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.
 A: 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...
A: Note that
$$t^{34}-1=(t^{33}+t^{31}+\cdots+t+1)(t-1)$$
and so, substituting $t=x^3$, we get
$$x^{102}-1=(x^{99}+x^{96}+\cdots+x^3+1)(x^3-1)$$
So any real root of $x^{99}+x^{96}+\cdots+x^3+1$ will be a real root of $x^{102}-1$ (and those  should be easy to find). But note that, for example, $1$ is a real root of $x^{102}-1$, but is not a root of $x^{99}+x^{96}+\cdots+x^3+1$, since $34=1+1+\cdots+1\neq0$. So, once you find the real roots of $x^{102}-1$ and determine which of them is in fact a root of $x^{99}+x^{96}+\cdots+x^3+1$, you can combine with the real roots of $x^2-3x+2=(x-1)(x-2)$ to get the answer.
To factorize $x^{99}+x^{96}+\cdots+x^3+1$ into irreducibles over $\mathbb{Z}$ (which, it turns out, is equivalent to factoring into irreducibles over $\mathbb{Q}$ in this case), we use the fact that
$$x^{99}+x^{96}+\cdots+x^3+1=\frac{x^{102}-1}{x^3-1}$$
combined with the fact that
$$x^{102}-1=\prod_{d\mid 102}\Phi_d(x)=\Phi_{102}(x)\Phi_{51}(x)\Phi_{34}(x)\Phi_{17}(x)\Phi_6(x)\Phi_3(x)\Phi_2(x)\Phi_1(x)$$
where $\Phi_d(x)$ is the $d$th cyclotomic polynomial. The cyclotomic polynomials are all irreducible over $\mathbb{Q}$. Any irreducible polynomial in $\mathbb{R}[x]$, though, is either a linear $x-a$ for $a\in\mathbb{R}$, or a quadratic $x^2+ax+b$ for which $a^2-4b<0$. The factorization into irreducibles over $\mathbb{R}$ is just $(x-1)$, $(x+1)$, and then a bunch of quadratics $$x^2-(\zeta_{102}^k+\overline{\zeta_{102}}^k)x+1=(x-\zeta_{102}^k)(x-\overline{\zeta_{102}}^k)$$
where $\zeta_{102}$ is a primitive $102$th root of unity and $0<k<51$.
Of course, the factorization into irreducibles over $\mathbb{C}$ is just 
$$(x-1)(x-\zeta_{102})(x-\zeta_{102}^2)\cdots(x-\zeta_{102}^{50})(x+1)(x-\zeta_{102}^{52})\cdots(x-\zeta_{102}^{101})$$
Wolfram Alpha has a nice printout with the conjugate pairs, it may be helpful.
