Question

Step 1 of 2: Red the rational expression to its lowest terms 2 - 2x^2/2x^2 - 2Step 2 of 2: Find the restricted values of X, if any, for the given rational expression.

Answer 1

$\frac{2-2x^2}{2x^2-2}$

Step1

we can factor the numerator and denominator because we have a square subtract

$\begin{gathered} \frac{(\sqrt[]{2}+\sqrt[]{2}x)(\sqrt[]{2}-\sqrt[]{2}x)}{(\sqrt[]{2}x-\sqrt[]{2})(\sqrt[]{2}x+\sqrt[]{2})} \\ \\ \frac{(\sqrt[]{2}+\sqrt[]{2}x)}{\sqrt[]{2}x+\sqrt[]{2})}\cdot\frac{(\sqrt[]{2}-\sqrt[]{2}x)}{(\sqrt[]{2}x-\sqrt[]{2})} \\ \\ 1\cdot-1 \\ \\ =-1 \end{gathered}$

step 2

restricted values of X are when the denominator is 0

then

$2x^2-2=0$

factor this

$(\sqrt[]{2}x-\sqrt[]{2})(\sqrt[]{2}x+\sqrt[]{2})=0$

and find the value of x what make each parenthesis 0

$\begin{gathered} \sqrt[]{2}x-\sqrt[]{2}=0 \\ \sqrt[]{2}x=\sqrt[]{2} \\ x=1 \end{gathered}$

first restricted value is x=1

$\begin{gathered} \sqrt[]{2}x+\sqrt[]{2}=0 \\ \sqrt[]{2}x=-\sqrt[]{2} \\ x=-1 \end{gathered}$

and the other restricted value is x=-1