On factorizing  and solving the polynomial: $x^{101} – 3x^{100} + 2x^{99} + x^{98} – 3x^{97} + 2x^{96} + \cdots + x^2 – 3x + 2 = 0$ 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.

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...
