How can other unknowns be calculated in polynomial division given there are no remainders? I’ve being practicing polynomial long division for the last week and have built some competence/confidence around the algorithm for performing the operation, but this is stumping me:
Given P($x$) = $(x^3-2x^2-x+2)/(x-k)$ has three values for k in which thee quotient has no remainder, what are the possible k values?
I’m not sure where to begin.  The idea of the modulus function comes to mind, but I’m not sure that’s the way either.  I’m posting in the hope someone could enlightenment me with the general strategy for this.
 A: A useful theorem you can use is the following:
Let $P\in\mathbb{R}\left[X\right]$ be a polynomial over the reals. Then $a$ is a root of $P$ if and only if $\left(x-a\right)$ divides $P(x)$. (This is true not only for the reals but it will suffice in this case).
The fact that $P$ is divisible by a polynomial of the form $x-k$ means that there exists a polynomial $Q$ with $\deg Q =2$ such that $P(x)=(x-k)Q(x)$.
So by finding a root of the numerator polynomial, you can represent the numerator like the product above, divide by $(x-k)$, i.e. you found one value you were looking for, and may proceed by finding other real roots if they exist (in this case they do).
A: After more research, I found a way to solve for x in a cubic.  Dividing all of the factors of d by the factors in a and adding to these the set of their selves to the unit negative (-1), thus reducing the trial space to, in this instance, a cardinality of four: -2, -1, 1 and 2. Of these candidate values, -1, 1 and 2 were valid values for k in the original post via either synthetic division or trial and error.
A: What they are basically trying to ask is:
What are the roots of the equation $x^{3}-2x^{2}-x+2$? (k=the roots)
This is because they are asking for k such that the quotient has no remainder. This implies that they want you to factor the cubic.
Answer spoiler: (try not to look to the right) k=2,1,-1
But you may be asking how do we factor a cubic.
There are 3 ways in which I know:
Cardano's formula (It's a "cubic" formula, like a quadratic formula)
Lagrange resolvents (Substitution to another variable to convert it to a quadratic)
Factorisation (Encouraged for this example).
I'm not going to take the fun out of factorisation, but here's how you would do it:
expand $(x+a)(x+b)(x+c)$, equate coefficients in your example, and solve for the variables, (assuming that a,b,c are integers would be helpful as they are commonly the case).
