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

Answer: 
It is a polynomial of degree 3, and it is the product of the quadratic factor $x^2-3$ and the linear factor $x+1$. But maybe I miss your point.
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 $\;x^2+ 0\mkern 1.5mux-3=x^2-3$.