How do you factor completely #25a^4-49b^2#?

1 Answer
Apr 19, 2017

#25a^4-49b^2 = (5a^2-7b)(5a^2+7b)#

Explanation:

Note that both #25a^4 = (5a^2)^2# and #49b^2=(7b)^2# are perfect squares.

The difference of squares identity can be written:

#A^2-B^2=(A-B)(A+B)#

Using this with #A=5a^2# and #B=7b# we find:

#25a^4-49b^2 = (5a^2)^2-(7b)^2#

#color(white)(25a^4-49b^2) = (5a^2-7b)(5a^2+7b)#

There are no simpler factors.