# Question #ec66c

Oct 26, 2016

On $R$ you can't.

#### Explanation:

If you can write it as product of 1 degree factors it should be
$P \left(x\right) = a \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$
where ${x}_{1}$ and ${x}_{2}$ are roots of $P \left(x\right)$, but if you try to solve the second degrre equation you will find
$\frac{\Delta}{4} = {8}^{2} - 3 \cdot 24 = - 8 < 0$
so there are no roots on $R$.

Oct 26, 2016

Complete the square or use the quadratic formula to find the roots, then find the factors.

#### Explanation:

$3 {x}^{2} - 16 x + 24 = 0$ has solutions

$x = \frac{8 \pm 2 \sqrt{2} i}{3}$

So

$x - \frac{8 + 2 \sqrt{2} i}{3}$, and $x - \frac{8 - 2 \sqrt{2} i}{3}$ are factors of the quadratic.

In order to get leading coefficient $3$, we need a constant factor of $3$.

So

$3 {x}^{2} - 16 x + 24 = 3 \left(x - \frac{8 + 2 \sqrt{2} i}{3}\right) \left(x - \frac{8 - 2 \sqrt{2} i}{3}\right)$

If you want strict standard for for the imaginaries, write

$3 {x}^{2} - 16 x + 24 = 3 \left(x - \left(\frac{8}{3} + \frac{2 \sqrt{2}}{3} i\right)\right) \left(x - \left(\frac{8}{3} - \frac{2 \sqrt{2}}{3} i\right)\right)$