# What are the possible rational roots of x^3-5x^2-4x+20=0 and then determine the rational roots?

Dec 12, 2016

The possible rational roots are $\pm 1 , \pm 2 , \pm 4 , \pm 5 , \pm 10 , \pm 20$. The rational roots are $x = - 2 , x = 2 , x = 5$.

#### Explanation:

$\textcolor{b l u e}{1} {x}^{3} - 5 {x}^{2} - 4 x + \textcolor{red}{20} = 0$

The possible rational roots are the factors of the constant $\textcolor{red}{20}$ divided by the factors of the leading coefficient $\textcolor{b l u e}{1}$. The factors of the constant are called $p$ and the factors of the leading coefficient are called $q$.

$\frac{p}{q} = \frac{\pm 1 , \pm 2 , \pm 4 , \pm 5 , \pm 10 , \pm 20}{\pm 1} =$

$\pm 1 , \pm 2 , \pm 4 , \pm 5 , \pm 10 , \pm 20$

The rational roots of this particular function can be found by factoring.

Factor by grouping.

First, group the first two terms and the second two terms.

$\left({x}^{3} - 5 {x}^{2}\right) \textcolor{w h i t e}{a} + \left(- 4 x + 20\right) = 0$

Factor out a GCF from each group.

${x}^{2} \left(x - 5\right) - 4 \left(x - 5\right) = 0$

Regroup.

$\left({x}^{2} - 4\right) \left(x - 5\right) = 0$

Factor ${x}^{2} - 4$ as the difference of squares.

$\left(x + 2\right) \left(x - 2\right) \left(x - 5\right) = 0$

Set each factor equal to zero and solve.

$x + 2 = 0 \textcolor{w h i t e}{a a} x - 2 = 0 \textcolor{w h i t e}{a a} x - 5 = 0$

$x = - 2 \textcolor{w h i t e}{a a a} x = 2 \textcolor{w h i t e}{a a a} x = 5$