How do you factorize #204x^2+100y^2+64#?

1 Answer
Feb 28, 2017

#204x^2+100y^2+64=4(51x^2+25y^2+16)#

Explanation:

In the polynomial #204x^2+100y^2+64#, we have no variable common between the three monomials forming the polynomial.

However, the coefficients #204# and #100# and constant term #64# have HCF as #4#

and hence #204x^2+100y^2+64#

= #4xx51x^2+4xx25y^2+4xx16#

= #4(51x^2+25y^2+16)#