Question: I know that the linear factor is $(x+1) (x^2-3),$ but how would I find the quadratic factor?

Answer: 
Simply calculate the roots of your polynomial (if possible).
You have a polynomial of degree 3. 
Normally you first try to find a root, by "guessing". We have to check the divesors of the constant -3. Which are $\pm 3, \pm 1$.

Easily we see, that x=-1 is a root and then make a long division

$(x^3+x^2-3x-3)\div (x+1)$ giving us the quadratic factor.

On other way is to compare coefficients by:

$(x^3+x^2-3x-3)=(ax+b)(cx^2+dx+e)$

Which should involve more calculation.
You can use *Horner's algorithm* for the division by $x+1$:
\begin{array}{r|rrrr}
&1&1&-3&-3\\
\times -1&\downarrow&\nearrow -1&0&3\\
\hline
&\color{red}1&\color{red}0&\color{red}{-3}&0
\end{array}
so the quotient is $\;x^2+ 0\mkern 1.5mux-3=x^2-3$.