How do you factor # 16y^2-25#?

2 Answers
Zachary
Mar 21, 2018

#(4y+5)(4y-5)#
You need to consider what multiplies to make 16(either #1*16, 2*8,#or #4*4#), and what multiplies to make 25( #5*5#). Also notice that this is a binomial, not a trinomial.

Explanation:

The only factor of #25# is #5*5=5^2#, so the factorization must be of the form #(a+5)(b-5)#, as a negative times a positive is negative. Now consider that there is no middle term, so it must have been canceled out. This implies that the coefficients of #y# are the same. This leaves only #(4y+5)(4y-5)#.

David Drayer
Mar 21, 2018

As the difference between two squares.

Explanation:

# 16y^2 - 25 = ( 4y -5) xx ( 4y +5) #

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