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

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Answer 1 · 2017-06-19T16:40:48

See a solution process below:

This is a special form of the quadratic:

#(x + y)(x - y) = x^2 - y^2#

Substitute:

#2a# for #x#

#5# for #y#

Giving:

#(2a + 5)(2a - 5) = (2a)^2 - 5^2#

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