Answer
$2x^2+x$, $(-\infty, \infty)$
$x+2$, $(-\infty, \infty)$
$x^4+x^3-x-1$, $(-\infty, \infty)$
$\frac{x^2+x+1}{x^2-1}$, $(-\infty,-1)\cup(-1,1)\cup(1, \infty)$
Work Step by Step
Step 1. Given $f(x)=x^2+x+1$ and $g(x)=x^2-1$, we have $f+g=2x^2+x$ with domain $(-\infty, \infty)$
Step 2. We have $f-g=x+2$ with domain $(-\infty, \infty)$
Step 3. We have $fg=(x^2+x+1)(x^2-1)=x^4-x^2+x^3-x+x^2-1=x^4+x^3-x-1$ with domain $(-\infty, \infty)$
Step 4. We have $\frac{f}{g}=\frac{x^2+x+1}{x^2-1}$ with domain $(-\infty,-1)\cup(-1,1)\cup(1, \infty)$
