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.

We notice that x = 1 is a multiple root. If the order of multiplicity is 2, it means that x = 1 is the root of the polynomial and the root of the first derivative of the polynomial.

p(1) = 0

p'(1) = 0

We'll compute p(1), substituting x by 1 in the expression of polynomial:

p(1) = a + b + 1

a + b + 1 = 0

a + b = -1 (1)

To compute p'(1), we'll have to determine the expression of the first derivative:

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

We'll substitute x by 1:

p'(1) = 4a + 3b

4a + 3b = 0 (2)

We'll multipy (1) by -3:

-3a - 3b = 3 (3)

We'll add (2) + (3):

4a + 3b - 3a - 3b = 3

We'll combine and eliminate like terms:

a = 3

3 + b = -1

b = -4

**The polynomial is: p(x) = 3x^4 - 4x^3 + 1**