# How do you find all the zeros of f(x)=x^3+x^2-7x+2?

Apr 16, 2016

$f \left(x\right)$ has zeros $x = 2$ and $x = - \frac{3}{2} \pm \frac{\sqrt{13}}{2}$

#### Explanation:

By the rational root theorem, any rational zeros of $f \left(x\right)$ must be 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:

$\pm 1$, $\pm 2$

We find:

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

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 factor the remaining quadratic expression by completing the square. I will multiply by $4$ first to cut down on the fractions involved:

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

$= 4 {x}^{2} + 12 x - 4$

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

$= {\left(2 x + 3\right)}^{2} - {\left(\sqrt{13}\right)}^{2}$

$= \left(\left(2 x + 3\right) - \sqrt{13}\right) \left(\left(2 x + 3\right) + \sqrt{13}\right)$

$= \left(2 x + 3 - \sqrt{13}\right) \left(2 x + 3 + \sqrt{13}\right)$

Hence $x = - \frac{3}{2} \pm \frac{\sqrt{13}}{2}$