Question

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

Answer 1

`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)`