Question

In the following exercises, evaluate the rational expression for the given values.

$$\begin{gathered} \frac{b^2+2}{b^2-3b-4} \\ \left(a\right)b=0 \\ \left(b\right)b=2 \\ \left(c\right)b=-2 \end{gathered}$$

Short answer

The required answers are

$$\begin{gathered} \mathrm{Part}\left(\mathrm{a}\right)\frac{-1}{2} \\ \mathrm{Part}\left(\mathrm{b}\right)-1 \\ \mathrm{Part}\left(\mathrm{c}\right)1 \end{gathered}$$

Step-by-step solution

Part(a) Step 1. Given Information  

The given expression is$\frac{b^2+2}{b^2-3b-4}$

Part (a) Step 2. Calculation   

Substitute 0 for b,

$$\begin{gathered} \frac{b^2+2}{b^2-3b-4}=\frac{{\left(0\right)}^2+2}{{\left(0\right)}^2-3\left(0\right)-4} \\ =\frac{2}{-4} \\ =-\frac{1}{2} \end{gathered}$$

Part (b) Step 1. Calculation    

Substitute 2 for b,

$$\begin{gathered} \frac{b^2+2}{b^2-3b-4}=\frac{{\left(2\right)}^2+2}{{\left(2\right)}^2-3\left(2\right)-4} \\ =\frac{4+2}{4-6-4} \\ =\frac{6}{-6} \\ =-1 \end{gathered}$$

Part (c) Step 1. Calculation  

Substitute $-2$ for b,

$$\begin{gathered} \frac{b^2+2}{b^2-3b-4}=\frac{{(-2)}^2+2}{{(-2)}^2-3(-2)-4} \\ =\frac{4+2}{4+6-4} \\ =\frac{6}{6} \\ =1 \end{gathered}$$