How do you factor #4x^2 -20xy+25y^2#?

2 Answers
Oct 15, 2015

Answer:

(2x-5y)(2x-5y).

Explanation:

#4x^2-20xy+25y^2#
#=4x^2-10xy-10xy+25y^2#
#=2x(2x-5y)-5y(2x-5y)#
#=(2x-5y)(2x-5y)#

Oct 15, 2015

Answer:

#4x^2+20xy+25y^2=(2x+5y)^2#

Explanation:

Use the formula for the square of a binomial: #(a+b)^2=a^2+2ab+b^2#.

Both #4# and #25#, the coefficient of #x^2# and #y^2#, are perfect squares. This makes us think that the whole expression could be a perfect square: #4# is #2^2#, and #25# is #5^2#. So, our claim is that

#4x^2-20xy+25y^2# is #(2x-5y)^2#. Is it true? The only term to verify is #-20xy#, and it is indeed twice the product of #2x# and #-5y#. So, the conjecture was right.