, 2016

The zeros of $f \left(x\right)$ are $x = - 1$ with multiplicity $3$, $x = 4$ and $x = - 3$.

#### Explanation:

$f \left(x\right) = {x}^{5} + 2 {x}^{4} - 12 {x}^{3} - 38 {x}^{2} - 37 x - 12$

$f \left(- 1\right) = - 1 + 2 + 12 - 38 + 37 - 12 = 0$

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

${x}^{5} + 2 {x}^{4} - 12 {x}^{3} - 38 {x}^{2} - 37 x - 12 = \left(x + 1\right) \left({x}^{4} + {x}^{3} - 13 {x}^{2} - 25 x - 12\right)$

Substituting $x = - 1$ in the remaining quartic factor, we get:

$1 - 1 - 13 + 25 - 12 = 0$

So the quartic also has $x = - 1$ as a zero and $\left(x + 1\right)$ as a factor:

${x}^{4} + {x}^{3} - 13 {x}^{2} - 25 x - 12 = \left(x + 1\right) \left({x}^{3} - 13 x - 12\right)$

Substituting $x = - 1$ in the remaining cubic factor, we get:

$- 1 + 13 - 12 = 0$

So the cubic also has $x = - 1$ as a zero and $\left(x + 1\right)$ as a factor:

${x}^{3} - 13 x - 12 = \left(x + 1\right) \left({x}^{2} - x - 12\right)$

To factor the remaining quadratic, note that $4 \cdot 3 = 12$ and $4 - 3 = 1$, so:

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

hence zeros $x = 4$ and $x = - 3$

So the zeros of $f \left(x\right)$ are $x = - 1$ with multiplicity $3$, $x = 4$ and $x = - 3$.