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

Nov 22, 2016

All possible rational zeros are $- 3$, $- 1$ and $1$.

#### Explanation:

If $a$, $b$ and $c$ are zeros of a polynomial of degree $3$ in $x$, then the function is $\left(x - a\right) \left(x - b\right) \left(x - c\right)$. Hence let us factorize ${x}^{3} + 3 {x}^{2} - x - 3$

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

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

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

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

Hence, all possible rational zeros are $- 3$, $- 1$ and $1$.