<s>[INST] How to divide the following polynomial and factor it? The question is 
 $$ (2x^3+3x^2-39x-20) / (x-4) $$
I divided the following and got this as the answer
$$ 2x^2+9x+3-8/(x-4))$$
I thought that this was the answer, but when i looked at the answer sheet the answer was this $$(x-4)(x+5)(2x+1)$$
I think the answer should've been factored but i don't know how to factor this 
 $$ (2x^3+3x^2-39x-20) / (x-4) $$
and also i have a general question about factoring these types of problems
$$ax^3+bx^2+cx+d$$
which method would i choose, because i should be getting 3 answers and I don't know how to do that.
 [/INST] $4$ is a root of $3x^3+3x^2-39x-20$, hence it is divisible by $x-4$. To obtain the quotient, you can use Horner's scheme:
$$\begin{array}{rrcrcrr}
&2&&3&&-39&&-20\\
\hline
&&&8&&44&&20\\
&&\!\!\nearrow\!\!&&\!\!\nearrow\!\!\!\!&&\!\!\nearrow\!\!\!\!\\
{}\times 4&\color{red}{2}&&\color{red}{11}&&\color{red}{5}&&0
\end{array}$$
Hence $\;2x^3+3x^2-39x-20=(2x^2+11x+5)(x-4)$.
The quadratic polynomial has discriminant equal to $81$, hence its roots are $-5$  and $-\frac12$, so it factors as $(2x+1)(x+5)$ and 
$$\frac{2x^3+3x^2-39x-20}{x-4}=\frac{(2x+1)(x+5)(x-4)}{x-4}=(2x+1)(x+5)$$
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