Q. 179 - Intermediate Algebra

Question

In the following exercises, multiply the monomials. 

$$\begin{gathered} \left(a\right) \left(-10x^5\right)\left(-3x^3\right) \\ \left(b\right) \left(\frac{5}{8}{x}^{3}y\right)\left(24x^{5}y\right) \end{gathered}$$

Step-by-Step Solution

Answer

The solution is 

$$\begin{gathered} \left(a\right) \left(-10x^5\right)\left(-3x^3\right)=30x^8 \\ \left(b\right) \left(\frac{5}{8}{x}^{3}y\right)\left(24x^{5}y\right)=15x^8{y}^2 \end{gathered}$$

Part (a) Step 1. Given information

The given expression is 

$\left(-10x^5\right)\left(-3x^3\right)$

Part (a) Step 2. Simplify

Using commutative property to rearrange the terms 

$$\begin{gathered} \left(-10x^5\right)\left(-3x^3\right) \\ =\left(-10\right)·\left(-3\right)·x^5·x^3 \\ =30x^{5+3} \\ =30x^8 \end{gathered}$$

Part (b) Step 1. Given information

The given expression is 

$\left(\frac{5}{8}{x}^{3}y\right)\left(24x^{5}y\right)$

Part (b) Step 2. Simplify

Using commutative property to rearrange the terms

$$\begin{gathered} \left(\frac{5}{8}{x}^{3}y\right)\left(24x^{5}y\right) \\ =\frac{5}{8}·24·x^3·x^5·y·y \\ =15x^{3+5}{y}^2 \\ =15x^8{y}^2 \end{gathered}$$