Problem 5 - College Algebra and Calculus: An Applied Approach

Question

Write the partial fraction decomposition for the expression. $$ \frac{4 x-13}{x^{2}-3 x-10} $$

Step-by-Step Solution

Answer

The partial fraction decomposition for the given expression is \[\frac{4x-13}{(x-5)(x+2)} = \frac{3}{x-5} + \frac{1}{x+2}\]

Step 1: Factorize Denominator

Factorize the denominator of the given function. The denominator is a quadratic equation \(x^{2}-3x-10\). To factorize it, we find two numbers that add up to -3 (the coefficient of \(x\)) and multiply to -10 (the constant term). The numbers -5 and 2 fit these criteria and hence, the factorization of the denominator is \((x-5)(x+2)\). So the expression becomes: \[\frac{4x-13}{(x-5)(x+2)}\]

Step 2: Set Up Partial Fraction Decomposition

Set up the partial fractions decomposition. This is achieved by expressing as a sum of two fractions where the denominators are the factors of the original denominator and each has a constant \(A\) and \(B\), unknown at this point, as a numerator. The partial fraction decomposition is set up as: \[\frac{4x-13}{(x-5)(x+2)} = \frac{A}{x-5} + \frac{B}{x+2}\]

Step 3: Solve for A and B

To solve for the constants \(A\) and \(B\), get rid of the denominators by multiplying through by \((x-5)(x+2)\). This gives: \[4x-13 = A(x+2) + B(x-5)\]On expanding the expressions on the right hand side and grouping like terms together, this equation becomes: \[4x - 13 = (A + B) x + 2A -5B\]Setting the coefficients equal gives the two equations: \(A + B = 4\) and \(2A - 5B = -13\). Solving this system of equations yields \(A = 3\) and \(B = 1\). This gives us the final partial fraction decomposition: \[\frac{4x-13}{(x-5)(x+2)} = \frac{3}{x-5} + \frac{1}{x+2}\]