Problem 11 - Precalculus: Functions and Graphs

Question

Find the partial fraction decomposition. \(\frac{19 x^{2}+50 x-25}{3 x^{3}-5 x^{2}}\)

Step-by-Step Solution

Answer

The partial fraction decomposition is \(\frac{-7}{x} + \frac{5}{x^2} + \frac{40}{3x - 5}\).

Step 1: Identify the type of partial fractions

First, we need to identify the type of decomposition required. The denominator is \(3x^3 - 5x^2 = x^2(3x - 5)\). This expression indicates a quadratic factor \(x^2\) and a linear factor \(3x - 5\). Thus, we should decompose into terms of the form \(\frac{A}{x} + \frac{B}{x^2} + \frac{C}{3x - 5}\).

Step 2: Write the partial fraction decomposition

Express the fraction as the sum of simpler fractions:\[\frac{19x^2 + 50x - 25}{3x^3 - 5x^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{3x - 5}\]

Step 3: Clear the denominators

Multiply both sides by the common denominator \(3x^3 - 5x^2\) to eliminate the fractions:\[19x^2 + 50x - 25 = A(x)(3x - 5) + B(3x - 5) + C(x^2)\]

Step 4: Expand and Combine Like Terms

Expand the right-hand side and collect like terms:\[Ax(3x - 5) = 3Ax^2 - 5Ax\]\[B(3x - 5) = 3Bx - 5B\]Thus, \[C(x^2) + 3Ax^2 + (3B - 5A)x - 5B = 19x^2 + 50x - 25\]

Step 5: Set up a system of equations

Equate the coefficients of the like terms from both sides:- For \(x^2\): \(C + 3A = 19\)- For \(x\): \(3B - 5A = 50\)- Constant term: \(-5B = -25\)

Step 6: Solve the system of equations

Start by solving for \(B\) from the constant term:- \(-5B = -25\) implies \(B = 5\).Substitute \(B = 5\) into the equation for \(x\):- \(3(5) - 5A = 50\) simplifies to \(15 - 5A = 50\).- Solve for \(A\): \(-5A = 35\) gives \(A = -7\).Finally, substitute \(A = -7\) into the equation for \(x^2\):- \(C + 3(-7) = 19\) simplifies to \(C - 21 = 19\).- Solve for \(C\): \(C = 40\).

Step 7: Substitute back into the decomposition

Substitute \(A = -7\), \(B = 5\), and \(C = 40\) back into the partial fractions:\[\frac{19x^2 + 50x - 25}{3x^3 - 5x^2} = \frac{-7}{x} + \frac{5}{x^2} + \frac{40}{3x - 5}\]

Key Concepts

Quadratic Factor

In partial fraction decomposition, identifying whether a factor is quadratic is crucial. A quadratic factor is any polynomial factor of degree two, distinguished by its highest power being two. In the given exercise, the factor \(x^2\) is a quadratic one. Quadratic factors can appear in the denominator of rational expressions, which are fractions involving polynomials. To decompose a rational expression with a quadratic factor, you represent it with a term like \(\frac{B}{x^2}\). This separation helps in simplifying the expression and solving it more easily. Making these distinctions is key to successful decomposition.

Linear Factor

A linear factor is a polynomial of degree one, meaning its highest exponent is one. In our problem, the linear factor is \(3x - 5\). Recognizing linear factors is a foundational step in breaking down complex rational expressions. When handling a rational expression, any linear factor in the denominator leads to a term in the partial fraction form like \(\frac{C}{3x-5}\). This decomposition into simpler terms lets us handle each part individually, simplifying the analytical or computational tasks that come afterward. Linear and quadratic factors together dictate the structure of our partial fractions.

System of Equations

After setting up a partial fraction decomposition, often we need to solve a system of equations. These equations derive from matching coefficients on both sides of an equation after clearing denominators. For example, our system setup involves:

• \(C + 3A = 19\)
• \(3B - 5A = 50\)
• \(-5B = -25\)

Solving these equations step-by-step, often begins with the simplest equation, allowing us to find one variable first. With one variable known, substitute back into other equations to solve for remaining variables. This systematic approach ensures every part of the partial fractions is accurately calculated.

Rational Expressions

Rational expressions consist of a ratio between two polynomials. In partial fraction decomposition, the goal is to express a complex rational expression as a sum of simpler fractions. For instance, \(\frac{19x^2+50x-25}{3x^3-5x^2}\) is simplified through partial fraction decomposition. Recognizing types of factors within the rational expression supports the set-up for decomposition. This structure within rational expressions provides a methodical path to break down and understand relationships between polynomials in a manageable way. Recognizing and converting complex rational expressions into a sum of simpler fractions, using partial fractions, extensively eases the difficulty in both solving and interpreting polynomial relationships.