# How do you find the number of complex zeros for the function f(x)=3x^3+2x^2-x?

Oct 5, 2016

x_1=0; x_2=-1; x_3=1/3

#### Explanation:

$f \left(x\right)$ is a polynomial with degree 3, therefore we have 3 solutions real or complex.

To simplify the computation we could factorize $f \left(x\right)$

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

We have to find the zeros, then:

$x \left(3 {x}^{2} + 2 x - 1\right) = 0$

A product is zero when the factors are zero:

$x = 0 \implies {x}_{1} = 0$

$3 {x}^{2} + 2 x - 1 = 0$

${x}_{2 , 3} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
with: a=3; b=2; c=-1
$\therefore {x}_{2 , 3} = \frac{- 2 \pm \sqrt{4 + 12}}{6} = \frac{- 2 \pm \sqrt{16}}{6} = \frac{- 2 \pm 4}{6}$
$\implies {x}_{2} = - {\cancel{6}}^{1} / {\cancel{6}}_{1} = - 1$
$\implies {x}_{3} = {\cancel{2}}^{1} / {\cancel{6}}_{3} = \frac{1}{3}$