# What are all the zeroes of f(x)= x^3 +2x^2 -3x +20 ?

Nov 28, 2015

$x = - 4$

$x = 1 + 2 i$

$x = 1 - 2 i$

#### Explanation:

By the rational root theorem, any rational zeros of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ in lowest terms, where $p , q \in \mathbb{Z}$, $p$ a divisor of the constant term $20$ and $q$ a divisor of the coefficient $1$ of the leading term.

So the possible rational roots are the divisors of $20$:

$\pm 1$, $\pm 2$, $\pm 4$, $\pm 5$, $\pm 10$, $\pm 20$

After a bit of trial and error, find:

$f \left(- 4\right) = - 64 + 32 + 12 + 20 = 0$

so $x = - 4$ is a zero and $\left(x + 4\right)$ is a factor of $f \left(x\right)$.

${x}^{3} + 2 {x}^{2} - 3 x + 20 = \left(x + 4\right) \left({x}^{2} - 2 x + 5\right)$

Find the Complex zeros of the remaining quadratic factor using the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{- 16}}{2}$

$= \frac{2 \pm 4 i}{2} = 1 \pm 2 i$