Question

If \(\frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}\), then the values of (A, B, C) are

• A. (1, -5, 4)
• B. (1, 5, 4)
• C. (4, 5, 1)
• D. (1, 4, 5)

Hint:

The "cover-up" method is extremely fast for finding coefficients in partial fractions when the denominator has distinct linear factors. To find the coefficient A for a term \(\frac{A}{x-a}\), "cover-up" the \((x-a)\) factor in the original fraction's denominator and substitute \(x=a\) into the rest of the expression. For example, to find B: cover up \((x-2)\) in \(\frac{3x-1}{(x-1)(x-2)(x-3)}\) to get \(\frac{3x-1}{(x-1)(x-3)}\), and substitute \(x=2\): \(\frac{3(2)-1}{(2-1)(2-3)} = \frac{5}{(1)(-1)} = -5\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the values of the constants A, B, and C in the partial fraction decomposition of the given rational function.

Step 2: Key Formula or Approach:
The problem is based on partial fraction decomposition. We start by setting up the identity:
\[ 3x-1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2) \]
The most efficient way to find A, B, and C is the "cover-up" method, where we substitute the roots of the denominator (\(x=1, x=2, x=3\)) into this identity.

Step 3: Detailed Explanation:
We use the identity: \(3x-1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)\).

To find A, set \(x=1\):
Substitute \(x=1\) into the identity. The terms with B and C will become zero.
\[ 3(1) - 1 = A(1-2)(1-3) + B(0) + C(0) \]
\[ 2 = A(-1)(-2) \]
\[ 2 = 2A \]
\[ A = 1 \]

To find B, set \(x=2\):
Substitute \(x=2\) into the identity. The terms with A and C will become zero.
\[ 3(2) - 1 = A(0) + B(2-1)(2-3) + C(0) \]
\[ 6 - 1 = B(1)(-1) \]
\[ 5 = -B \]
\[ B = -5 \]

To find C, set \(x=3\):
Substitute \(x=3\) into the identity. The terms with A and B will become zero.
\[ 3(3) - 1 = A(0) + B(0) + C(3-1)(3-2) \]
\[ 9 - 1 = C(2)(1) \]
\[ 8 = 2C \]
\[ C = 4 \]

Step 4: Final Answer:
The values are \(A=1\), \(B=-5\), and \(C=4\). So, (A, B, C) = (1, -5, 4).