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

##### 2 Answers

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

#2x^2-3xy-2y^2 = (2x+y)(x-2y)#

#### Explanation:

If you divide the quadratic through by

#(2x^2)/y^2-(3xy)/y^2-(2y^2)/y^2 = 2(x/y)^2-3(x/y)-2#

Let

To factor

#t = (-b+-sqrt(b^2-4ac))/(2a) = (3+-sqrt((-3)^2-(4xx2xx-2)))/(2xx2)#

#=(3+-sqrt(9+16))/4 = (3+-sqrt(25))/4 = (3+-5)/4#

That is

So

#2t^2-3t-2 = (2t+1)(t-2)#

Substitute

#2(x/y)^2-3(x/y)-2 = (2(x/y)+1)((x/y)-2)#

Then multiply through by

#2x^2-3xy-2y^2 = (2x+y)(x-2y)#

2

= (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.