# What are all the possible rational zeros for f(x)=x^3+2x^2-5x-6 and how do you find all zeros?

Aug 22, 2016

Zeros are $\left\{- 1 , 2 , - 3\right\}$

#### Explanation:

As coefficient of highest power of $x$ is $1$, the product of all the zeros will be $6$.

A trial indicates that $1$ is not a zero, but $- 1$ is as (-1)^3+2×(-1)^2-5(-1)-6=-1+2+5-6=0. Hence one factor of ${x}^{3} + 2 {x}^{2} - 5 x - 6$ is $\left(x + 1\right)$.

Dividing former by latter

${x}^{3} + 2 {x}^{2} - 5 x - 6$

= ${x}^{2} \left(x + 1\right) + x \left(x + 1\right) - 6 \left(x + 1\right)$

= $\left(x + 1\right) \left({x}^{2} + x - 6\right)$

We can further factorize second by splitting middle term as follows

$\left(x + 1\right) \left({x}^{2} + x - 6\right)$

= $\left(x + 1\right) \left({x}^{2} + 3 x - 2 x - 6\right)$

= $\left(x + 1\right) \left(x \left(x + 3\right) - 2 \left(x + 3\right)\right)$

= $\left(x + 1\right) \left(x - 2\right) \left(x + 3\right)$

Hence, Zeros are $\left\{- 1 , 2 , - 3\right\}$