Question

Find the partial fraction decomposition of each rational expression. $$ \frac{x^{2}-x-8}{(x+1)\left(x^{2}+5 x+6\right)} $$

Short answer

\( \frac{1}{x+1} - \frac{3}{x+2} + \frac{3}{x+3} \)

Step-by-step solution

Factor the Denominator

Factor the denominator \( (x+1)(x^{2}+5x+6) \). The quadratic term can be factored as \( (x+1)(x+2)(x+3) \). Thus, the expression becomes \( \frac{x^{2}-x-8}{(x+1)(x+2)(x+3)} \).

Set Up the Partial Fractions

Express the rational expression as a sum of partial fractions: \( \frac{x^{2}-x-8}{(x+1)(x+2)(x+3)} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x+3} \).

Clear the Denominator

Multiply both sides by \( (x+1)(x+2)(x+3) \), the common denominator, to clear the fractions: \( x^{2}-x-8 = A(x+2)(x+3) + B(x+1)(x+3) + C(x+1)(x+2) \).

Expand and Combine Like Terms

Expand the right-hand side and combine like terms: \[ x^{2} - x - 8 = A(x^{2} + 5x + 6) + B(x^{2} + 4x + 3) + C(x^{2} + 3x + 2) \] \[ = (A+B+C)x^{2} + (5A + 4B + 3C)x + (6A + 3B + 2C) \].

Set Up the System of Equations

Set up a system of equations by equating the coefficients from both sides: \[ A + B + C = 1 \] \[ 5A + 4B + 3C = -1 \] \[ 6A + 3B + 2C = -8 \].

Solve the System of Equations

Solve the system of equations for \( A, B \), and \( C \). Using various methods (substitution, elimination, matrix methods), find that \[ A = 1 \] \[ B = -3 \] \[ C = 3 \].

Write the Final Partial Fraction Decomposition

Substitute \( A, B \), and \( C \) back into the partial fraction setup: \[ \frac{x^{2}-x-8}{(x+1)(x+2)(x+3)} = \frac{1}{x+1} - \frac{3}{x+2} + \frac{3}{x+3} \].

Key concepts

rational expressions

When dealing with partial fraction decomposition, understanding rational expressions is crucial. A rational expression is a fraction where both the numerator and the denominator are polynomials. In our exercise, the rational expression is \( \frac{x^2 - x - 8}{(x+1)(x^2 + 5x + 6)} \). Notice that both the top \(x^2 - x - 8\) and bottom \( (x+1)(x^2 + 5x + 6)\) are polynomial expressions.
Rational expressions can often be simplified or decomposed into simpler parts, making them easier to work with, especially when performing operations like integration or solving certain equations.

system of equations

When decomposing a rational expression into partial fractions, you frequently set up a system of equations to solve for unknown coefficients. In our example, after setting up the partial fractions and expanding, we match coefficients of like terms on both sides of the equation. This results in a system of linear equations:
\[ \ A + B + C = 1 \] \[ 5A + 4B + 3C = -1 \] \[ 6A + 3B + 2C = -8 \].
These equations represent the relationships between the coefficients. Solving this system is essential for finding the specific values of \( A, B, \) and \( C \), allowing us to accurately decompose the original rational expression.

algebraic factoring

A key step in partial fraction decomposition is algebraic factoring. Factoring simplifies the denominator into linear or irreducible quadratic factors. In this problem, we first factor the quadratic polynomial in the denominator:
\( x^2 + 5x + 6 = (x+2)(x+3)\).
Combining this with the linear factor \(x+1\), the denominator becomes \( (x+1)(x+2)(x+3) \).
Factoring is critical because it allows us to break down complex fractions into simpler ones. Each factor in the denominator corresponds to a term in the partial fraction decomposition, making it easier to solve for the unknown coefficients.

coefficients comparison

Comparing coefficients is the method used to solve the system of equations derived from the partial fraction setup. After multiplying out and combining like terms, we equate the coefficients of corresponding powers of \( x \) from both the numerator of the original rational expression and its expanded version. This process gives us a set of simultaneous equations:
\[ 1 \text{ (constant term) is the sum of the constant coefficients in A, B, and C's expansions} \]
\[ -1 \text{ (linear coefficient) is the sum of the linear coefficients in A, B, and C's expansions} \]
By solving these equations, we find the values of coefficients \( A, B, \) and \( C \). This technique ensures that the decomposed fractions accurately represent the original rational expression.