# How do you use the rational root theorem to find the roots of x^4 + x^3 -2x^2 + 0x -290 = 0?

Aug 4, 2018

The "possible" rational roots are:

$\pm 1 , \pm 2 , \pm 5 , \pm 10 , \pm 29 , \pm 58 , \pm 145 , \pm 290$

but this quartic has no rational roots.

#### Explanation:

Given:

${x}^{4} + {x}^{3} - 2 {x}^{2} + 0 x - 290 = 0$

Assuming there's no typographical error in the question, this can be written:

${x}^{4} + {x}^{3} - 2 {x}^{2} - 290 = 0$

By the rational roots theorem, any rational roots of this quartic are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 290$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational roots are:

$\pm 1 , \pm 2 , \pm 5 , \pm 10 , \pm 29 , \pm 58 , \pm 145 , \pm 290$

${\left(\textcolor{b l u e}{5}\right)}^{4} = 629 > 465 = {\left(\textcolor{b l u e}{5}\right)}^{3} + 2 {\left(\textcolor{b l u e}{5}\right)}^{2} + 290$

Hence there are no zeros with $\left\mid x \right\mid \ge 5$.

Note also that:

${\left(\textcolor{b l u e}{2}\right)}^{4} + {\left(\textcolor{b l u e}{2}\right)}^{3} + 2 {\left(\textcolor{b l u e}{2}\right)}^{2} = 32 < 290$

Hence there are no zeros with $\left\mid x \right\mid \le 2$.

So the given quartic has no rational roots.

It has one negative irrational root in $\left(- 5 , - 2\right)$, one positive irrational root in $\left(2 , 5\right)$ and two non-real complex roots.

graph{x^4+x^3-2x^2-290 [-10, 10, -500, 500]}