# How do you solve 4x^3 + 3x^2 + x + 2 = 0?

Apr 29, 2017

$x = - 1 \text{ }$ or $\text{ } x = \frac{1}{8} \pm \frac{\sqrt{31}}{8} i$

#### Explanation:

Given:

$4 {x}^{3} + 3 {x}^{2} + x + 2 = 0$

Note that:

$- 4 + 3 - 1 + 2 = 0$

So we can deduce that $x = - 1$ is a solution and $\left(x + 1\right)$ a factor:

$0 = 4 {x}^{3} + 3 {x}^{2} + x + 2$

$\textcolor{w h i t e}{0} = \left(x + 1\right) \left(4 {x}^{2} - x + 2\right)$

The remaining quadratic is in the form $a {x}^{2} + b x + c$ with $a = 4$, $b = - 1$ and $c = 2$

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(\textcolor{b l u e}{- 1}\right)}^{2} - 4 \left(\textcolor{b l u e}{4}\right) \left(\textcolor{b l u e}{2}\right) = 1 - 32 = - 31$

Since $\Delta < 0$ this quadratic has no real zeros.

We can find Complex zeros for it using the quadratic formula:

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

$\textcolor{w h i t e}{x} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{1 \pm \sqrt{- 31}}{8}$

$\textcolor{w h i t e}{x} = \frac{1}{8} \pm \frac{\sqrt{31}}{8} i$

where $i$ is the imaginary unit.