Q. 181 - Intermediate Algebra

Question

In the following exercises, multiply the monomials. 

$$\begin{gathered} \left(a\right) \left(-6c^4\right)\left(-12c\right) \\ \left(b\right) \left(\frac{3}{5}{m}^3{n}^2\right)\left(\frac{5}{9}{m}^2{n}^3\right) \end{gathered}$$

Step-by-Step Solution

Answer

The solution is 

$$\begin{gathered} \left(a\right) \left(-6c^4\right)\left(-12c\right)=72c^5 \\ \left(b\right) \left(\frac{3}{5}{m}^3{n}^2\right)\left(\frac{5}{9}{m}^2{n}^3\right)=\frac{m^5{n}^5}{3} \end{gathered}$$

Part (a) Step 1. Given information

The given expression is 

$\left(-6c^4\right)\left(-12c\right)$

Part (a) Step 2. Simplify

Using commutative property to rearrange the terms

$$\begin{gathered} \left(-6c^4\right)\left(-12c\right) \\ =\left(-6\right)·\left(-12\right)·c^4·c \\ =72c^{4+1} \\ =72c^5 \end{gathered}$$

Part (b) Step 1. Given expression

The given expression is 

$\left(\frac{3}{5}{m}^3{n}^2\right)\left(\frac{5}{9}{m}^2{n}^3\right)$

Part (b) Step 2. Simplify

Using commutative property to rearrange the terms

$$\begin{gathered} \left(\frac{3}{5}{m}^3{n}^2\right)\left(\frac{5}{9}{m}^2{n}^3\right) \\ =\frac{3}{5}·\frac{5}{9}·m^3·m^2·n^2·n^3 \\ =\frac{m^{3+2}{n}^{2+3}}{3} \\ =\frac{m^5{n}^5}{3} \end{gathered}$$