Two roots of the polynomial $x^4+x^3-7x^2-x+6$ are $2$ and $-3$. Find two other roots. I have divided this polynomial first with $(x-2)$ and then divided with $(x+3)$ the quotient. The other quotient I have set equal to $0$ and have found the other two roots. Can you explain to me if these actions are correct and why?

Yes, provided the computings are OK, what you have done is what must be done.
Given a polynomial $P(x)$, and a root of its, $a$, we have that $P(a)=0$. But if you divide $P(x)$ by $x-a$, you will get a quotient $Q(x)$ and a remainder $r$, which is a number because the degree of the divisor is $1$.
Then
$$P(x)=Q(x)(x-a)+r$$
and therefore
$$0=P(a)=Q(a)(a-a)+r=r$$
that is, the division has remainder $0$.
Now, we know that $r=0$, so if $b$ is another root of $P$,
$$0=P(b)=Q(b)(b-a)$$
and since $b-a\neq 0$, $Q(b)=0$, that is, $b$ is a root of $Q$.
So yes, if you divide the given polynomial $P$ by the factors associated to the known roots (as you have done), the roots of the quotient are the remaining roots of $P$.
