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

Sep 22, 2015

Use the rational zero theorem to find one root. Then factor (or divide) and find the other roots.

#### Explanation:

Possible rational roots: $\pm 1 , \pm 2 , \pm \frac{1}{2} , \pm \frac{1}{4}$

$1$ is a root, so $x - 1$ is a factor on the polynomial.
By division or by stepwise analysis,

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

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

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

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

The other two roots are roots of $4 {x}^{2} + x + 2 = 0$, which may be found by completing the square or by the quadratic formula.

The roots of the original equation are:

$1 , \frac{- 1 + \sqrt{31} i}{8} , \frac{- 1 - \sqrt{31} i}{8}$