As the title says, I'm supposed to show $1-999x^{888}\in \mathbb{Q}[x]$ is irreducible.

In a previous part of the question I had to show $x^{888} -999\in \mathbb{Q}[x]$ was irreducible which I did using Eisenstein's criterion, but the same can't be done here and I can't see any way to use the previous result.

Any help you could offer would be appreciated.


To see that the reducibility of a polynomial implies the reducibility of its reciprocal:

Consider the polynomial $P(x)=\sum_{i=0}^da_ix^i=a_0+a_ax+\cdots +a_dx^d$. Then the "reciprocal" polynomial $Q(x)$ is defined by $$Q(x)=\sum_{i=0}^d a_ix^{d-i}=a_d+a_{d-1}x+\cdots + a_0x^d$$.

That's the situation you have, with $P(x)=1-999x^{888}$.

We want to show that $P(x)$ reducible implies that $Q(x)$ is reducible as well (given what you have already shown, that will complete the proof).

The key point is to note that the reciprocal is given by $$Q(x)=x^dP\left( \frac 1x\right)$$

Suppose $P(x)$ factors as $F_1(x)\times F_2(x)$ with degree $F_i$ = $d_i$ with $1<d_i<d$, and $d=d_1+d_2$. Then $$Q(x)=x^{d_1+d_2}F_1\left( \frac 1x\right)F_2\left(\frac 1x\right)=\left(x^{d_1}F_1\left( \frac 1x\right)\right)\times \left(x^{d_2}F_2\left( \frac 1x\right)\right)$$

From which we deduce that $Q(x)$, the reciprocal of $P(x)$,factors as the product of the reciprocals of the factors of $P(x)$, and we are done.


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