# How do you find all the zeros of P(x) = x^3 + 5x^2 - 2x - 24 where one zero is 2?

Aug 7, 2016

$P \left(x\right)$ has zeros $2$, $- 3$ and $- 4$.

#### Explanation:

$P \left(x\right) = {x}^{3} + 5 {x}^{2} - 2 x - 24$

SInce we are told that $x = 2$ is one of the zeros, $\left(x - 2\right)$ must be a factor:

${x}^{3} + 5 {x}^{2} - 2 x - 24$

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

The remaining quadratic can be factored by noticing that $3 + 4 = 7$ and $3 \times 4 = 12$. So:

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

Hence the other zeros are $x = - 3$ and $x = - 4$