How do you factor completely #45x^4 + 43x^2 y^2 - 28y^4#?

1 Answer
Apr 6, 2016

Answer:

#(3x+2y)(3x-2y)(5x^2+7y^2)#

Explanation:

Given
#color(white)("XXX")45x^4+43x^2y^2-28y^4#

Replacing #p=x^2# and #q=y^2#
#color(white)("XXX")=45p^2+43pq-28q^2#

Factoring by examining factors of #45# and #28#
or by using the quadratic formula:
#color(white)("XXX")=(9p-4q)(5p+7q)#

Replacing #p# and #q# with #x^2# and #y^2# respectively
and
Noting that the first of these factors is the difference of squares
#color(white)("XXX")=(3x+2y)(3x+2y)(5x^2+7y^2)#