Answer
$f+g,f-g,fg$: all real numbers; $\frac{f}{g}$: all real numbers except $-5$ and $3$
Work Step by Step
For all but $\frac{f}{g}$, the domain includes all real numbers. To find the domain of $\frac{f}{g}$, set the denominator equal to $0$.
$f+g=(3-x^2)+(x^2+2x-15)$
$f+g=2x-12$
$(-∞,∞)$
$f-g=(3-x^2)-(x^2+2x-15)$
$f-g=-2x^2-2x+18$
$(-∞,∞)$
$fg=(3-x^2)(x^2+2x-15)$
$fg=(-x^4-2x^3+18x^2+6x-45$
$(-∞,∞)$
$\frac{f}{g}=\frac{3-x^2}{(x+5)(x-3)}$
$x\ne-5,3$