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

Apr 2, 2017

"Possible" rational roots: $\pm 1$, $\pm 2$

Rational root: $1$

Irrational roots: $\frac{3}{2} \pm \frac{\sqrt{17}}{2}$

#### Explanation:

Given:

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

By the rational roots theorem, any rational roots of this cubic are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $2$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational roots are:

$\pm 1$, $\pm 2$

In addition, notice that the sum of the coefficients is zero. That is:

$1 - 4 + 1 + 2 = 0$

Hence we can tell that $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

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

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

The remaining roots can be found using the quadratic formula.

$x = \frac{3 \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(1\right) \left(- 2\right)}}{2 \cdot 1} = \frac{3}{2} \pm \frac{\sqrt{17}}{2}$