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?
 A: Your idea is correct. Here is another way, without the division.
The sum of all the roots is $-1$. The sum of the given roots is $-1$. So the sum of the missing roots is $0$.
The product of all the roots is $-6$. The product of the given roots is $-6$. So the product of the missing roots is $-1$.
The missing roots satisfy the equation $x^2-1=0$.
A: 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 $P$.
A: Your factorisation 
$(x-2)(x+3)(x-1)(x+1)$ 
is zero if $x \in \{2,-3,1,-1\}$ since multiplication by zero gives zero, 
but it is non-zero for all other values of $x$ since the product of four non-zero numbers is non-zero. 
A: $(x-2)(x+3)=x^2+x-6$
$x^4+x^3-7x^2-x+6\\=(x^4+x^3-6x^2)-(x^2+x-6)\\= x^2(x^2+x-6)-1(x^2+x-6)\\=(x^2-1)(x^2+x-6)$
