QuestionAugust 27, 2025

The partial fraction decomposition of (6)/((x-1)(x+1)) can be written in the form of (f(x))/(x-1)+(g(x))/(x+1) where f(x)=square g(x)=square

The partial fraction decomposition of (6)/((x-1)(x+1)) can be written in the form of (f(x))/(x-1)+(g(x))/(x+1) where f(x)=square g(x)=square
The partial fraction decomposition of (6)/((x-1)(x+1)) can be written in the form of (f(x))/(x-1)+(g(x))/(x+1) where f(x)=square g(x)=square

Solution
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Answer

f(x) = 3 ### g(x) = -3 Explanation 1. Set up the partial fraction decomposition Assume \frac{6}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}. 2. Clear the denominators Multiply both sides by (x-1)(x+1) to get 6 = A(x+1) + B(x-1). 3. Expand and collect like terms Expand: 6 = Ax + A + Bx - B. Combine: 6 = (A + B)x + (A - B). 4. Equate coefficients For x term: A + B = 0. For constant term: A - B = 6. 5. Solve the system of equations From A + B = 0, we have B = -A. Substitute into A - B = 6: A - (-A) = 6 \Rightarrow 2A = 6 \Rightarrow A = 3. Thus, B = -3.

Explanation

1. Set up the partial fraction decomposition<br /> Assume $\frac{6}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$.<br /><br />2. Clear the denominators<br /> Multiply both sides by $(x-1)(x+1)$ to get $6 = A(x+1) + B(x-1)$.<br /><br />3. Expand and collect like terms<br /> Expand: $6 = Ax + A + Bx - B$. Combine: $6 = (A + B)x + (A - B)$.<br /><br />4. Equate coefficients<br /> For $x$ term: $A + B = 0$. For constant term: $A - B = 6$.<br /><br />5. Solve the system of equations<br /> From $A + B = 0$, we have $B = -A$. Substitute into $A - B = 6$: $A - (-A) = 6 \Rightarrow 2A = 6 \Rightarrow A = 3$. Thus, $B = -3$.
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