Resolve `(3x^(3)-8x^(2)+10)/((x-1)^(4))` into partial fractions.

Answer

To resolve the expression \(\frac{3x^3 - 8x^2 + 10}{(x-1)^4}\) into partial fractions, we will follow these steps:

### Step 1: Set up the Partial Fraction Decomposition
The denominator \((x-1)^4\) suggests that we can express the fraction as:
\[ \frac{3x^3 - 8x^2 + 10}{(x-1)^4} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3} + \frac{D}{(x-1)^4} \]
where \(A\), \(B\), \(C\), and \(D\) are constants we need to determine.

### Step 2: Multiply through by the Denominator
To eliminate the denominator, we multiply both sides by \((x-1)^4\):
\[ 3x^3 - 8x^2 + 10 = A(x-1)^3 + B(x-1)^2 + C(x-1) + D \]

### Step 3: Expand the Right Side
Now we need to expand the right side:
1. \(A(x-1)^3 = A(x^3 - 3x^2 + 3x - 1)\)
2. \(B(x-1)^2 = B(x^2 - 2x + 1)\)
3. \(C(x-1) = C(x - 1)\)
4. \(D\) is just \(D\)

Combining these, we have:
\[ 3x^3 - 8x^2 + 10 = Ax^3 + (-3A + B)x^2 + (3A - 2B + C)x + (-A + B - C + D) \]

### Step 4: Compare Coefficients
Now we will compare the coefficients of \(x^3\), \(x^2\), \(x\), and the constant term on both sides.

1. Coefficient of \(x^3\):
\[ A = 3 \]

2. Coefficient of \(x^2\):
\[ -3A + B = -8 \implies -3(3) + B = -8 \implies B = 1 \]

3. Coefficient of \(x\):
\[ 3A - 2B + C = 0 \implies 3(3) - 2(1) + C = 0 \implies C = -7 \]

4. Constant term:
\[ -A + B - C + D = 10 \implies -3 + 1 - (-7) + D = 10 \implies D = 5 \]

### Step 5: Write the Partial Fraction Decomposition
Now that we have determined the values of \(A\), \(B\), \(C\), and \(D\):
- \(A = 3\)
- \(B = 1\)
- \(C = -7\)
- \(D = 5\)

We can write the partial fraction decomposition as:
\[ \frac{3}{x-1} + \frac{1}{(x-1)^2} - \frac{7}{(x-1)^3} + \frac{5}{(x-1)^4} \]

### Final Answer
Thus, the resolved partial fractions are:
\[ \frac{3}{x-1} + \frac{1}{(x-1)^2} - \frac{7}{(x-1)^3} + \frac{5}{(x-1)^4} \]

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