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

Sep 28, 2016

The possible rational zeros are $1 , - 1 , 5 , - 5$.
The zeros are $x = - 5 , x = 2 + \sqrt{3} , x = 2 - \sqrt{3}$.

#### Explanation:

$f \left(x\right) = \textcolor{b l u e}{1} {x}^{3} + {x}^{2} - 19 x + \textcolor{red}{5}$

The possible rational zeros are the factors of the constant term $\textcolor{red}{5}$ divided by the factors of the leading coefficient $\textcolor{b l u e}{1}$.

$x = \frac{\pm 1 , \pm 5}{\pm 1} = 1 , - 1 , 5 , - 5$

To find all the zeros, choose one of the possible zeros and use synthetic division. If the remainder is zero, you have verified that the possible zero is in fact a zero.

Choosing $- 5$ as a possible zero:

$\textcolor{w h i t e}{-} - - - - - - - -$
-5]color(white)a1color(white)(aaaa)1color(white)(aa)-19color(white)(ax^111)5
$\textcolor{w h i t e}{{a}^{1} {a}^{1}} \downarrow \textcolor{w h i t e}{1} - 5 \textcolor{w h i t e}{{a}^{1} +} 20 \textcolor{w h i t e}{{x}^{1}} - 5$
$\textcolor{w h i t e}{-} - - - - - - - -$
$\textcolor{w h i t e}{{a}^{1} {a}^{11}} 1 \textcolor{w h i t e}{{x}^{2}} - 4 \textcolor{w h i t e}{a a a a a} 1 \textcolor{w h i t e}{a a {x}^{11}} 0$

The remainder is zero, so $- 5$ is confirmed as a zero.

The coefficients of the result of synthetic division give
${x}^{2} - 4 x + 1$. Because this is not factorable, use the quadratic formula to find the last two zeros.

$x = \frac{4 \pm \sqrt{{\left(- 4\right)}^{2} - 4 \cdot 1 \cdot 1}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2 \sqrt{3}}{2}$

$x = 2 \pm \sqrt{3}$

The zeros are $x = - 5 , x = 2 + \sqrt{3} , x = 2 - \sqrt{3}$