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

Jul 28, 2015

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

Explanation:

(Use a version of the ac factoring method.)

$3 {x}^{2} - x y - 24 {y}^{2}$

The product of the first and last coefficients is
$3 \times - 24 = - 72$

We want to split the middle term $- x y$ into two terms. The coefficient of this middle term is $- 1$

So, we need two numbers whose product is $- 72$ and whose sum is $- 1$.

$1 \times - 72$ no, they do not add to $- 1$
$2 \times - 36$ no
$3 \times - 24$ no
$4 \times - 18$ no
$5 \times$ no
$6 \times - 12$ no
$7 \times$ no
$8 \times - 9$ yes! the product is $- 72$ and the sum is $- 1$

Now split the middle term using $8 x y$ and $- 9 x y$ (In the end it won't matter which order you write them in, so let's just use the one above.)

$3 {x}^{2} - x y - 24 {y}^{2}$

$3 {x}^{2} + 8 x y - 9 x y - 24 {y}^{2}$ Now factor by grouping.

$\left(3 {x}^{2} + 8 x y\right) + \left(- 9 x y - 24 {y}^{2}\right)$

$= x \left(3 x + 8 y\right) + \left(- 3 y\right) \left(3 x + 8 y\right)$

$= \left(x - 3 y\right) \left(3 x + 8 y\right)$

Check by multiplying:

$3 {x}^{2} + 8 x y - 9 x y - 24 {y}^{2}$ looks good, so we are finished.