# What are all the possible rational zeros for f(x)=5x^3+x^2-5x-1 and how do you find all zeros?

Sep 24, 2016

The possible rational zeros are $- 1 , - \frac{1}{5} , \frac{1}{5} , 5$ and the zeros are $- 1 , - \frac{1}{5} , 1$

#### Explanation:

$f \left(x\right) = \textcolor{red}{5} {x}^{3} + {x}^{2} - 5 x - \textcolor{b l u e}{1}$

To find all the possible rational zeros, divide all the factors $p$ of the constant term by all the factors $q$ of the leading coefficient. The list of possible rational zeros is given by $\frac{p}{q}$.

The constant term $= \textcolor{b l u e}{1}$ and the leading coefficient $= \textcolor{red}{5}$.

The factors $p$ of the constant term $\textcolor{b l u e}{1}$ are $\pm 1$.
The factors $q$ of the leading coefficient $\textcolor{red}{5}$ are $\pm 1$ and $\pm 5$

$\frac{p}{q}$=$\frac{\pm 1}{\pm 1 , \pm 5} = 1 , - 1 , \frac{1}{5} , - \frac{1}{5}$

The possible rational zeros are $- 1 , - \frac{1}{5} , \frac{1}{5} , 1$

To find all the zeros, factor the polynomial and set each factor equal to zero.

$f \left(x\right) = 5 {x}^{3} + {x}^{2} - 5 x - 1$

Factor by grouping.
$\left(5 {x}^{3} + {x}^{2}\right) - \left(5 x + 1\right)$
${x}^{2} \left(5 x + 1\right) - 1 \left(5 x + 1\right)$
$\left({x}^{2} - 1\right) \left(5 x + 1\right)$

Use the difference of squares to factor ${x}^{2} - 1$
$\left(x + 1\right) \left(x - 1\right) \left(5 x + 1\right)$

Set each factor equal to zero and solve.
$x + 1 = 0 \textcolor{w h i t e}{a a a} x - 1 = 0 \textcolor{w h i t e}{a a a} 5 x + 1 = 0$

$x = - 1 \textcolor{w h i t e}{a a a} x = 1 \textcolor{w h i t e}{a a a} x = - \frac{1}{5}$