# How do you find the GCF of 42x^2,  56x^3?

Oct 13, 2015

$G C F \left(42 {x}^{2} , 56 {x}^{3}\right) = 14 {x}^{2}$

#### Explanation:

As for the numerical part:

$G C F \left(42 , 56\right) = G C F \left(2 \cdot 3 \cdot 7 , {2}^{3} \cdot 7\right)$

The rule is to look for common primes, and take them with the smallest exponents. $2$ is a common prime. It occours once as $2$, then as ${2}^{3}$. So, the one with the smallest exponent is $2$. $3$ is not a common prime, since it only divides $42$. $7$ is a common prime, and it appears in both factorization as $7$, so there's nothing to choose. The GCF is thus $2 \cdot 7 = 14$

For the variables, it is easier: ${x}^{2}$ divides ${x}^{3}$, so $G C F \left({x}^{2} , {x}^{3}\right) = {x}^{2}$.