Q. 190 - Intermediate Algebra

Question

In the following exercises, multiply. 

a.$\left(4\sqrt{2k^5}\right)\left(-3\sqrt{32k^6}\right)$

b.$\left(-\sqrt[4]{6b^3}\right)\left(3\sqrt[4]{8b^3}\right)$

Step-by-Step Solution

Answer

On multiplication, we get $-48z^5\sqrt{4z},-6b\sqrt[4]{3b^2}$ respectively

Part (a) Step 1: Given information

Given the expression $\left(4\sqrt{2k^5}\right)\left(-3\sqrt{32k^6}\right),\left(-\sqrt[4]{6b^3}\right)\left(3\sqrt[4]{8b^3}\right)$

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

Multiplying the terms and simplifying

$$\begin{gathered} \left(4\sqrt{2k^5}\right)\left(-3\sqrt{32k^6}\right)=[4·(-3\left)\right]\left(\sqrt{2k^5}·\sqrt{32k^6}\right) \\ \left(4\sqrt{2k^5}\right)\left(-3\sqrt{32k^6}\right)=-12·\sqrt{64k^{11}} \\ \left(4\sqrt{2k^5}\right)\left(-3\sqrt{32k^6}\right)=-12·\sqrt{16z^{10}·4z} \\ \left(4\sqrt{2k^5}\right)\left(-3\sqrt{32k^6}\right)=-12·\sqrt{16z^{10}}·\sqrt{4z} \\ \left(4\sqrt{2k^5}\right)\left(-3\sqrt{32k^6}\right)=-12·\sqrt{{\left(4z^5\right)}^2}·\sqrt{4z} \\ \left(4\sqrt{2k^5}\right)\left(-3\sqrt{32k^6}\right)=-12·4z^5·\sqrt{4z} \\ \left(4\sqrt{2k^5}\right)\left(-3\sqrt{32k^6}\right)=-48z^5\sqrt{4z} \end{gathered}$$

Part (b) Step 1: Given information

Given expression $\left(-\sqrt[4]{6b^3}\right)\left(3\sqrt[4]{8b^3}\right)$

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

 Multiplying the terms and simplifying, we get

$$\begin{gathered} \left(-\sqrt[4]{6b^3}\right)\left(3\sqrt[4]{8b^3}\right)=\left[\right(-1)·(3\left)\right]\left(\sqrt[4]{6b^3}·\sqrt[4]{8b^3}\right) \\ \left(-\sqrt[4]{6b^3}\right)\left(3\sqrt[4]{8b^3}\right)=-3·\sqrt[4]{48b^6} \\ \left(-\sqrt[4]{6b^3}\right)\left(3\sqrt[4]{8b^3}\right)=-3·\sqrt[4]{16b^4·3b^2} \\ \left(-\sqrt[4]{6b^3}\right)\left(3\sqrt[4]{8b^3}\right)=-3·\sqrt[4]{(2b{)}^4}·\sqrt[4]{3b^2} \\ \left(-\sqrt[4]{6b^3}\right)\left(3\sqrt[4]{8b^3}\right)=-3·2b·\sqrt[4]{3b^2} \\ \left(-\sqrt[4]{6b^3}\right)\left(3\sqrt[4]{8b^3}\right)=-6b\sqrt[4]{3b^2} \end{gathered}$$