# How do you solve 24x^3+18x^2-168x = 0?

Jun 11, 2015

First off, notice that you can divide out $x$. That means one of the solutions is $x = 0$.

$0 = 24 {x}^{2} + 18 x - 168$

Now you can divide by 3.

$0 = 8 {x}^{2} + 6 x - 56$

You can't really factor it into integers though. The factors of 56 include 1, 2, 4, 7, 8, 14, 28, and 56. The ones that could work with $8 x$ and $x$ to give $6$ are $7$ with $8$ or $4$ with $14$; the rest are too far apart. Regardless, neither of those give $6 x$ in the end. So:

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

$= \frac{- 6 \pm \sqrt{36 - 4 \cdot 8 \cdot - 56}}{16}$

$= \frac{- 6 \pm \sqrt{1828}}{16}$

$= \frac{- 6 \pm 2 \sqrt{457}}{16}$

Overall:

$x = 0$
$x \approx 2.3125$
$x \approx - 3.0625$