# How do you find all the zeros of P(X) = x^3 - 6x^2 + 4x + 16?

Mar 26, 2016

the roots or zeros are: ${x}_{1 , 2} = 1 \pm \sqrt{5} \mathmr{and} x = 4$ graph{x^3-6x^2+4x+16 [-2.603, 6.86, -1.507, 3.23]}

#### Explanation:

Using Zero Rational Theorem you see that the potential roots $\frac{p}{q}$ are $\left\{1 , 2 , 4 , 8 , 16\right\}$.
First root that we will try will $x - 4$
$\frac{{x}^{3} - 6 {x}^{2} + 4 x + 16}{x - 4}$ apply long division

4 |bar(1 - 6 + 4 + 16
$\text{ } | 1 + 4 - 8 - 16$
$\text{ } | 1 - 2 - 4 - 0$

Thus $\left(x - 4\right)$ divides our polynomial without remainder and $\therefore$ it is a factor.
$\left({x}^{3} - 6 {x}^{2} + 4 x + 16\right) = \textcolor{b r o w n}{\left({x}^{2} - 2 x - 4\right)} \left(x - 4\right)$
Now factor the binomial $\textcolor{b r o w n}{{x}^{2} - 2 x - 4}$
${x}_{1 , 2} = - \frac{b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{2 \pm \sqrt{{\left(- 2\right)}^{2} - 4 \left(1\right) \left(- 4\right)}}{2}$
${x}_{1 , 2} = \frac{2 \pm \sqrt{4 + 16}}{2} = \frac{2 \pm 2 \sqrt{5}}{2} = 1 \pm \sqrt{5}$
So the roots or zeros are: ${x}_{1 , 2} = 1 \pm \sqrt{5} \mathmr{and} x = 4$