Express the rational function as a sum or difference of two simpler fractions. Use a system of equations.
$\frac{4x^2-21x-40}{x^3-x^2-20x}$

• A. $\frac{2}{x}-\frac{1}{x-5}-\frac{3}{x+4}$
• B. $\frac{1}{x}-\frac{2}{x^2-x-20}$
• C. $\frac{2}{x}+\frac{1}{x+5}-\frac{3}{x-4}$
• D. $\frac{2}{x}-\frac{1}{x-5}+\frac{3}{x+4}$

Answer: (D)

Verified step by step guidance

1. Factorize the denominator of the given rational function \( x^3 - x^2 - 20x \). Start by factoring out \( x \), which gives \( x(x^2 - x - 20) \). Then, factorize \( x^2 - x - 20 \) into \( (x - 5)(x + 4) \). The denominator becomes \( x(x - 5)(x + 4) \).
2. Rewrite the given rational function \( \frac{4x^2 - 21x - 40}{x^3 - x^2 - 20x} \) as \( \frac{4x^2 - 21x - 40}{x(x - 5)(x + 4)} \). The goal is to express this as a sum of simpler fractions: \( \frac{A}{x} + \frac{B}{x - 5} + \frac{C}{x + 4} \).
3. Set up the equation \( \frac{4x^2 - 21x - 40}{x(x - 5)(x + 4)} = \frac{A}{x} + \frac{B}{x - 5} + \frac{C}{x + 4} \). Multiply through by the common denominator \( x(x - 5)(x + 4) \) to eliminate the denominators, resulting in \( 4x^2 - 21x - 40 = A(x - 5)(x + 4) + Bx(x + 4) + Cx(x - 5) \).
4. Expand each term on the right-hand side: \( A(x - 5)(x + 4) = A(x^2 - x - 20) \), \( Bx(x + 4) = B(x^2 + 4x) \), and \( Cx(x - 5) = C(x^2 - 5x) \). Combine all terms to get \( 4x^2 - 21x - 40 = A(x^2 - x - 20) + B(x^2 + 4x) + C(x^2 - 5x) \).
5. Group like terms (\( x^2 \), \( x \), and constant terms) on the right-hand side and equate coefficients with the left-hand side. Solve the resulting system of equations for \( A \), \( B \), and \( C \). Substitute these values back into \( \frac{A}{x} + \frac{B}{x - 5} + \frac{C}{x + 4} \) to express the original function as a sum of simpler fractions.