Irreducibility of $1+x+x^2+\cdots+x^{96}+x^{97}+x^{98}$ The question I encounter is

Determine whether the polynomial $1+x+x^2+\cdots+x^{96}+x^{97}+x^{98}$ is irreducible in $\mathbb Q[x]$ or not

All the methods that I learnt (Eisenstein Criterion, Rational Root Theorem and reduce the polynomial to $\mathbb F_p$) fails. I have no idea towards the question.
 A: Here’s the secret that nobody is telling you: $X^{99}-1$ is the polynomial whose roots are the ninety-ninth roots of unity, of which one is $1$, so that $X-1$ divides $X^{99}-1$, with quotient your polynomial. But (moderately) advanced mathematics tells you that $X^n-1$ is the product of the “cyclotomic polynomials” $\Phi_d(X)$, where $d$ ranges through the divisors of $n$. $\Phi_d$ is the polynomial whose roots are the $d$-th roots of unity that aren’t $d'$-th roots of unity for any $d'<d$, so that $\Phi_1(X)=X-1$, $\Phi_2(X)=X+1$, and $\Phi_3(X)=X^2+X+1$. All the cyclotomic polynomials have their coefficients in $\Bbb Z$.
As a particular example of what I said above, we have:
$$
X^{99}-1=\Phi_1\Phi_3\Phi_9\Phi_{11}\Phi_{33}\Phi_{99}\,,
$$
and your polynomial is the above product with the factor $X-1=\Phi_1$ omitted. The factors that people have exhibited are $\Phi_3$, $\Phi_9$, and $\Phi_{11}$.
A: Your polynomial is divisible by $1+x+x^2$. One can verify this in many ways, for example polynomial division. Or else divide into groups of $3$, the first being $1+x+x^2$, the second $x^3(1+x+x^2)$, and so on. Or else one can see that $e^{\pm 2\pi i/3}$ are roots.
A: For those keeping score at home: THIS JUST POPPED INTO MY HEAD. Multiply it out.
$$ (1 + x + x^2 + \cdots + x^{10})(1 + x^{11}  + x^{22} + \cdots  + x^{88}  )  $$
OR
$$ (1 + x + x^2 + \cdots + x^{8})(1 + x^{9}  + x^{18} + \cdots  + x^{90}  )  $$
