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.