Comparing coefficients to find the other roots of a polynomial? Suppose I am given that $(x - \alpha)$ is a factor of $p(x) = a_0x^n + a_1x^{n-1} + .... a_n$
That obviously means $(x - \alpha)$ is a factor of $p(x)$ i.e $p(x) = (x - \alpha)q(x)$. My question is, if I want to find $q(x)$, and therefore the other roots, can I just multiply $(x - \alpha)$ by $q(x)$ and compare coefficients? Because theoretically, $q(x)$ being of degree $n - 1$ would have $n$ coefficients and if I compare them to $p(x)$ with $n+1$ coefficients, I would have n variables and n+1 equations, meaning a solution isn't guaranteed. Is this correct? If so, what does this mean, and how do I find the other roots if this happens?
 A: Theorem 3,4 from this page comes in handy. The theorem is quite general, but we actually only need a special case to answer your question:

Suppose $p(x)$ is a polynomial of degree $n\geq1$ and $d(x)$ is a linear function of the form $d(x)=ax+b$. Then, there exists a unique polynomial $q(x)$ of degree $n-1$ such that
$$
p(x)=q(x)d(x)+r \, ,
$$
where $r$, the remainder, is a constant.

Let's see this theorem in action. Suppose $p(x)=2x^3-3x^2-3x+7$ and $d(x)=x-2$. Then, by Theorem 3.4, we know that there exists a unique polynomial $q(x)$ such that
$$
2x^3-3x^2-3x+7=q(x)(x-2)+r \, .
$$
Since the degree of $p(x)$ is $3$, the degree of $q(x)$ is $2$, and so $q(x)$ must be of the form $Ax^2+Bx+C$. Hence,
$$
2x^3-3x^2-3x+7=(Ax^2+Bx+C)(x-2)+r \, .
$$
If we expand the RHS, then we get
$$
2x^3-3x^2-3x+7 \equiv Ax^3+(B-2A)x^2 + (C-2B)x + (-2C+r) \, .
$$
From this obtain
\begin{align}
A &= 2 \\
B-2A &= -3 \\
C-2B &= -3 \\
-2C + r &= 7 
\end{align}
So that's four equations with four unknowns. Note however that not every system of equations with four equations and four unknowns has a solution. Consider the system
\begin{align}
w+x+2y+5z &= 2 \\
w+x+2y+5z &= 3 \\
4w+2x+4y+10z &= 6 \\
7w+4x-5y+z &= -11 \, .
\end{align}
Since the first two equations contradict each other, there are no solutions. If we change the second equation  to $2w+x+2y+5z = 3$, then we get the opposite problem: since equations $2$ and $3$ are multiples of each other, there are infinitely many solutions.
However, in our case Theorem 3.4 guarantees that there is a unique polynomial $q(x)=Ax^2+Bx+C$ such that
$$
2x^3-3x^2-3x+7=(Ax^2+Bx+C)(x-2)+r \, .
$$
If the solutions for $A$, $B$, and $C$ didn't exist, then $q(x)$ wouldn't exist, contradicting Theorem 3.4. Likewise, if $A$, $B$, and $C$ had infinitely many solutions, then $q(x)$ wouldn't be unique, also contradicting Theorem 3.4.
I'll end this by saying that it is good that you were sceptical about whether or not a solution was guaranteed. As Marc van Leeuwen puts it in this post,

If you take a docile attitude and accept the statement to be proved from the onset, you will never be able to understand what all this reasoning was needed for in the first place.

However, in this particular case, you have nothing to worry about.
