Problem 30 - Precalculus

Question

Find the partial-fraction decomposition for each rational function. $$\frac{14 x^{2}+8 x+40}{(x+5)\left(2 x^{2}-3 x+5\right)}$$

Step-by-Step Solution

Answer

The partial-fraction decomposition is \(\frac{2}{x+5} + \frac{10x + 6}{2x^2 - 3x + 5}\).

Step 1: Recognize the Fraction Type

Since the denominator \((x+5)(2x^2 - 3x + 5)\) is a product of a linear factor \((x+5)\) and an irreducible quadratic factor \((2x^2 - 3x + 5)\), we can decompose our rational function into a sum of partial fractions.

Step 2: Set Up the Partial Fraction Decomposition

The partial fraction decomposition will be of the form: \( \frac{14x^2 + 8x + 40}{(x+5)(2x^2-3x+5)} = \frac{A}{x+5} + \frac{Bx + C}{2x^2 - 3x + 5} \). Our goal is to determine the constants \(A\), \(B\), and \(C\).

Step 3: Multiply Through by the Denominator

Multiply both sides by \((x+5)(2x^2 - 3x + 5)\) to clear the fractions: \[ 14x^2 + 8x + 40 = A(2x^2 - 3x + 5) + (Bx + C)(x + 5) \].

Step 4: Expand and Collect Like Terms

Expand the right side: \[ 14x^2 + 8x + 40 = 2Ax^2 - 3Ax + 5A + Bx^2 + 5Bx + Cx + 5C \]. Combine like terms: \[ (2A + B)x^2 + (-3A + 5B + C)x + (5A + 5C) \].

Step 5: Equate Coefficients

Equate the coefficients on both sides for each power of \(x\): 1. \(2A + B = 14\)2. \(-3A + 5B + C = 8\)3. \(5A + 5C = 40\)

Step 6: Solve the System of Equations

Solve the system of equations:1. \(2A + B = 14\)2. \(-3A + 5B + C = 8\)3. \(5A + 5C = 40\)From equation 3, solve for \(C\): \[ C = 8 - A \].Substitute \(C\) in equation 2 and solve for \(A\) and \(B\):After solving, we find \(A = 2\), \(B = 10\), and \(C = 6\).

Step 7: Write the Final Decomposition

Substitute \(A\), \(B\), and \(C\) back into the partial fractions: \[ \frac{14x^2 + 8x + 40}{(x+5)(2x^2-3x+5)} = \frac{2}{x+5} + \frac{10x + 6}{2x^2 - 3x + 5} \].

Key Concepts

Rational Functions

A rational function is essentially a fraction in which both the numerator and the denominator are polynomials. It can be expressed in the form \( \frac{P(x)}{Q(x)} \), where both \(P(x)\) and \(Q(x)\) are polynomials, and \(Q(x) eq 0\). In this exercise, our rational function is \( \frac{14x^2 + 8x + 40}{(x+5)(2x^2 - 3x + 5)} \). Understanding rational functions is crucial, as they often arise in calculus and algebra problems.

When working with rational functions, if \(Q(x)\), the polynomial in the denominator, can be factored, this allows us to decompose the function into simpler parts, which can be helpful in integration or solving equations. The decomposition results in partial fractions, a form that can make complex functions easier to handle in many mathematical contexts.

Linear Factors

Linear factors are expressions of the form \(x - r\), where \(r\) represents a root or zero of the polynomial. An example of a linear factor in the given problem is \(x + 5\). It's often more intuitive to deal with linear factors because they break down a polynomial into simpler, linear equations that are easier to handle.

Each linear factor corresponds to a potential vertical asymptote of the rational function, except in cases where the factor is canceled by a similar term in the numerator. In partial-fraction decomposition, each linear factor in the denominator becomes a separate fraction itself, simplifying the overall expression. This process allows solving systems involving integration, especially when linear factors are involved.

Quadratic Factors

Quadratic factors take the form \(ax^2 + bx + c\). They are part of polynomial expressions where the highest exponent of \(x\) is two. In our problem, \(2x^2 - 3x + 5\) is an irreducible quadratic factor, meaning it cannot be factored further into real linear factors.

When dealing with quadratic factors in partial-fraction decomposition, we express them as fractions like \(\frac{Bx + C}{2x^2 - 3x + 5}\), where \(B\) and \(C\) are constants to be determined.

Quadratic factors are generally handled differently than linear factors because of their potential complexity. They can sometimes represent parabolic asymptotic behavior in rational functions. In decomposition, irreducible quadratic factors can introduce complex roots, which need careful manipulation, like in this step-by-step solution.

System of Equations

Solving partial-fraction decomposition often involves forming and solving a system of equations. This process helps to determine the unknown constants in the decomposed fractions. In our example, after aligning the form \(A/(x+5) + (Bx + C)/(2x^2 - 3x + 5)\), we need to solve for the constants \(A\), \(B\), and \(C\).

The step-by-step approach given leads us to a system of equations derived from equating coefficients of like powers of \(x\) from both sides of the equation:

• \(2A + B = 14\)
• \(-3A + 5B + C = 8\)
• \(5A + 5C = 40\)

Using these equations, we can apply methods such as substitution or elimination to solve for the constants.
Systems of equations allow the methodical find of precise values needed in the decomposition. This is an essential aspect, as inaccurate or incomplete systems lead to erroneous decompositions.