# How do you use the rational root theorem to find the roots of 2x^4 -7x^3 -35x^2 +13x +3?

Sep 2, 2015

Use rational root theorem to find possible rational roots and try them to find rational roots $- 3$ and $\frac{1}{2}$.

Then divide $f \left(x\right)$ by $\left(x + 3\right) \left(2 x - 1\right)$ and solve to get roots:

$3 \pm \sqrt{10}$

#### Explanation:

Let $f \left(x\right) = 2 {x}^{4} - 7 {x}^{3} - 35 {x}^{2} + 13 x + 3$

By the rational root theorem, any rational roots of $f \left(x\right) = 0$ must be expressible as $\frac{p}{q}$ where $p , q \in \mathbb{Z}$, $q > 0$, $\text{hcf} \left(p , q\right) = 1$, $p | 3$, $q | 2$

So the only possible rational roots are:

$\pm 3$, $\pm \frac{3}{2}$, $\pm 1$, $\pm \frac{1}{2}$

$f \left(- 3\right) = 162 + 189 - 315 - 39 + 3 = 0$

$f \left(\frac{1}{2}\right) = \frac{1}{8} - \frac{7}{8} - \frac{35}{4} + \frac{13}{2} + 3$

$= \frac{1 - 7 - 70 + 52 + 24}{8} = 0$

So $x = - 3$ and $x = \frac{1}{2}$ are roots and $f \left(x\right)$ is divisible by

$\left(x + 3\right) \left(2 x - 1\right) = 2 {x}^{2} + 5 x - 3$

Use synthetic division to find:

$2 {x}^{4} - 7 {x}^{3} - 35 {x}^{2} + 13 x + 3$

$= \left(2 {x}^{2} + 5 x - 3\right) \left({x}^{2} - 6 x - 1\right)$

The remaining quadratic factor $\left({x}^{2} - 6 x - 1\right)$ has irrational roots given by the quadratic formula as:

$x = \frac{6 \pm \sqrt{36 + 4}}{2} = 3 \pm \sqrt{10}$