How to divide the following polynomial and factor it?

Question

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.

Answer 1

Dividing $2x^3 +3x^2 -39x -20 $ by $(x-4)$ should give you $2x^2 +11x+5$

Then factorise $2x^2 +11x+5$ to get the remaining factors:

$$2x^2 +11x+5 = (2x + 1)(x+5)$$

For more on dividing polynomials, see the following khan academy video

Answer 2

$4$ 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 $$\frac{2x^3+3x^2-39x-20}{x-4}=\frac{(2x+1)(x+5)(x-4)}{x-4}=(2x+1)(x+5)$$