# How do you use the rational root theorem to find the roots of x^4 + 3x^3 - x^2 - 9x - 6 = 0?

Dec 10, 2015

The roots are $- 1$, $- 2$, $\sqrt{3}$ and $- \sqrt{3}$

#### Explanation:

By the rational roots theorem, the possible rational roots of ${x}^{4} + 3 {x}^{3} - {x}^{2} - 9 x - 6 = 0$ are $\pm 1$, $\pm 2$, $\pm 3$, or $\pm 6$.

By trial (or division) $- 1$ is a root, so (by the factor theorem) $x - \left(- 1\right) = x + 1$ is a factor.

By division or trial and error,

${x}^{4} + 3 {x}^{3} - {x}^{2} - 9 x - 6 = \left(x + 1\right) \left({x}^{3} + 2 {x}^{2} - 3 x - 6\right)$

Possible rational roots of ${x}^{3} + 2 {x}^{2} - 3 x - 6 = 0$ are the same as for the original.

By trial, $1 , - 1 , 2$ are not roots, but $- 2$ is a root. So $x + 2$ is a factor. Division gets us

${x}^{3} + 2 {x}^{2} - 3 x - 6 = \left(x + 2\right) \left({x}^{2} - 3\right)$

The roots of ${x}^{2} - 3 = 0$ are $\pm \sqrt{3}$.