# What are the possible rational roots of 2x^4-x^3-6x+3=0 and then determine the rational roots?

Dec 30, 2016

The "possible" rational roots are:

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

The only actual rational root is:

$\frac{1}{2}$

The other roots are the three cube roots of $3$.

#### Explanation:

Given:

$2 {x}^{4} - {x}^{3} - 6 x + 3 = 0$

By the rational roots theorem, any rational zeros of this quartic are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $6$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational roots are:

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

Note that the ratio of the first and second terms is the same as that between the third and fourth terms. So this quartic will factor by grouping:

$0 = 2 {x}^{4} - {x}^{3} - 6 x + 3$

$\textcolor{w h i t e}{0} = \left(2 {x}^{4} - {x}^{3}\right) - \left(6 x - 3\right)$

$\textcolor{w h i t e}{0} = {x}^{3} \left(2 x - 1\right) - 3 \left(2 x - 1\right)$

$\textcolor{w h i t e}{0} = \left({x}^{3} - 3\right) \left(2 x - 1\right)$

Hence the only rational zero is $x = \frac{1}{2}$ and the remaining three zeros are cube roots of $3$:

$x = \sqrt{3}$

$x = \omega \sqrt{3}$

$x = {\omega}^{2} \sqrt{3}$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.