# What are the possible rational roots of x^3-2x^2+x+18=0 and then determine the rational roots?

May 7, 2017

The "possible" rational roots are:

$\pm 1 , \pm 2 , \pm 3 , \pm 6 , \pm 9 , \pm 18$

The actual roots are:

$- 2 \text{ }$ and $\text{ } 2 \pm \sqrt{5} i$

#### Explanation:

Given:

${x}^{3} - 2 {x}^{2} + x + 18 = 0$

By the rational roots theorem, any rational roots of this polynomial are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $18$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational roots are:

$\pm 1 , \pm 2 , \pm 3 , \pm 6 , \pm 9 , \pm 18$

With $x = - 2$, we find:

${x}^{3} - 2 {x}^{2} + x + 18 = {\left(\textcolor{b l u e}{- 2}\right)}^{2} - 2 {\left(\textcolor{b l u e}{- 2}\right)}^{2} + \left(\textcolor{b l u e}{- 2}\right) + 18$

${x}^{3} - 2 {x}^{2} + x + 18 = - 8 - 8 - 2 + 18$

${x}^{3} - 2 {x}^{2} + x + 18 = 0$

So $x = - 2$ is a root and $\left(x + 2\right)$ a factor:

${x}^{3} - 2 {x}^{2} + x + 18 = \left(x + 2\right) \left({x}^{2} - 4 x + 9\right)$

The remaining quadratic has no real zeros, which we can tell by examining its discriminant:

${x}^{2} - 4 x + 9$

is in the form:

$a {x}^{2} + b x + c$

with $a = 1$, $b = - 4$ and $c = 9$

Its discriminant $\Delta$ is given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(\textcolor{b l u e}{- 4}\right)}^{2} - 4 \left(\textcolor{b l u e}{1}\right) \left(\textcolor{b l u e}{9}\right) = 16 - 36 = - 20$

Since $\Delta < 0$ there are no real zeros and no linear factors with real (let alone rational) coefficients.

We can find the roots by completing the square:

$0 = {x}^{2} - 4 x + 9$

$\textcolor{w h i t e}{0} = {x}^{2} - 4 x + 4 + 5$

$\textcolor{w h i t e}{0} = {\left(x - 2\right)}^{2} - {\left(\sqrt{5} i\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(x - 2 - \sqrt{5} i\right) \left(x - 2 + \sqrt{5} i\right)$

Hence roots:

$x = 2 \pm \sqrt{5} i$