How do you factor 21x^3 - 18x^2y + 24xy^2?

Sep 19, 2016

$21 {x}^{3} - 18 {x}^{2} y + 24 x {y}^{2} = 3 x \left(7 {x}^{2} - 6 x y + 8 {y}^{2}\right)$

Explanation:

Note that all of the terms are divisible by $3 x$, so we can separate that out as a factor first:

$21 {x}^{3} - 18 {x}^{2} y + 24 x {y}^{2} = 3 x \left(7 {x}^{2} - 6 x y + 8 {y}^{2}\right)$

The remaining quadratic is in the form $a {x}^{2} + b x y + c {y}^{2}$ with $a = 7$, $b = - 6$ and $c = 8$.

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 6\right)}^{2} - 4 \cdot 7 \cdot 8 = 36 - 224 = - 188$

Since $\Delta < 0$ this quadratic has no linear factors with Real coefficients.

$\textcolor{w h i t e}{}$
Complex coefficients

If we are determined to factor it, we can find some Complex coefficients using the quadratic formula to find zeros:

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

$\textcolor{w h i t e}{\frac{x}{y}} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$\textcolor{w h i t e}{\frac{x}{y}} = \frac{6 \pm \sqrt{- 188}}{14}$

$\textcolor{w h i t e}{\frac{x}{y}} = \frac{6 \pm 2 \sqrt{47} i}{14}$

$\textcolor{w h i t e}{\frac{x}{y}} = \frac{3 \pm \sqrt{47} i}{7}$

Hence:

$7 {x}^{2} - 6 x y + 8 {y}^{2} = 7 \left(x - \left(\frac{3 + \sqrt{47} i}{7}\right) y\right) \left(x - \left(\frac{3 - \sqrt{47} i}{7}\right) y\right)$

$\textcolor{w h i t e}{7 {x}^{2} - 6 x y + 8 {y}^{2}} = \frac{1}{7} \left(7 x - \left(3 + \sqrt{47} i\right) y\right) \left(7 x - \left(3 - \sqrt{47} i\right) y\right)$