How do you find the roots of x^3-5x^2-x+5=0?

Oct 13, 2016

Use factoring by grouping to get $x = - 1 , x = 1 , x = 5$.

Explanation:

${x}^{3} - 5 {x}^{2} - x + 5 = 0$

Use a method called "factoring by grouping", which only works when the polynomial has four terms.

$\textcolor{red}{{x}^{2}} \left(x - 5\right) - x + 5 = 0 \textcolor{w h i t e}{a a a}$Factor out an $\textcolor{red}{{x}^{2}}$ from the first two $\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a}$terms.

$\textcolor{red}{{x}^{2}} \left(x - 5\right) \textcolor{b l u e}{- 1} \left(x - 5\right) = 0 \textcolor{w h i t e}{a a a}$Factor out a $\textcolor{b l u e}{- 1}$ from the 2nd two $\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a}$ terms.

$\left(\textcolor{red}{{x}^{2}} \textcolor{b l u e}{- 1}\right) \left(x - 5\right) = 0 \textcolor{w h i t e}{a a a}$Regroup

$\left(x + 1\right) \left(x - 1\right) \left(x - 5\right) = 0 \textcolor{w h i t e}{a a a}$Factor the $\left({x}^{2} - 1\right)$ term using the $\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a}$ difference of squares formula
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a} \left({a}^{2} - {b}^{2}\right) = \left(a + b\right) \left(a - b\right)$

Set each factor equal to zero and solve.

$x + 1 = 0 , x - 1 = 0 , x - 5 = 0$

$x = - 1 , x = 1 , x = 5$