# How do you use the rational root theorem to find the roots of 2x^4 + 3x^3 - 7x^2 + 3x - 9 = 0 ?

May 23, 2015

Any rational root $\frac{p}{q}$ - if written in lowest terms, so that $p$ and $q$ have no common factor other than 1, satisfies the property that $p$ is a divisor of the constant term $- 9$ and $q$ is a divisor of the coefficient $2$ of the highest order term ($2 {x}^{4}$).

So the only possible rational roots are:

$\pm \frac{1}{2}$, $\pm 1$, $\pm \frac{3}{2}$, $\pm 3$, $\pm \frac{9}{2}$ or $\pm 9$.

Rustling up a quick spreadsheet to help, I found that $- 3$ and $\frac{3}{2}$ are roots, hence $\left(x + 3\right)$ and $\left(2 x - 3\right)$ are factors of $2 {x}^{4} + 3 {x}^{3} - 7 {x}^{2} + 3 x - 9$.

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

Use synthetic division to find:

$2 {x}^{4} + 3 {x}^{3} - 7 {x}^{2} + 3 x - 9 = \left(2 {x}^{2} + 3 x - 9\right) \left({x}^{2} + 1\right)$

So the other 2 roots of $2 {x}^{4} + 3 {x}^{3} - 7 {x}^{2} + 3 x - 9 = 0$ are $\pm i$