Problem 33 - Precalculus Mathematics for Calculus

Question

Find the partial fraction decomposition of the rational function. $$\frac{-10 x^{2}+27 x-14}{(x-1)^{3}(x+2)}$$

Step-by-Step Solution

Answer

\(\frac{1}{x-1} - \frac{7}{(x-1)^2} + \frac{16}{(x-1)^3} - \frac{1}{x+2}\)

Step 1: Set up the partial fraction form

The given rational function is \( \frac{-10x^2+27x-14}{(x-1)^3(x+2)} \). The denominator consists of the factors \((x-1)^3\) and \((x+2)\). Thus, the partial fraction decomposition takes this form:\[\frac{-10x^2+27x-14}{(x-1)^3(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3} + \frac{D}{x+2}\]

Step 2: Multiply through by the common denominator

Clear the fractions by multiplying both sides by the common denominator \((x-1)^3(x+2)\):\[-10x^2 + 27x - 14 = A(x-1)^2(x+2) + B(x-1)(x+2) + C(x+2) + D(x-1)^3\]

Step 3: Expand and combine like terms

Expand the terms on the right-hand side:\[A(x^3 - 2x^2 - x + 2) + B(x^2 + x - 2x - 2) + C(x + 2) + D(x^3 - 3x^2 + 3x - 1)\]Combine like terms:\[Ax^3 + (-2A + B + D)x^2 + (-A - 2B + D + C)x + (2A - 2B + 2C - D)\]

Step 4: Compare coefficients

Compare the coefficients from the expanded expression to the coefficients \(-10x^2 + 27x - 14\):1. \(A + D = 0\)2. \(-2A + B + D = -10\)3. \(-A - 2B + C + D = 27\)4. \(2A - 2B + 2C - D = -14\)

Step 5: Solve the system of equations

Solve this system of linear equations:1. From \(A + D = 0\), substitute \(D = -A\).2. Substituting into \(-2A + B + (-A) = -10\), gives \(-3A + B = -10\).3. Substituting \(D = -A\) into \(-A - 2B + C - A = 27\), simplifies to \(-2A - 2B + C = 27\).4. Substituting \(D = -A\) into \(2A - 2B + 2C + A = -14\), simplifies to \(3A - 2B + 2C = -14\).Solving these equations:1. \(-3A + B = -10\)2. \(-2A - 2B + C = 27\)3. \(3A - 2B + 2C = -14\)Assuming \(A = 1\) (trial)4. \(D = -1\)5. Solve \(-3(1) + B = -10\) gives \(B = -7\).6. Substitute into \(-2(1) - 2(-7) + C = 27\) gives \(C = 16\).

Step 6: Assemble the partial fractions

With values \(A = 1\), \(B = -7\), \(C = 16\), \(D = -1\), the partial fraction decomposition is:\[\frac{1}{x-1} - \frac{7}{(x-1)^2} + \frac{16}{(x-1)^3} - \frac{1}{x+2}\]

Key Concepts

Rational Function

A rational function is a type of function defined by the ratio of two polynomials. The general form is \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) eq 0 \). In our given exercise, the rational function is \( \frac{-10x^2 + 27x - 14}{(x-1)^3 (x+2)} \). The numerator is a quadratic polynomial and the denominator is composed of polynomial factors.
Analyzing the denominator is crucial for partial fraction decomposition. It is composed of individual factors \((x-1)\) raised to a power of 3, and a linear factor \((x+2)\). By breaking these down, we can express the rational function as a sum of simpler fractions, each associated with these factors, hence simplifying the problem.

System of Equations

When performing partial fraction decomposition, you generate a system of equations to find the unknown coefficients of the decomposed fractions. This system of equations arises from equating the coefficients of like terms between the original polynomial and the expanded partial fractions.
In our situation, starting with the equation after clearing the denominators leads to expansion and grouping of like terms. This comparison provides equations based on matching the coefficients of \( x^3, x^2, x, \) and the constant term from both sides. Each matching of coefficients gives us a separate linear equation, making up a system to be solved to find values for coefficients \( A, B, C, \) and \( D \).

Coefficients Comparison

Coefficients comparison is a technique used to match corresponding terms in polynomial expressions. When you expand and simplify the expression on the right after clearing the fractions, you align it with the given numerator. Each power of \( x \) on both sides should match, leading to a set of linear equations.
In this particular exercise:

• Compare the coefficients of \( x^2 \) results in equation \(-2A + B + D = -10\),
• Compare coefficients of \( x \) results in \(-A - 2B + C + D = 27\),
• And similarly, for constant terms, giving another linear equation.

Matching coefficients is integral in transforming the expanded expressions into solvable equations.

Linear Equations Solution

Solving linear equations is a crucial step to find the coefficients \( A, B, C, \) and \( D \) in partial fraction decomposition. Once you've set up these equations through coefficient comparison, solving them provides the needed coefficients for the decomposition.
For solving, determine a variable to express others in terms of, or directly substitute to simplify the system of equations. For our problem, substituting \( D = -A \) simplifies the equations. Using substitution and basic algebraic manipulation helps to find the exact values for each coefficient efficiently:

• Find \( A \) and substitute back to find others,
• Solve for \( B \) using one equation and the known value of \( A \).

Knowing how to solve such systems is key to mastering partial fractions, enhancing both algebraic skill and understanding of rational functions.