Resolve into partial fractions the expression `(8x^(3)-5x^(2)+2x+4)/((2x-1)^(2)(3x^(2)+4))`

Answer

To resolve the expression \(\frac{8x^3 - 5x^2 + 2x + 4}{(2x - 1)^2(3x^2 + 4)}\) into partial fractions, we will follow these steps:

### Step 1: Set up the partial fraction decomposition
We can express the given rational function as:
\[ \frac{8x^3 - 5x^2 + 2x + 4}{(2x - 1)^2(3x^2 + 4)} = \frac{A}{2x - 1} + \frac{B}{(2x - 1)^2} + \frac{Cx + D}{3x^2 + 4} \]
where \(A\), \(B\), \(C\), and \(D\) are constants to be determined.

### Step 2: Multiply through by the denominator
Multiply both sides by \((2x - 1)^2(3x^2 + 4)\) to eliminate the denominators:
\[ 8x^3 - 5x^2 + 2x + 4 = A(2x - 1)(3x^2 + 4) + B(3x^2 + 4) + (Cx + D)(2x - 1)^2 \]

### Step 3: Expand the right-hand side
Now, we will expand the right-hand side:
1. Expand \(A(2x - 1)(3x^2 + 4)\):
\[ A(6x^3 + 8x - 3x^2 - 4) = A(6x^3 - 3x^2 + 8x - 4) \]

2. Expand \(B(3x^2 + 4)\):
\[ B(3x^2 + 4) = 3Bx^2 + 4B \]

3. Expand \((Cx + D)(2x - 1)^2\):
\[ (Cx + D)(4x^2 - 4x + 1) = 4Cx^3 + (4D - 4C)x^2 + Dx \]

Combining all these expansions:
\[ 8x^3 - 5x^2 + 2x + 4 = (6A + 4C)x^3 + (-3A + 3B + 4D - 4C)x^2 + (8A + 4B + D)x + (-4A + 4B) \]

### Step 4: Set up equations by comparing coefficients
Now we can equate the coefficients of like powers of \(x\):
1. For \(x^3\): \(6A + 4C = 8\)
2. For \(x^2\): \(-3A + 3B + 4D - 4C = -5\)
3. For \(x^1\): \(8A + 4B + D = 2\)
4. For the constant term: \(-4A + 4B = 4\)

### Step 5: Solve the system of equations
From the constant term equation:
\[ -4A + 4B = 4 \implies B = A + 1 \]

Substituting \(B\) into the other equations:
1. Substitute \(B = A + 1\) into \(6A + 4C = 8\):
\[ 6A + 4C = 8 \tag{1} \]

2. Substitute \(B = A + 1\) into \(-3A + 3B + 4D - 4C = -5\):
\[ -3A + 3(A + 1) + 4D - 4C = -5 \implies 4D - 4C = -5 \tag{2} \]

3. Substitute \(B = A + 1\) into \(8A + 4B + D = 2\):
\[ 8A + 4(A + 1) + D = 2 \implies D = 2 - 12A \tag{3} \]

Now we can substitute \(D\) from equation (3) into equation (2):
\[ 4(2 - 12A) - 4C = -5 \implies 8 - 48A - 4C = -5 \implies -4C = -5 - 8 + 48A \implies C = \frac{13 - 12A}{4} \]

Substituting \(C\) back into equation (1):
\[ 6A + 4\left(\frac{13 - 12A}{4}\right) = 8 \implies 6A + 13 - 12A = 8 \implies -6A = -5 \implies A = \frac{5}{6} \]

Now substituting \(A\) back to find \(B\), \(C\), and \(D\):
- \(B = \frac{5}{6} + 1 = \frac{11}{6}\)
- \(C = \frac{13 - 12 \cdot \frac{5}{6}}{4} = \frac{13 - 10}{4} = \frac{3}{4}\)
- \(D = 2 - 12 \cdot \frac{5}{6} = 2 - 10 = -8\)

### Final Step: Write the partial fractions
Thus, the partial fraction decomposition is:
\[ \frac{5/6}{2x - 1} + \frac{11/6}{(2x - 1)^2} + \frac{(3/4)x - 8}{3x^2 + 4} \]