We have to find the remainder when f=3x^4-2x^3+x^2+ax-1 is divided by (x-1)^2

(x-1)^2

=> x^2 - 2x + 1

We use long division to divide 3x^4-2x^3+x^2+ax-1 by x^2 - 2x + 1.

x^2 - 2x + 1 | 3x^4-2x^3+x^2+ax-1 | 3x^2 + 4x+6

......................3x^4-6x^3+3x^2

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..............................4x^3 - 2x^2 +ax - 1

..............................4x^3 - 8x^2 + 4x

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........................................6x^2 +(a-4)x - 1

........................................6x^2 - 12x + 6

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.................................................. +(a + 8)x - 7

**Therefore the remainder is (a+8)x -7**.

We'll write the division with reminder:

f(x) = (x-1)^2*C(x) + R(x)

Since the divisor is a polynomial of second order, the reminder is a polynomial of first order.

R(x) = mx + n

We'll calculate f(1):

f(1) = (1-1)^2*C(1) + R(1)

f(1) = R(1)

We'll determine f(1), substituting x by 1 in the given expression of polynomial f:

f(1) = 3-2+1+a-1

f(1) = 1 + a

R(1) = m + n

m + n = 1+a (1)

Now, we'll calculate the first derivative of f(x):

f'(x) = (3x^4-2x^3+x^2+ax-1)'

f'(x) = 12x^3 - 6x^2 + 2x + a

f'(x) = [(x-1)^2*C(x) + R(x)]'

f'(x) = 2(x-1)*C(x) + (x-1)^2*C'(x) + R'(x)

f'(x) = 2(x-1)*C(x) + (x-1)^2*C'(x) + m

f'(1) = m

f'(1) = 12 - 6 + 2 + a

m = 8 + a (2)

We'll substitute (2) inĀ (1):

1+a = 8 + a + n

We'll subtract 8 + a:

n = 1 + a - 8 - a

We'll eliminate and combine like terms:

n = -7

**The reminder R(x) is:**

**R(x) = (8 + a)x - 7**