Question

find the values of a and b so that the polynomial p(x)=x^4+x^3+8x^3-ax+b is exactly divisible by x^2-1.

Answer 1

$(x^4+9x^3-ax+b):(x^2-1)=x^2+9x+1\\-x^4+9x^3+x^2-ax+b\\-9x^3+x^2+9x-ax+b\\-x^2+9x-ax+b+1\\9x-ax+b+1$

The rest of divide is 9x - ax + b + 1 and it must be equal to 0:
(9 - a)x + b + 1 = 0
For all x:
(9 - a)x + b + 1 = 0    ⇔  9 - a = 0 and b + 1 = 0 ⇔ a = 9 and b = -1
Check now the result of dividing a polynomial by binomial:
$(x^4+9x^3-9x-1):(x^2-1)$
This polynomial is exatly (no rest) divisible by binomial!

Answer: a = 9,  b = -1