TITLE: On factorizing  and solving the polynomial: $x^{101} – 3x^{100} + 2x^{99} + x^{98} – 3x^{97} + 2x^{96} + \cdots + x^2 – 3x + 2 = 0$
QUESTION [6 upvotes]: The actual problem is to find the product of all the real roots of this equation,I am stuck with his factorization:
$$x^{101} – 3x^{100} + 2x^{99} + x^{98} – 3x^{97} + 2x^{96} + \cdots + x^2 – 3x + 2 = 0$$
By just guessing I noticed that $(x^2 – 3x + 2)$ is one factor and then dividing that whole thing we get $(x^{99}+x^{96}+x^{93} + \cdots + 1)$ as the other factor , but I really don't know how to solve in those where wild guessing won't work! Do we have any trick for factorizing this kind of big polynomial?
Also I am not sure how to find the roots of  $(x^{99}+x^{96}+x^{93} + \cdots + 1)=0$,so any help in this regard will be appreciated.

REPLY [6 votes]: In regard to the first part of your question ("wild guessing"),  the point was to note that the polynomial can be expressed as the sum of three polynomials, grouping same coefficients:
$$  P(x)= x^{101} – 3x^{100} + 2x^{99} + x^{98} – 3x^{97} + 2x^{96} + \cdots + x^2 – 3x + 2 
 = A(x)+B(x)+C(x)$$
with
$$\begin{eqnarray}
 A(x) &= x^{101} + x^{98} + \cdots + x^2  &= x^2 (x^{99} + x^{96} + \cdots + 1) \\
 B(x) &= - 3 x^{100} -3 x^{97} - \cdots -3 x &= - 3 x (x^{99} + x^{96} + \cdots + 1)\\
 C(x) &= 2 x^{99} + 2 x^{96} + \cdots + 2 &= 2 (x^{99} + x^{96} + \cdots + 1) \\
\end{eqnarray}
$$
so $$P(x) = (x^2 - 3x +2) (x^{99} + x^{96} + \cdots + 1) $$ 
and applying the geometric finite sum formula:
$$P(x)=(x^2 - 3x +2) ({(x^{3})}^{33} + {(x^{3})}^{32} + \cdots + 1) = (x^2 - 3x +2) \frac{x^{102}-1}{x^3-1} $$
As Andre notes in the comments, your "guessing" was dictated by the very particular structure of the polynomial, you can't hope for some general guessing recipe...
{"set_name": "stack_exchange", "score": 6, "question_id": 62655}