# How do you multiply 5u ^ { - 4} x ^ { 6} \cdot 8u ^ { 4} y \cdot 2x ^ { 2} y ^ { - 1}?

May 10, 2017

$80 {x}^{8}$

#### Explanation:

$5 {u}^{-} 4$ just means $5 \cdot {u}^{-} 4$ and each group of terms and coefficients is multiplied so we can just multiply everything.

When multiplying terms with exponents, the exponents for the same term are added, so the exponents for $u$ and $y$ cancel and the exponents $6$ and $2$ for $x$ add to get ${x}^{8}$.

u^4 * u^(-4) = u^((4 + (-4)) = u^0 = 1

$y \cdot {y}^{- 1} = {y}^{0} = 1$

${x}^{6} \cdot {x}^{2} = {x}^{\left(6 + 2\right)} = {x}^{8}$

The coefficients $5$, $8$, and $2$ multiply to get $80$.