Question

Determine the H.C.F and L.C.M of the polynomial $81x^3{y}^2,27x^4{y}^2$.

Answer 1

Let $f\left(x\right)=$ $81x^3{y}^2$, $g\left(x\right)=$ $27x^4{y}^2$
Factors of $81$ are: $1,3,9,9,27,81$
Factors of $27$ are: $1,3,9,27$
So, H.C.F of $81$ and $27$ is $27$.
Therefore, H.C.F $=$ $27x^3{y}^2$
L.C.M of $81$ and $27$ is $81$.
L.C.M $=$ $81x^4{y}^2$