By either long or synthetic division, it's easy to show that
The quartic will be exactly divisible by
when the numerator of the remainder term vanishes, or for those values of 
such that
I'm not sure how to count the number of solutions (software tells me it should be 80), but hopefully this is a helpful push in the right direction.
The quartic will be exactly divisible by
I'm not sure how to count the number of solutions (software tells me it should be 80), but hopefully this is a helpful push in the right direction.
