Are You a Good Critic?

When looking for integer roots, we must first determine if the polynomial is reducible over the integers. One of the well-known irreducibility criterion is the $Perron's$ $Irreducibility$ $Criterion$ . The criterion states that for a polynomial $P(x)$ ,

$P(x) = x^n+ a_{n-1}x^{n-1} + a_{n-2}x^{n-2}+$ $...$ $+ a_1x+a_0$ with $a_0 \neq 0$

If $|a_{n-1}| > 1+|a_{n-2}|+|a_{n-3}|+$ $...$ $+|a_1|+|a_0|$ , then $P(x)$ is irreducible over the integers.

Since in the given polynomial,

$|-1024| > 1 + |512| + |-256| +$ $...$ $+ |8| + |-4| = 1021$

$1024 > 1021$

Then, the polynomial is irreducible over the integers, hence, it has $\boxed 0$ integer roots.

Useful link: Perron's Irreducibility Criterion starts on page 2 of this (It also contains the proof of this and as well as other criterion, I believe.)

Using rational root theorem, rational roots must be one of the following: $-4, -2, -1, 1, 2, 4$ . The negative roots can easily be discarded, as all the terms will be negative if a negative number is plugged in. 1 is also easy to discard, as the sum of the coefficients are not 0. Plugging in 2 and 4 and evaluating results in a non zero number, meaning that there are no integer roots.