Express the rational function as a sum or difference or simpler rational expressions.
$\frac{13x+2}{2x^2-x-1}$

• A. $\frac{13}{x-1}+\frac{2}{2-x}$
• B. $\frac{13}{x+1}-\frac{2}{2-x}$
• C. $\frac{5}{x-1}+\frac{3}{2x+1}$
• D. $\frac{5}{x+1}+\frac{3}{2x-1}$

Answer: (C)

Verified step by step guidance

1. Factorize the denominator of the given rational function \( \frac{13x+2}{2x^2 - x - 1} \). Start by factoring \( 2x^2 - x - 1 \) into two binomials. Look for two numbers that multiply to \(-2\) (the product of \(2\) and \(-1\)) and add to \(-1\) (the middle coefficient). This gives \( (2x + 1)(x - 1) \).
2. Rewrite the rational function as \( \frac{13x+2}{(2x+1)(x-1)} \). The goal is to express this as a sum of simpler fractions, \( \frac{A}{2x+1} + \frac{B}{x-1} \), where \(A\) and \(B\) are constants to be determined.
3. Set up the equation \( \frac{13x+2}{(2x+1)(x-1)} = \frac{A}{2x+1} + \frac{B}{x-1} \). Combine the right-hand side over a common denominator: \( \frac{A(x-1) + B(2x+1)}{(2x+1)(x-1)} \).
4. Equate the numerators: \( 13x + 2 = A(x-1) + B(2x+1) \). Expand the right-hand side: \( A(x-1) = Ax - A \) and \( B(2x+1) = 2Bx + B \). Combine terms: \( Ax - A + 2Bx + B = (A + 2B)x + (-A + B) \).
5. Match coefficients of \(x\) and the constant terms from both sides of the equation. From \( 13x + 2 = (A + 2B)x + (-A + B) \), we get two equations: \( A + 2B = 13 \) (coefficient of \(x\)) and \( -A + B = 2 \) (constant term). Solve this system of equations to find \(A\) and \(B\). Substitute these values back into \( \frac{A}{2x+1} + \frac{B}{x-1} \) to get the final expression.