Q. 8.81 - Intermediate Algebra

Question

Simplify:

1. $\left(6\sqrt{6x^2}\right)\left(8\sqrt{30x^4}\right)$
   
2. $(-4\sqrt[4]{12y^3})(-\sqrt[4]{8y^3})$
   

Step-by-Step Solution

Answer

Part a. $288x^3\sqrt{5}$

Part b. $\begin{array}{rcl}& & 8y\sqrt[4]{6y^2}\end{array}$

Part (a) Step 1. Multiply the radicals and simplify

Using the product property of roots that is $\sqrt[n]{a}·\sqrt[n]{b}=\sqrt[n]{ab}$, the expression can be simplified to

$\begin{array}{rcl}\left(6\sqrt{6x^2}\right)\left(8\sqrt{30x^4}\right)& =& 6·8·\sqrt{6x^2·30x^4}\\ & =& 48\sqrt{180x^6}\end{array}$

Now, the radical can be simplified as

$\begin{array}{rcl}48\sqrt{180x^6}& =& 48\sqrt{36x^6·5}\\ & =& 48\sqrt{36x^6}·\sqrt{5}\\ & =& 48·6x^3\sqrt{5}\\ & =& 288x^3\sqrt{5}\end{array}$

Part (b) Step 1. Multiply the radicals and simplify

Using the product property of roots that is $\sqrt[n]{a}·\sqrt[n]{b}=\sqrt[n]{ab}$, the expression can be simplified to

$\begin{array}{rcl}(-4\sqrt[4]{12y^3})(-\sqrt[4]{8y^3})& =& (-4)(-1)(\sqrt[4]{12y^3}·\sqrt[4]{8y^3})\\ & =& 4\sqrt[4]{12y^3·8y^3}\\ & =& 4\sqrt[4]{96y^6}\end{array}$

Now, the radical can be simplified as

$\begin{array}{rcl}4\sqrt[4]{96y^6}& =& 4\sqrt[4]{16y^4·6y^2}\\ & =& 4·\sqrt[4]{16y^4}·\sqrt[4]{6y^2}\\ & =& 4·2y\sqrt[4]{6y^2}\\ & =& 8y\sqrt[4]{6y^2}\end{array}$