Question

Evaluate $\frac{x^2+1}{x^2-3x+2}$for each value:

$ⓐx=0ⓑx=-1ⓒx=3$

Short answer

$$\begin{gathered} ⓐ\mathrm{The}\mathrm{value}\mathrm{of}\frac{x^2+1}{x^2-3x+2}\mathrm{for}x=0\mathrm{is}\frac{1}{2}. \\ ⓑ\mathrm{The}\mathrm{value}\mathrm{of}\frac{x^2+1}{x^2-3x+2}\mathrm{for}x=-1\mathrm{is}\frac{1}{3}. \\ ⓒ\mathrm{The}\mathrm{value}\mathrm{of}\frac{x^2+1}{x^2-3x+2}\mathrm{for}x=3\mathrm{is}5. \end{gathered}$$

Step-by-step solution

Step 1. Given Information 

Evaluate $\frac{x^2+1}{x^2-3x+2}$for each value:

$ⓐx=0ⓑx=-1ⓒx=3$

Part (a) Step 1. Finding the value of x2+1x2−3x+2 for x=0

Putting $x=0$on the given expression

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

Part (b) Step 1. Finding the value of x2+1x2−3x+2 for x=-1.

Putting $x=-1$on the given expression

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

Part (c) Step 1. Finding the value of x2+1x2−3x+2 for x=3

Putting $x=3$on the given expression

localid="1648640957993" $$\begin{gathered} \frac{x^2+1}{x^2-3x+2}=\frac{{\left(3\right)}^2+1}{{\left(3\right)}^2-3\times 3+2} \\ \frac{x^2+1}{x^2-3x+2}=\frac{9+1}{9-9+2} \\ \frac{x^2+1}{x^2-3x+2}=\frac{10}{2} \\ \frac{x^2+1}{x^2-3x+2}=5 \end{gathered}$$