# How do you factor 3x^2 - 12x - 4?

May 11, 2016

Note that there are no "pretty" factors for this expression.
The best I could come up with is
$\textcolor{w h i t e}{\text{XXX}} 3 \left(x - 2 + \frac{2 \sqrt{12}}{3}\right) \left(x - 2 - \frac{2 \sqrt{12}}{3}\right)$

#### Explanation:

The rational root theorem reveals no rational roots
so we can apply the quadratic formula to find zeros for this expression.

For a quadratic of the form $a {x}^{2} + b x + c$
zeros occur at $x = \frac{- b \pm \sqrt{{b}^{2} + 4 a c}}{2 a}$

Applying this to the given equation results in
$\textcolor{w h i t e}{\text{XXX}} x = 2 \pm \frac{2 \sqrt{12}}{3}$
So
$\textcolor{w h i t e}{\text{XXX}} \left(x - \left(2 + \frac{2 \sqrt{12}}{3}\right)\right) \mathmr{and} \left(x - \left(2 - \frac{2 \sqrt{12}}{3}\right)\right)$ are both factors of the given expression.

Dividing the original expression by these factors leaves an additional factor of $3$
So the complete factorization is
$\textcolor{w h i t e}{\text{XXX}} 3 \left(x - 2 + \frac{2 \sqrt{12}}{3}\right) \left(x - 2 - \frac{2 \sqrt{12}}{3}\right)$

May 11, 2016

$3 \left(x - 2 - \frac{4 \sqrt{3}}{3}\right) \left(x - 2 + \frac{4 \sqrt{3}}{3}\right)$

#### Explanation:

Use the improved quadratic formula (Socratic Search)
$D = {d}^{2} = {b}^{2} - 4 a c = 144 + 48 = 192 = 64 \left(3\right)$ --> $d = \pm 8 \sqrt{3}$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{12}{6} \pm \frac{8 \sqrt{3}}{6} = 2 \pm \frac{4 \sqrt{3}}{3}$
$x 1 = 2 + \frac{4 \sqrt{3}}{3}$ and $x 2 = 2 - \frac{4 \sqrt{3}}{3}$
Factored form:
$y = a \left(x - x 1\right) \left(x - x 2\right) = 3 \left(x - 2 - 4 \frac{\sqrt{3}}{3}\right) \left(x - 2 + 4 \frac{\sqrt{3}}{3}\right)$