Question

EXERCISE - 2
1. HCF of the polynomials 20x’2y(x2-y2) and
35 xy2(x2-y2) is:​

Answer 1

had of any polynomial is the tem with lowest and common degree with variable

so the hcf of first poly is.

hcf(20x4y2-20x2y2and 35x2y2-35xy4)

is

5xy(x2-y2)