# How do you find the roots of x^3+x^2-7x+2=0?

Nov 11, 2016

The roots are $2$ and $- \frac{3}{2} \pm \frac{\sqrt{13}}{2}$

#### Explanation:

Given:

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

By the rational roots theorem, any rational zeros of $f \left(x\right)$ 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 zeros are:

$\pm 1 , \pm 2$

Note that:

$f \left(2\right) = 8 + 4 - 14 + 2 = 0$

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

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

We can find the zeros of the remaining quadratic factor using the quadratic formula with $a = 1$, $b = 3$ and $c = - 1$ :

$x = \frac{- \textcolor{b l u e}{3} \pm \sqrt{{\textcolor{b l u e}{3}}^{2} - 4 \left(\textcolor{b l u e}{1}\right) \left(\textcolor{b l u e}{- 1}\right)}}{2 \left(\textcolor{b l u e}{1}\right)}$

$\textcolor{w h i t e}{x} = \frac{- 3 \pm \sqrt{9 + 4}}{2}$

$\textcolor{w h i t e}{x} = - \frac{3}{2} \pm \frac{\sqrt{13}}{2}$