Number of integral roots?

Question

The number of integral roots of the equation

$x^8-24x^7-18x^5+39x^2+1155=0$

I tried to figure out by considering the changes in signs which tell us that there are atmost $2$ positive real roots and no negative real root.

But that does not help out with integer roots.

Answer 1

There's another way. Rather simple and old school. Take the 1155 to the other side. (1155=3.5.7.11) Also take x^2 common from the expression. x^2 is not a factor of any of the primes of the 1155 (3.5.7.11) Hence no integral roots.

Answer 2

If $a$ is an integer root of $x^8-24x^7-18x^5+39x^2+1155$ then $a$ must be a factor of $1155=3\cdot5\cdot7\cdot11$ so there are $(1+1)(1+1)(1+1)(1+1)=16$ positive possible factors and a way to see that any of them is a root would be division by $(x-a)$ for possible $a$.

But is much better to remark that criterium of Eisenstein apply with the prime $3$ so the given polynomial is irreducible and there are not root even rational.

Answer 3

In my view the easiest approach is to use even-odd parity.
If $a$ is an odd number we would have 3 odd integers and 2 even integers which can never result in 0 ( an even integer ).
Similarly, if we say that $a$ is an even integer then only constant term is an odd term which means we can't get a 0 again.