what is the gcf of 48m^5n and 81m^2n^2?\n3mn\n3m^2n\n48m^2n\n129m^5n^2

Explanation:

Step1: Find GCF of coefficients

Find GCF of 48 and 81. Prime - factorize: $48 = 2^4\times3$, $81 = 3^4$. The GCF of 48 and 81 is 3.

Step2: Find GCF of variables with m

For $m^5$ and $m^2$, using the rule $GCF(m^a,m^b)=m^{\min(a,b)}$, the GCF of $m^5$ and $m^2$ is $m^2$.

Step3: Find GCF of variables with n

For $n$ and $n^2$, using the rule $GCF(n^a,n^b)=n^{\min(a,b)}$, the GCF of $n$ and $n^2$ is $n$.

Step4: Combine GCFs

Multiply the GCF of the coefficients and the GCFs of the variables: $3\times m^2\times n = 3m^2n$.

Answer:

3m^2n