Domain of a quotient of two polynomials

A quotient $\frac{h(x)}{g(x)}$ is defined wherever both polynomials are defined, which is everywhere, except where the denominator is 0. So only the zeros of $g$ are excluded.

Factor $g$: $$g(x)=x^3-3x^2-18x=x(x^2-3x-18)=x(x-6)(x+3)$$ Check the quadratic: $(x-6)(x+3)=x^2+3x-6x-18=x^2-3x-18$.

So $g(x)=0$ at $x=0$, $x=6$ and $x=-3$, and the domain of $k$ is all real numbers except $-3$, $0$ and $6$. That is choice C.

The numerator does not matter for the domain. $h(x)=x^2-2x-35=(x-7)(x+5)$, so $h$ is 0 at $7$ and $-5$, which only makes $k$ equal to 0 there.

Each choice

• A. Only the factor $x$ was pulled out, and the quadratic left over was not factored. $x^2-3x-18$ also equals zero at 6 and at $-3$, and both of those must be excluded.
• B. These are the zeros of the numerator $h$. A zero on top makes the quotient equal 0, which is allowed. It is easy to mix up which polynomial cannot be zero.
• C (correct). The denominator factors as $x(x-6)(x+3)$, so exactly $-3$, $0$ and $6$ make it zero.
• D. Both polynomials' zeros were removed here. The values $-5$ and $7$ make $k$ equal 0, not a zero denominator, so they belong in the domain.