# How do you find the zeroes for -2x^6 + 2x^5 - 4x^4 - 44x^3 + 120x^2?

May 30, 2015

$- 2 {x}^{6} + 2 {x}^{5} - 4 {x}^{4} - 44 {x}^{3} + 120 {x}^{2}$

$= - 2 {x}^{2} \left({x}^{4} - {x}^{3} + 2 {x}^{2} + 22 x - 60\right)$

The $- 2 {x}^{2}$ factor means that $\textcolor{red}{x = 0}$ is a zero of the polynomial.

Using the rational root theorem, rational roots of

${x}^{4} - {x}^{3} + 2 {x}^{2} + 22 x - 60 = 0$

must be factors of $60$

So possible roots to try are $\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 5$, $\pm 6$, $\pm 10$, $\pm 15$, $\pm 20$, $\pm 30$, $\pm 60$.

It's easy to see that $\pm 1$ are not roots. How about $2$?

If $x = 2$

${x}^{4} - {x}^{3} + 2 {x}^{2} + 22 x - 60$

$= {2}^{4} - {2}^{3} + 2 \cdot {2}^{2} + 22 \cdot 2 - 60$

$= 16 - 8 + 8 + 44 - 60 = 0$

So $\textcolor{red}{x = 2}$ is another zero of the polynomial.

So $\left(x - 2\right)$ is a factor of ${x}^{4} - {x}^{3} + 2 {x}^{2} + 22 x - 60$

${x}^{4} - {x}^{3} + 2 {x}^{2} + 22 x - 60$

$= \left(x - 2\right) \left({x}^{3} + {x}^{2} + 4 x + 30\right)$

Rational roots of ${x}^{3} + {x}^{2} + 4 x + 30 = 0$ must be factors of $30$ and negative, so $- 1$, $- 2$, $- 3$, $- 5$, $- 6$, $- 10$, $- 15$ or $- 30$.

We have already eliminated $- 1$ as a possibility.
Trying $- 2$ does not work.
How about $- 3$?

${\left(- 3\right)}^{3} + {\left(- 3\right)}^{2} + 4 \left(- 3\right) + 30$

$= - 27 + 9 - 12 + 30 = 0$

So $\textcolor{red}{x = - 3}$ is a zero and $\left(x + 3\right)$ is a factor.

${x}^{3} + {x}^{2} + 4 x + 30 = \left(x + 3\right) \left({x}^{2} - 2 x + 10\right)$

The discriminant of ${x}^{2} - 2 x + 10$ is:

$\Delta = {\left(- 2\right)}^{2} - \left(4 \times 1 \times 10\right) = 4 - 40 = - 36 < 0$

So ${x}^{2} - 2 x + 10 = 0$ has no real roots.

So the only real roots of the original polynomial are $x = 0$, $x = 2$ and $x = - 3$.