Answer
$\{x\in R: x \neq -3, x\neq 2 \}=(-\infty , -3) \cup (-3, 2) \cup (2, \infty)$
Work Step by Step
The domain of $f(x)=\frac{x^4}{x^2 +x-6}$ is the set of all real numbers $x$ for which $\frac{x^4}{x^2 +x-6}$ is defined as a real number. We know a rational expression is not defined where the denominator is zero, so we must find all values of $x$ for which $$x^2 +x-6 =0.$$ Factoring the left hand side gives $$(x+3)(x-2)=0.$$ Thus $$ x+3=0 \hspace{ .5 cm} or \hspace{.5 cm}x-2=0.$$
We subtract $3$ from both sides of the equation on the left and add $2$ to both sides of the equation on the right to get $$x=-3\hspace{ .5 cm} or \hspace{.5 cm}x=2.$$ Thus $\frac{x^4}{x^2 +x -6}$ is not defined when $x=-3$ or $x =2$. This means the domain of $f(x)=\frac{x^4}{x^2 +x -6}$ is the set of all real numbers $x$ for which $x\neq -3$ and $x \neq 2$.
In set notation, the domain is $\{x\in R: x \neq -3, x\neq 2 \}.$
In interval notation, the domain is $(-\infty , -3) \cup (-3, 2) \cup (2, \infty).$
