Making $R(x)=\frac{x^3+8x^2+18x+15}{x+5}$ continuous from $x = -5$ point The function $R(x)=\frac{x^3+8x^2+18x+15}{x+5}$ is not defined at the point $x = -5$. How should it be defined to make it continuous at this point.
I'm pretty sure you have to factor the top but for some reason I'm having a hard time with this problem.
Please Help.

This is where polynomial long division comes in handy.
We want to remove the removable singularity at $x = -5$, so we test to determine the quotient when dividing the numerator by the factor $(x + 5)$.
Doing so gives us: $$x^3 + 8x^2 + 18x + 15 = (x+5) (x^2 + 3x + 3)$$
This gives us $$\frac{x^3+8x^2+18x+15}{x+5} \implies  \frac{(x+5)(x^2 + 3x + 3)}{(x + 5)} = x^2 + 3x + 3$$
which is continuous at $x = -5$.
