# How do you find all the real and complex roots of 3x^4+8x^3+6x^2+3x-2?

Jun 2, 2016

$x = \frac{1}{3}$, $x = - 2$, $x = - \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$

#### Explanation:

$f \left(x\right) = 3 {x}^{4} + 8 {x}^{3} + 6 {x}^{2} + 3 x - 2$

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 hte constant term $- 2$ and $q$ a divisor of the coefficient $3$ of the leading term.

So the only possible rational zeros are:

$\pm \frac{1}{3}$, $\pm \frac{2}{3}$, $\pm 1$, $\pm 2$

We find:

$f \left(\frac{1}{3}\right) = \frac{1}{27} + \frac{8}{27} + \frac{2}{3} + 1 - 2 = 0$

$f \left(- 2\right) = 48 - 64 + 24 - 6 - 2 = 0$

So $x = \frac{1}{3}$ and $x = - 2$ are zeros and $\left(3 x - 1\right)$ and $\left(x + 2\right)$ are factors:

$3 {x}^{4} + 8 {x}^{3} + 6 {x}^{2} + 3 x - 2$

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

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

The zeros of ${x}^{2} + x + 1$ are the Complex cube roots of $1$, since $\left(x - 1\right) \left({x}^{2} + x + 1\right) = {x}^{3} - 1$

We can find them using the quadratic formula or whatever your preferred method is to get:

$x = - \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$