# How do you find the zeroes of f(x)=x^4+x^3+x^2-9x-10?

Jul 23, 2015

Use the rational root theorem and the quadratic formula to find roots:

$x = - 1$, $x = 2$, $x = - 1 \pm 2 i$

#### Explanation:

By the rational root theorem any rational roots of $f \left(x\right) = 0$ must be of the form $\frac{p}{q}$ where $p$ and $q$ are integers, $q \ne 0$, $p$ a divisor of the constant term $- 10$ and $q$ a divisor of the coefficient $1$ of the term of highest degree ${x}^{4}$.

So the only possible rational roots are:

$\pm 1$, $\pm 2$, $\pm 5$, $\pm 10$

We find:

$f \left(1\right) = 1 + 1 + 1 - 9 - 10 = - 16$

$f \left(- 1\right) = 1 - 1 + 1 + 9 - 10 = 0$

$f \left(2\right) = 16 + 8 + 4 - 18 - 10 = 0$

So at least $x = - 1$ and $x = 2$ are roots of $f \left(x\right) = 0$ and $\left(x + 1\right)$ and $\left(x - 2\right)$ are factors of $f \left(x\right)$.

Divide $f \left(x\right)$ by $\left(x + 1\right) \left(x - 2\right) = {x}^{2} - x - 2$ to find:

$f \left(x\right) = {x}^{4} + {x}^{3} + {x}^{2} - 9 x - 10$

$= \left({x}^{2} - x - 2\right) \left({x}^{2} + 2 x + 5\right)$

Using the quadratic formula, the roots of ${x}^{2} + 2 x + 5 = 0$ are:

$x = \frac{- 2 \pm \sqrt{{2}^{2} - \left(4 \times 1 \times 5\right)}}{2 \cdot 1}$

$= \frac{- 2 \pm \sqrt{4 - 20}}{2}$

$= \frac{- 2 \pm \sqrt{- 16}}{2}$

$= - 1 \pm 2 i$