`(3x^(2)+x+1)/(x-1)^(4)=a/(x-1)+b/(x-1)^(2)+c/(x-1)^(3)+d/(x-1)^(4) rArr [{:(a,b), (c, d):}]=`

Options

A. `[(3,7),(5,0)]`
B. `[(0,3),(7,5)]`
C. `[(0,7),(3,5)]`
D. `[(3,5),(7, 0)]`

Answer

To solve the problem, we need to express the rational function \(\frac{3x^2 + x + 1}{(x-1)^4}\) in terms of partial fractions. We will set it equal to the form:

\[ \frac{a}{x-1} + \frac{b}{(x-1)^2} + \frac{c}{(x-1)^3} + \frac{d}{(x-1)^4} \]

### Step 1: Set up the equation
We start with the equation:

\[ \frac{3x^2 + x + 1}{(x-1)^4} = \frac{a}{x-1} + \frac{b}{(x-1)^2} + \frac{c}{(x-1)^3} + \frac{d}{(x-1)^4} \]

### Step 2: Multiply through by \((x-1)^4\)
To eliminate the denominators, we multiply both sides by \((x-1)^4\):

\[ 3x^2 + x + 1 = a(x-1)^3 + b(x-1)^2 + c(x-1) + d \]

### Step 3: Expand the right-hand side
Now we will expand the right-hand side:

1. \(a(x-1)^3 = a(x^3 - 3x^2 + 3x - 1)\)
2. \(b(x-1)^2 = b(x^2 - 2x + 1)\)
3. \(c(x-1) = cx - c\)
4. \(d = d\)

Combining these, we get:

\[ 3x^2 + x + 1 = ax^3 + (-3a + b)x^2 + (3a - 2b + c)x + (-a + b - c + d) \]

### Step 4: Equate coefficients
Now we equate the coefficients of the polynomial on both sides:

- Coefficient of \(x^3\): \(a = 0\)
- Coefficient of \(x^2\): \(-3a + b = 3\)
- Coefficient of \(x\): \(3a - 2b + c = 1\)
- Constant term: \(-a + b - c + d = 1\)

### Step 5: Solve the equations
From \(a = 0\), we substitute into the other equations:

1. From \(-3(0) + b = 3\) → \(b = 3\)
2. From \(3(0) - 2(3) + c = 1\) → \(-6 + c = 1\) → \(c = 7\)
3. From \(-0 + 3 - 7 + d = 1\) → \(-4 + d = 1\) → \(d = 5\)

### Step 6: Compile the results
Now we have:

- \(a = 0\)
- \(b = 3\)
- \(c = 7\)
- \(d = 5\)

Thus, the values are:

\[ (a, b) = (0, 3) \quad \text{and} \quad (c, d) = (7, 5) \]

### Step 7: Form the array
The array \(\{(a, b), (c, d)\}\) is:

\[ \begin{pmatrix} 0 & 3 \\ 7 & 5 \end{pmatrix} \]

### Final Answer
The answer is:

\[ \{(0, 3), (7, 5)\} \]