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

2 Answers
Oct 13, 2015

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 y^2, then you get:

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

Let t = x/y and f(t) = 2t^2-3t-2

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

f(t) = 2t^2-3t-2 is of the form at^2+bt+c, with a=2, b=-3 and c = -2. So f(t) = 0 has roots given by the quadratic formula:

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 t = -1/2 or t = 2

So f(t) has factors (2t+1) and (t - 2):

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

Substitute t = x/y to find:

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

Then multiply through by y^2 to get:

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

Oct 13, 2015

2x^2 - 3xy - 2y^2 = 2x^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.