# How do you factor 2x^2 - 3xy - 2y^2?

Oct 13, 2015

Reformulate as a quadratic in a single variable, use the quadratic formula, then reformulate back to find:

$2 {x}^{2} - 3 x y - 2 {y}^{2} = \left(2 x + y\right) \left(x - 2 y\right)$

#### Explanation:

If you divide the quadratic through by ${y}^{2}$, then you get:

$\frac{2 {x}^{2}}{y} ^ 2 - \frac{3 x y}{y} ^ 2 - \frac{2 {y}^{2}}{y} ^ 2 = 2 {\left(\frac{x}{y}\right)}^{2} - 3 \left(\frac{x}{y}\right) - 2$

Let $t = \frac{x}{y}$ and $f \left(t\right) = 2 {t}^{2} - 3 t - 2$

To factor $f \left(t\right)$, find roots of $f \left(t\right) = 0$ using the quadratic formula.

$f \left(t\right) = 2 {t}^{2} - 3 t - 2$ is of the form $a {t}^{2} + b t + c$, with $a = 2$, $b = - 3$ and $c = - 2$. So $f \left(t\right) = 0$ has roots given by the quadratic formula:

$t = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{3 \pm \sqrt{{\left(- 3\right)}^{2} - \left(4 \times 2 \times - 2\right)}}{2 \times 2}$

$= \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm \sqrt{25}}{4} = \frac{3 \pm 5}{4}$

That is $t = - \frac{1}{2}$ or $t = 2$

So $f \left(t\right)$ has factors $\left(2 t + 1\right)$ and $\left(t - 2\right)$:

$2 {t}^{2} - 3 t - 2 = \left(2 t + 1\right) \left(t - 2\right)$

Substitute $t = \frac{x}{y}$ to find:

$2 {\left(\frac{x}{y}\right)}^{2} - 3 \left(\frac{x}{y}\right) - 2 = \left(2 \left(\frac{x}{y}\right) + 1\right) \left(\left(\frac{x}{y}\right) - 2\right)$

Then multiply through by ${y}^{2}$ to get:

$2 {x}^{2} - 3 x y - 2 {y}^{2} = \left(2 x + y\right) \left(x - 2 y\right)$

Oct 13, 2015

2${x}^{2}$ - 3xy - 2${y}^{2}$ = 2${x}^{2}$ - 4xy + xy - 2 ${y}^{2}$ = 2x(x - 2y) + y (x - 2y)
= (x - 2y) (2x + y)

#### Explanation:

Here we write 3xy as -4xy + xy because the product of 2 and -4 the extreme coefficients equals the -4, so that the grouping of terms is made possible.