Q. 189 - Intermediate Algebra

Question

In the following exercises, simplify.

a.$\left(-2\sqrt{7z^3}\right)\left(3\sqrt{14z^8}\right)$

b.$\left(2\sqrt[4]{8y^2}\right)\left(-2\sqrt[4]{12y^3}\right)$

Step-by-Step Solution

Answer

On multiplication, we get $-42z^5\sqrt{2z},-8y\sqrt[4]{6y}$ respectively

Part (a) Step 1:Given information

Given the expression $\left(-2\sqrt{7z^3}\right)\left(3\sqrt{14z^8}\right),\left(2\sqrt[4]{8y^2}\right)\left(-2\sqrt[4]{12y^3}\right)$

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

Multiplying the terms and simplifying, we get

$$\begin{gathered} \left(-2\sqrt{7z^3}\right)\left(3\sqrt{14z^8}\right)=(-2·3)\left(\sqrt{7z^3}·\sqrt{14z^8}\right) \\ \left(-2\sqrt{7z^3}\right)\left(3\sqrt{14z^8}\right)=-6·\sqrt{98z^{11}} \\ \left(-2\sqrt{7z^3}\right)\left(3\sqrt{14z^8}\right)=-6·\sqrt{49z^{10}·2z} \\ \left(-2\sqrt{7z^3}\right)\left(3\sqrt{14z^8}\right)=-6·\sqrt{49z^{10}}·\sqrt{2z} \\ \left(-2\sqrt{7z^3}\right)\left(3\sqrt{14z^8}\right)=-6·\sqrt{{\left(7z^5\right)}^2}·\sqrt{2z} \\ \left(-2\sqrt{7z^3}\right)\left(3\sqrt{14z^8}\right)=-6·7z^5·\sqrt{2z} \\ \left(-2\sqrt{7z^3}\right)\left(3\sqrt{14z^8}\right)=-42z^5\sqrt{2z} \end{gathered}$$

Part (b) Step 1: Given information

Given expression $\left(2\sqrt[4]{8y^2}\right)\left(-2\sqrt[4]{12y^3}\right)$

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

Multiplying the terms and simplifying, we get

$$\begin{gathered} \left.\left(2\sqrt[4]{8y^2}\right)\left(-2\sqrt[4]{12y^3}\right)=\left[\right(2)·(-2\left)\right]\sqrt[4]{8y^2}·\sqrt[4]{12y^3}\right) \\ \left(2\sqrt[4]{8y^2}\right)\left(-2\sqrt[4]{12y^3}\right)=-4·\sqrt[4]{96y^5} \\ \left(2\sqrt[4]{8y^2}\right)\left(-2\sqrt[4]{12y^3}\right)=-4·\sqrt[4]{16y^4·6y} \\ \left(2\sqrt[4]{8y^2}\right)\left(-2\sqrt[4]{12y^3}\right)=-4·\sqrt[4]{16y^4}·\sqrt[4]{6y} \\ \left(2\sqrt[4]{8y^2}\right)\left(-2\sqrt[4]{12y^3}\right)=-4·\sqrt[4]{(2y{)}^4}·\sqrt[4]{6y} \\ \left(2\sqrt[4]{8y^2}\right)\left(-2\sqrt[4]{12y^3}\right)=-4·2y·\sqrt[4]{6y} \\ \left(2\sqrt[4]{8y^2}\right)\left(-2\sqrt[4]{12y^3}\right)=-8y\sqrt[4]{6y} \end{gathered}$$