Q. 180 - Intermediate Algebra

Question

In the following exercises, multiply the monomials. 

$$\begin{gathered} \left(a\right) \left(-8u^6\right)\left(-9u\right) \\ \left(b\right) \left(\frac{2}{3}{x}^{2}y\right)\left(\frac{3}{4}x{y}^2\right) \end{gathered}$$

Step-by-Step Solution

Answer

The solution is 

$$\begin{gathered} \left(a\right) \left(-8u^6\right)\left(-9u\right)=72u^7 \\ \left(b\right) \left(\frac{2}{3}{x}^{2}y\right)\left(\frac{3}{4}x{y}^2\right)=\frac{x^3{y}^3}{2} \end{gathered}$$

Part (a) Step 1. Given information

The given expression is 

$\left(-8u^6\right)\left(-9u\right)$

Part (a) Step 2. Simplify

Using commutative property to rearrange the terms 

$$\begin{gathered} \left(-8u^6\right)\left(-9u\right) \\ =\left(-8\right)·\left(-9\right)·u^6·u \\ =72u^{6+1} \\ =72u^7 \end{gathered}$$

Part (b) Step 1. Given information

The given expression is 

$\left(\frac{2}{3}{x}^{2}y\right)\left(\frac{3}{4}x{y}^2\right)$

Part (b) Step 2. Simplify

Using commutative property to rearrange the terms

$$\begin{gathered} \left(\frac{2}{3}{x}^{2}y\right)\left(\frac{3}{4}x{y}^2\right) \\ =\frac{2}{3}·\frac{3}{4}·x^2·x·y·y^2 \\ =\frac{x^{2+1}{y}^{2+1}}{2} \\ =\frac{x^3{y}^3}{2} \end{gathered}$$