What are a and b if p(x)=ax^4+bx^3+1 is divided by (x-1)^2?

Given that ax^4+bx^3+1 is divisible by
(x-1)^2.

Let p(x) =
x^4+bx^3+1.

Since (x-1)  is  a factor of p(x), by remainder
theorem, p(1) = 0.

=> p(1) = a*1^4+b*1^3+1 =
0.

So a+b +1 = 0.

Or a+b =
-1.....(1)

Since (x-1)^2  is factor of p(x),  x-1 is a
factor of p'(x) also.

p'(x) = (ax^4+bx^3+c)' = 4ax^3+3bx^2
.

So p'(1) = 0 => {(aX^4+bX^3+1) at x =
1} = 0.

=> 4a+3b =
0...(1).

From (1) a = -(b+1). We put a= -(b+1) in
(2):

-4*(b+1)+3b = 0.

-b-4 =
0.

b = -4.

So a= -(b+1) =
-(-4+1) = 3.

Therefore a= 3 and b = -4 in ax^4+bx^3+1 to
have a factor (x-1)

Therefore ax^4+bx^3+1 =
3x^4 -4x^3+1 = 0.