# How do you factor completely: 3x^2 + 12x + 7?

Aug 15, 2015

It depends on what kinds of numbers we are using.

#### Explanation:

$3 {x}^{2} + 12 x + 7$

Using integers (or rational numbers) this quadratic cannot be factored any further. It is already factored completely.
(The only possibilities are $\left(3 x + 1\right) \left(x + 7\right)$ and $\left(3 x + 7\right) \left(x + 1\right)$, neither of which works.)

If we are using Real numbers or Complex numbers, we can factor the quadratic by finding its zeros.

Solve
$3 {x}^{2} + 12 x + 7 = 0$ using your choice of completing the square of the quadratic formula.

$x = \frac{- \left(12\right) \pm \sqrt{{\left(12\right)}^{2} - 4 \left(3\right) \left(7\right)}}{2 \left(3\right)}$

$= \frac{- 12 \pm \sqrt{144 - 84}}{6} = \frac{- 12 \pm \sqrt{60}}{6} = \frac{- 12 \pm 2 \sqrt{15}}{6}$

$= - 2 \pm \frac{\sqrt{15}}{3}$

The zeros are:

${z}_{1} = - 2 + \frac{\sqrt{15}}{3}$ and z_2 = -2 - sqrt 15/3

The factors are:

$\left(x - {z}_{1}\right) \left(x - {z}_{2}\right) = \left(x - \left(- 2 + \frac{\sqrt{15}}{3}\right)\right) \left(x - \left(- 2 - \frac{\sqrt{15}}{3}\right)\right)$

 = (x + 2 - sqrt 15/3))(x + 2 + sqrt 15/3))#