Q. 188 - Intermediate Algebra

Question

In the following exercises, simplify.

a.$\left(3\sqrt{2x^3}\right)\left(7\sqrt{18x^2}\right)$

b. $\left(-6\sqrt[3]{20a^2}\right)\left(-2\sqrt[3]{16a^3}\right)$

Step-by-Step Solution

Answer

On multiplication, we get $63x^2\sqrt{4x},60a\sqrt[3]{5a^2}$ respectively

Part (a) Step 1: Given information

Given the expression $\left(3\sqrt{2x^3}\right)\left(7\sqrt{18x^2}\right),\left(-6\sqrt[3]{20a^2}\right)\left(-2\sqrt[3]{16a^3}\right)$

Part (a) Step 2: Multiplying the terms and simplifying

Multiplying the terms and simplifying, we get

$$\begin{gathered} \left(3\sqrt{2x^3}\right)\left(7\sqrt{18x^2}\right)=(3·7)\left(\sqrt{2x^3}·\sqrt{18x^2}\right) \\ \left(3\sqrt{2x^3}\right)\left(7\sqrt{18x^2}\right)=21·\sqrt{36x^5} \\ \left(3\sqrt{2x^3}\right)\left(7\sqrt{18x^2}\right)=21·\sqrt{9x^4·4x} \\ \left(3\sqrt{2x^3}\right)\left(7\sqrt{18x^2}\right)=21·\sqrt{9x^4}·\sqrt{4x} \\ \left(3\sqrt{2x^3}\right)\left(7\sqrt{18x^2}\right)=21·\sqrt{{\left(3x^2\right)}^2}·\sqrt{4x} \\ \left(3\sqrt{2x^3}\right)\left(7\sqrt{18x^2}\right)=21·3x^2·\sqrt{4x} \\ \left(3\sqrt{2x^3}\right)\left(7\sqrt{18x^2}\right)=63x^2\sqrt{4x} \end{gathered}$$

Part (b) Step 1: Given information

Given expression $\left(-6\sqrt[3]{20a^2}\right)\left(-2\sqrt[3]{16a^3}\right)$

Part (b) Step 2: Multiplying the terms and simplifying

Simplifying, we get

$$\begin{gathered} \left(-6\sqrt[3]{20a^2}\right)\left(-2\sqrt[3]{16a^3}\right)=\left[\right(-6)·(-2\left)\right]\left(\sqrt[3]{20a^2}\right..\left.\sqrt[3]{16a^3}\right) \\ \left(-6\sqrt[3]{20a^2}\right)\left(-2\sqrt[3]{16a^3}\right)=12·\sqrt[3]{320a^5} \\ \left(-6\sqrt[3]{20a^2}\right)\left(-2\sqrt[3]{16a^3}\right)=12·\sqrt[3]{64a^3·5a^2} \\ \left(-6\sqrt[3]{20a^2}\right)\left(-2\sqrt[3]{16a^3}\right)=12·\sqrt[3]{64a^3}·\sqrt[3]{5a^2} \\ \left(-6\sqrt[3]{20a^2}\right)\left(-2\sqrt[3]{16a^3}\right)=12·\sqrt[3]{(4a{)}^3}·\sqrt[3]{5a^2} \\ \left(-6\sqrt[3]{20a^2}\right)\left(-2\sqrt[3]{16a^3}\right)=15·4a·\sqrt[3]{5a^2} \\ \left(-6\sqrt[3]{20a^2}\right)\left(-2\sqrt[3]{16a^3}\right)=60a\sqrt[3]{5a^2} \end{gathered}$$