The partial fractions of `(x^(3)-5)/(x^(2)-3x+2)` are

Options

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

Answer

To find the partial fractions of the expression \(\frac{x^3 - 5}{x^2 - 3x + 2}\), we will follow these steps:

### Step 1: Factor the Denominator
First, we need to factor the denominator \(x^2 - 3x + 2\).

\[ x^2 - 3x + 2 = (x - 1)(x - 2) \]

### Step 2: Set Up the Partial Fraction Decomposition
We can express the fraction as a sum of partial fractions:

\[ \frac{x^3 - 5}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2} \]

where \(A\) and \(B\) are constants that we need to determine.

### Step 3: Clear the Denominator
Multiply both sides by the denominator \((x - 1)(x - 2)\) to eliminate the fraction:

\[ x^3 - 5 = A(x - 2) + B(x - 1) \]

### Step 4: Expand the Right Side
Now, expand the right side:

\[ x^3 - 5 = Ax - 2A + Bx - B \]
\[ = (A + B)x - (2A + B) \]

### Step 5: Equate Coefficients
Now, we equate the coefficients from both sides of the equation. The left side has a cubic term, so we need to ensure that the coefficients match:

1. Coefficient of \(x^3\): \(1 = 0\) (since there is no \(x^3\) term on the right)
2. Coefficient of \(x^2\): \(0 = 0\) (no \(x^2\) term on the right)
3. Coefficient of \(x\): \(0 = A + B\)
4. Constant term: \(-5 = -2A - B\)

### Step 6: Solve the System of Equations
From the equations, we have:

1. \(A + B = 0\) (Equation 1)
2. \(-2A - B = -5\) (Equation 2)

Substituting \(B = -A\) from Equation 1 into Equation 2:

\[ -2A - (-A) = -5 \]
\[ -2A + A = -5 \]
\[ -A = -5 \implies A = 5 \]

Now substitute \(A = 5\) back into Equation 1:

\[ 5 + B = 0 \implies B = -5 \]

### Step 7: Write the Partial Fraction Decomposition
Now that we have \(A\) and \(B\), we can write the partial fraction decomposition:

\[ \frac{x^3 - 5}{(x - 1)(x - 2)} = \frac{5}{x - 1} - \frac{5}{x - 2} \]

### Final Answer
Thus, the partial fractions of \(\frac{x^3 - 5}{x^2 - 3x + 2}\) are:

\[ \frac{5}{x - 1} - \frac{5}{x - 2} \]

---