# How do you factor x^2+3x+12=0?

Mar 5, 2018

There are no Real factors of ${x}^{2} + 3 x + 12$

#### Explanation:

If you are willing to include Complex values in your solution
then you can use the quadratic formula to get
$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- 3 \pm \sqrt{{3}^{2} - \left(4 \cdot 1 \cdot 12\right)}}{2 \cdot 1} = \frac{- 3 \pm \sqrt{39} i}{2}$
for the factoring:
$\left(x + \frac{3 + \sqrt{39} i}{2}\right) \left(x + \frac{3 - \sqrt{39} i}{2}\right) = 0$

Mar 5, 2018

$\left[x + \frac{3}{2} + \frac{\sqrt{39} \textcolor{w h i t e}{. .} i}{2}\right] \textcolor{w h i t e}{. .} \left[x + \frac{3}{2} - \frac{\sqrt{39} \textcolor{w h i t e}{. .} i}{2}\right] = 0$

#### Explanation:

First of all there are no whole number factors of 12 that can be use to sum to 3 from the $3 x$. Consequently we are looking at fractional factors.

Lets check to see if the plot actually crosses the x-axis. If it does then the determinate part of $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ will be positive.

Determinate $\to {b}^{2} - 4 a c \textcolor{w h i t e}{\text{d") ->color(white)("d}} {3}^{2} - 4 \left(1\right) \left(12\right) = 9 - 48 < 0$

Thus the roots do not belong to the set of real numbers ( $\mathbb{R}$ )

They belong to the set of numbers called complex ( $\mathbb{C}$ )
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$x = \frac{- 3 \pm \sqrt{- 39}}{2} = - \frac{3}{2} \pm \frac{\left(\sqrt{39}\right) \textcolor{w h i t e}{. .} i}{2}$

$x = \left(- \frac{3}{2} - \frac{\left(\sqrt{39}\right) \textcolor{w h i t e}{. .} i}{2}\right)$

$x = \left(- \frac{3}{2} + \frac{\left(\sqrt{39}\right) \textcolor{w h i t e}{. .} i}{2}\right)$

$\left[x + \frac{3}{2} + \frac{\left(\sqrt{39}\right) \textcolor{w h i t e}{. .} i}{2}\right] \textcolor{w h i t e}{. .} \left[x + \frac{3}{2} - \frac{\left(\sqrt{39}\right) \textcolor{w h i t e}{. .} i}{2}\right] = 0$