# What are all the possible rational zeros for f(x)=5x^3+29x^2+19x-5 and how do you find all zeros?

Oct 24, 2016

All the rational zeros arefor $x = - 1$ , $x = - 5$ and $x = \frac{1}{5}$

#### Explanation:

By trial and error, we can see that

$f \left(- 1\right) = - 5 + 29 - 19 - 5 = 0$
So $\left(x + 1\right)$ is a factor
So we do a long division to find the other factors
$5 {x}^{3} + 29 {x}^{2} + 19 x - 5$$\textcolor{w h i t e}{a a a a a a}$∣x+1
$5 {x}^{3} + 5 {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a a a a a}$∣$5 {x}^{2} + 24 x - 5$
$\textcolor{w h i t e}{a}$ $0 + 24 {x}^{2}$
$\textcolor{w h i t e}{a a a a a}$ $24 {x}^{2} + 19 x$
$\textcolor{w h i t e}{a a a a a a a}$ $0 + 24 x$
$\textcolor{w h i t e}{a a a a a a a a a a a}$ $- 5 x - 5$
$\textcolor{w h i t e}{a a a a a a a a a a a}$ $- 5 x - 5$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a}$ $0 + 0$
Now we have to factorize
$5 {x}^{2} + 24 x - 5 = \left(5 x - 1\right) \left(x + 5\right)$

So all the rational zeros arefor $x = - 1$ , $x = - 5$ and $x = \frac{1}{5}$