# How do you list all possible roots and find all factors of 3x^2+2x+2?

Aug 29, 2016

This quadratic has zeros: $- \frac{1}{3} \pm \frac{\sqrt{5}}{3} i$ and hence can be factored as:

$3 {x}^{2} + 2 x + 2$

$= 3 \left(x + \frac{1}{3} - \frac{\sqrt{5}}{3} i\right) \left(x + \frac{1}{3} + \frac{\sqrt{5}}{3} i\right)$

$= \frac{1}{3} \left(3 x + 1 - \sqrt{5} i\right) \left(3 x + 1 + \sqrt{5} i\right)$

#### Explanation:

Given:

$3 {x}^{2} + 2 x + 2$

Note that this is in standard quadratic form $a {x}^{2} + b x + c$ with $a = 3$, $b = 2$ and $c = 2$.

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

$\Delta = {b}^{2} - 4 a c = {2}^{2} - \left(4 \cdot 3 \cdot 2\right) = 4 - 24 = - 20 = - {2}^{2} \cdot 5$

Since $\Delta < 0$, this quadratic has no Real zeros, but we can find the Complex zeros using the quadratic formula:

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

$= \frac{- b \pm \sqrt{\Delta}}{2 a}$

$= \frac{- 2 \pm \sqrt{- 20}}{6}$

$= \frac{- 2 \pm 2 \sqrt{5} i}{6}$

$= \frac{1}{3} \left(- 1 \pm \sqrt{5} i\right)$

That is:

${x}_{1} = \frac{1}{3} \left(- 1 + \sqrt{5} i\right)$

${x}_{2} = \frac{1}{3} \left(- 1 - \sqrt{5} i\right)$

Both $\left(x - {x}_{1}\right)$ and $\left(x - {x}_{2}\right)$ are factors of our quadratic, so also are:

$3 \left(x - {x}_{1}\right) = \left(3 x + 1 - \sqrt{5} i\right)$

$3 \left(x - {x}_{2}\right) = \left(3 x + 1 + \sqrt{5} i\right)$

If we multiply these together we will get a leading term $9 {x}^{2}$, so we need to divide by $3$ somewhere. So the factorisation can be expressed:

$3 {x}^{2} + 2 x + 2 = \frac{1}{3} \left(3 x + 1 - \sqrt{5} i\right) \left(3 x + 1 + \sqrt{5} i\right)$

Alternatively, if we prefer monic factors, we can write:

$3 {x}^{2} + 2 x + 2 = 3 \left(x + \frac{1}{3} - \frac{\sqrt{5}}{3} i\right) \left(x + \frac{1}{3} + \frac{\sqrt{5}}{3} i\right)$