Find the complex $a$,such $x^2-x+a|x^{13}+x+90$ If $a$ be complex number,such 
$$x^2-x+a|x^{13}+x+90$$find $a$
I think  it is equality if $r^2-r+a=0$  $$\Longrightarrow r^{13}+r+90=0$$,find the $a$
 A: $$x^2-x+a|x^{13}+x+90$$
Hoping you may get lucky finding an 'integer' $a$, here is a simple shortcut
When $x=0$, you want $a | 90$
When $x=1$, you want $a | 92$
Can you guess $a$ that satisfies both above requirements 
A: As remarked by @rsadhvika, $a=\pm 1 $ or $\pm 2$. If $a=-2$ then the polynomial has $-1$ as a root, this gives a contradiction. If $a=+1$ let $x=2$ this gives a non divisiblity, and if $a=-1$, let $x=-2$, another non divisibility. So $a=2$.
A: Is cheating allowed?
>> P=[1 0 0 0 0 0 0 0 0 0 0 0 1 90]

P =

     1     0     0     0     0     0     0     0     0     0     0     0     1    90

>> r=roots(P)

r =

   1.3740 + 0.3391i
   1.3740 - 0.3391i
   1.0583 + 0.9391i
   1.0583 - 0.9391i
   0.5000 + 1.3229i
   0.5000 - 1.3229i
  -0.1721 + 1.4029i
  -0.1721 - 1.4029i
  -0.8036 + 1.1618i
  -0.8036 - 1.1618i
  -1.4119 + 0.0000i
  -1.2507 + 0.6555i
  -1.2507 - 0.6555i

>> r(5)*r(6)

ans =

    2.0000

So we found all the roots of the polynomial and selected the $2$ roots that added to $1$ so as to match the linear term of the quadratic factor. Then their product was $a=2$.
