Question

The H.C.F of 3(x2−9)(x+4) and 12(x−3)23(x2−9)(x+4) and 12(x−3)2  is​

Answer 1

Answer:

x2−3x+2=x2−2x−x+2

x(x−2)−1(x−2)=(x−2)(x−1)

Now

x2−4x+3=x2−3x−x+3

x(x−3)−1(x−3)=(x−3)(x−1)

Thus, the only common factor is (x-1