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

2 Answers
George C.
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)#

Chandra S.
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.

Related questions
Impact of this question
11642 views around the world
You can reuse this answer
Creative Commons License