Question

In the following exercises, multiply the monomials.

$$\begin{gathered} \left(a\right)\left(6y^7\right)\left(-3y^4\right) \\ \left(b\right)\left(\frac{4}{7}r{s}^2\right)\left(14r{s}^3\right) \end{gathered}$$

Short answer

The solution is

$$\begin{gathered} \left(a\right)\left(6y^7\right)\left(-3y^4\right)=-18y^{11} \\ \left(b\right)\left(\frac{4}{7}r{s}^2\right)\left(14r{s}^3\right)=8r^2{s}^5 \end{gathered}$$

Step-by-step solution

Part (a) Step 1. Given information

The given expression is

$\left(6y^7\right)\left(-3y^4\right)$

Part (a) Step 2. Simplify

Using commutative property to rearrange the terms

$$\begin{gathered} \left(6y^7\right)\left(-3y^4\right) \\ =6·\left(-3\right)·y^7·y^4 \\ =-18y^{7+4} \\ =-18y^{11} \end{gathered}$$

Part (b) Step1. Given information

The given expression is

$\left(\frac{4}{7}r{s}^2\right)\left(14r{s}^3\right)$

Part (b) Step 2. Simplify

Using commutative property to rearrange the terms

$$\begin{gathered} \left(\frac{4}{7}r{s}^2\right)\left(14r{s}^3\right) \\ =\frac{4}{7}·14·r·r·s^2·s^3 \\ =8r^2{s}^{3+2} \\ =8r^2{s}^5 \end{gathered}$$