How to factor this polynomial? I was trying to factor this polynomial:

$x^3 + x^2 - 16x + 20$

There are four options in this question:

*

*(a) It could be factored in the following form: $(x^2 + b)(x+c)$;

*(b) It could be factored in the following form: $(x+b)(x+c)(x+d)$, assuming that $b \neq c \neq d$

*(c) It could not be factored.

*(d) It could be factored in the following form: $(x+b)^2  (x+c) $
Here's how I've tried to do it:
I've tried to factor by grouping the x, therefore I've obtained:
$x(x^2 + x - 16) + 20$. Now, I've put the $x$ and the $20$ together:
$(x+20)(x^2 + x - 16)$.
Then, I've tried to factor the second term: $(x+20)(x-16)(x+1)$. So, the answer would be "b", according to this algorithm.
I've completed the test (it's a simulation for the admission test I'm going to do), I submit the answers, and I've noticed that this question isn't correct.
 A: As @Fernis pointed out in the comments,

You cannot factor out the $(x+20)$ as you have done. There is no common factor of $(x+20)$ between $x^2+x−16$ and $20$.

Using the Rational Root Theorem, you can know that the possible rational roots are $\pm 1, \pm2, \pm4, \pm5, \pm10, \pm20$.
Through inspection and polynomial/synthetic division, you can get $(x-2)^2(x-5)$, as @saulspatz said. Therefore, (d) is your answer.
A: The easiest to try is (d), because it says that the polynomial has a double root. We will look for a root of the derivative and check if it cancels the polynomial.
$$3x^2+2x-16=0\iff x=2\text{ or }x=-\dfrac83.$$
Now $p(2)=0$, bingo !
A: This one is easy to factor. Let's see how.
Start by plugging in $x=0,1,-1,2$ and so on.
You will find by inspection that $x=2$ is a zero of the polynomial. Therefore,$(x-2)$ is its factor.
Now factor the polynomial in such a way that $(x-2)$ gets common.
$$x^3+x^2-16x+20$$
$$=x^3-2x^2+3x^2-6x-10x+20$$
$$=x^2(x-2)+3x(x-2)-10(x-2)$$
$$=(x-2)(x^2+3x-10)$$
$$=(x-2)(x^2+5x-2x-10)$$
$$=(x-2)[x(x+5)-2(x+5)]$$
$$=(x-2)^2(x+5)$$
