By either long or synthetic division, it's easy to show that
[tex]\dfrac{x^4+ax^3+bx^2+cx+d}{x+1}=x^3+(a-1)x^2+(-a+b+1)x+(a-b+c-1)-\dfrac{a-b+c-d-1}{x+1}[/tex]
The quartic will be exactly divisible by [tex]x+1[/tex] when the numerator of the remainder term vanishes, or for those values of [tex]a,b,c,d[/tex] such that
[tex]a-b+c-d-1=0[/tex]
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.
