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

Options

A. `(3)/(x-1)-(1)/((x-1)^(2))-(7)/((x-1)^(3))+(5)/((x-1)^(4))`
B. `(3)/(x-1)-(7)/((x-1)^(3))+(5)/((x-1)^(4))`
C. `(3)/(x-1)+(1)/((x-1)^(2))-(7)/((x-1)^(3))+(5)/((x-1)^(4))`
D. `(3)/(x-1)-(1)/((x-1)^(2))+(7)/((x-1)^(3))-(5)/((x-1)^(4))`

Answer

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

### Step 1: Set Up the Partial Fraction Decomposition
We express the given fraction in terms of its partial fractions. Since the denominator is \((x - 1)^4\), we can write:

\[ \frac{3x^2 - 8x + 10}{(x - 1)^4} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{(x - 1)^3} + \frac{D}{(x - 1)^4} \]

### Step 2: Clear the Denominator
Multiply both sides by \((x - 1)^4\) to eliminate the denominator:

\[ 3x^2 - 8x + 10 = A(x - 1)^3 + B(x - 1)^2 + C(x - 1) + D \]

### Step 3: Expand the Right Side
Now, we 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. \(D = D\)

Combining these gives:

\[ 3x^2 - 8x + 10 = Ax^3 + (-3A + B)x^2 + (3A - 2B + C)x + (-A + B + C + D) \]

### Step 4: Equate Coefficients
Now we equate the coefficients of the corresponding powers of \(x\) on both sides:

- For \(x^3\): \(A = 0\)
- For \(x^2\): \(-3A + B = 3\)
- For \(x^1\): \(3A - 2B + C = -8\)
- For the constant term: \(-A + B + C + D = 10\)

### Step 5: Solve the System of Equations
Substituting \(A = 0\) into the equations:

1. From \(-3(0) + B = 3\), we get \(B = 3\).
2. From \(3(0) - 2(3) + C = -8\), we have \(-6 + C = -8\) which gives \(C = -2\).
3. From \(-0 + 3 - 2 + D = 10\), we have \(1 + D = 10\) which gives \(D = 9\).

### Step 6: Write the Final Answer
Now we can substitute \(A\), \(B\), \(C\), and \(D\) back into the partial fraction form:

\[ \frac{3x^2 - 8x + 10}{(x - 1)^4} = \frac{0}{x - 1} + \frac{3}{(x - 1)^2} - \frac{2}{(x - 1)^3} + \frac{9}{(x - 1)^4} \]

Thus, the final answer is:

\[ \frac{3}{(x - 1)^2} - \frac{2}{(x - 1)^3} + \frac{9}{(x - 1)^4} \]