Find all integers $a$ for which $x^2-x+a$ divide $x^{13}+x+90$. Find all integers $a$ for which $x^2-x+a$ divide $x^{13}+x+90$.
The answer is $a=2$. 

If $a$ is negative or zero, then the quadratic has two real roots.
But we can easily check that the other polynomial has derivative everywhere positive
and hence only one real root.
So $a$ must be positive. 
If $x^2-x+a$ divides $x^{13}+x+90$, then $x^{13}+x+90=f(x)(x^2-x+a)$,
where $f(x)$ is a polynomial with integer coefficients.
Let $x=0$, we see that $a$ must divide $90$. Let $x=1$, we see that it must divide $92.$
Hence it must divide $92-90=2.$ So the only possibilities are $1$ and $2.$
Suppose $a=1$, then putting $x=2$, we have that $3$ divides $2^{13}+92$ but $2^{\text{odd}}$ is congruent to $2 \mod 3,$ so $2^{13}+92$ is congruent to $1 \mod 3.$ 
So $a$ cannot be $1.$ 
To see that $a=2$ is possible, we write
$(x^2-x+2)(x^{11}+x^{10}-x^9-3x^8-x^7+5x^6+7x^5-3x^4-17x^3-11x^2+23x+45)=x^{13}+x+90$.
