Problem 22 - Precalculus: Mathematical for Calculus

Question

Find the partial fraction decomposition of the rational function. $$\frac{7 x-3}{x^{3}+2 x^{2}-3 x}$$

Step-by-Step Solution

Answer

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

Step 1: Factor the Denominator

First, factor the denominator of the rational expression \(x^3 + 2x^2 - 3x\). Factor out the greatest common factor, which is \(x\). This gives \(x(x^2 + 2x - 3)\). Next, factor the quadratic \(x^2 + 2x - 3\). It factors into \((x + 3)(x - 1)\). So, the complete factorization of the denominator is \(x(x + 3)(x - 1)\).

Step 2: Set Up the Partial Fraction Decomposition

Write the rational function \(\frac{7x - 3}{x(x + 3)(x - 1)}\) as the sum of partial fractions. We can express it as: \(\frac{A}{x} + \frac{B}{x + 3} + \frac{C}{x - 1}\).

Step 3: Clear the Denominator

Multiply both sides of the equation by the common denominator \(x(x + 3)(x - 1)\) to eliminate the denominators. This gives: \[ 7x - 3 = A(x + 3)(x - 1) + Bx(x - 1) + Cx(x + 3) \]

Step 4: Expand and Combine Like Terms

Expand the right side of the equation: \[ A(x^2 + 2x - 3) + B(x^2 - x) + C(x^2 + 3x) \] Combine like terms to form a single polynomial: \[ (A + B + C)x^2 + (2A - B + 3C)x - 3A \]

Step 5: Solve for Coefficients

Set the coefficients of corresponding powers of \(x\) equal on both sides of the equation: - For \(x^2\): \(A + B + C = 0\)- For \(x\): \(2A - B + 3C = 7\)- Constant term: \(-3A = -3\)Solve the system: \[ A = 1, \quad B + C = -1, \quad 2 - B + 3C = 7 \] From \(-3A = -3\), we get \(A = 1\). Substitute \(A = 1\) into the other equations to solve for \(B\) and \(C\). - \(1 + B + C = 0\) gives \(B + C = -1\)- \(2 - B + 3C = 7\) gives \(B - 3C = -5\)Solving these equations gives \(B = -2\) and \(C = 1\).

Step 6: Write the Partial Fraction Decomposition

Substitute the values of \(A\), \(B\), and \(C\) back into the partial fractions. Thus, the partial fraction decomposition of \(\frac{7x - 3}{x(x + 3)(x - 1)}\) is: \[ \frac{1}{x} - \frac{2}{x + 3} + \frac{1}{x - 1} \]

Key Concepts

Rational Function

A rational function is a fraction where both the numerator and the denominator are polynomials. For example, in the expression \( \frac{7x - 3}{x^3 + 2x^2 - 3x} \), both \(7x - 3\) and \(x^3 + 2x^2 - 3x\) are polynomials.

One key feature of rational functions is that they can be decomposed into partial fractions. This is useful particularly in integral calculus, as it simplifies the integration process. The idea is to express a complex rational function as the sum of simpler fractions.

To decompose a rational function, you need to make sure its degree (highest power of \(x\)) in the numerator is less than the degree in the denominator. If not, you might need to perform polynomial division first. Understanding how to manipulate and decompose rational functions into simpler components is an essential skill in algebra and calculus.

Factoring Polynomials

Factoring is the process of breaking down a polynomial into simpler terms, known as factors, that when multiplied together give the original polynomial. In the context of partial fraction decomposition, factoring is crucial for simplifying the denominator.

Consider the polynomial \(x^3 + 2x^2 - 3x\). The first step is to factor out the greatest common factor. Here, the GCF is \(x\), resulting in \(x(x^2 + 2x - 3)\). This step makes the polynomial manageable for further factoring.

• Factor each part separately: Quadratics like \(x^2 + 2x - 3\) are often factored into two binomials. In this case, it becomes \((x + 3)(x - 1)\).
  
• Factoring polynomials simplifies rational functions, allowing them to be expressed in terms of simpler fractions. This is crucial when decomposing into partial fractions.
  

Understanding how to correctly factor polynomials helps in simplifying complex algebraic expressions, making them easier to work with in both algebra and calculus.

Solving Systems of Equations

After setting up the partial fraction decomposition, we often find a system of equations that needs to be solved to find the coefficients of the partial fractions. This is usually done by equating coefficients on both sides of the expanded polynomial.

For instance, once we multiply through by the common denominator and collect like terms, we get equations like:

• For \(x^2\): \(A + B + C = 0\)
• For \(x\): \(2A - B + 3C = 7\)
• Constant term: \(-3A = -3\)

These equations represent a system that can be solved using algebraic methods like substitution or elimination. For example, from \(-3A = -3\), we immediately get \(A = 1\). By substituting \(A\) in other equations, you can solve for \(B\) and \(C\).

Solving systems of equations is a key step in manipulating algebraic expressions and is a foundational skill in many mathematical problems.

Algebraic Expressions

An algebraic expression like \(7x - 3\) encapsulates numbers, variables, and arithmetic operations. These play a central role in mathematics as they form the building blocks for more complex structures such as rational functions and systems of equations.

Manipulating algebraic expressions is crucial when it comes to partial fraction decomposition. You need to expand and simplify expressions diligently. For example, during the partial fraction decomposition, we expand expressions with multiple terms, such as \(A(x^2 + 2x - 3) + B(x^2 - x) + C(x^2 + 3x)\), and then combine like terms.

• Like terms have the same variable raised to the same power. For instance, terms involving \(x^2\) or terms without any \(x\) (i.e., constants).
  
• The goal is to simplify them so that systems of equations can form, from which the coefficients can be determined.
  
• Clear understanding of how to expand and simplify algebraic expressions is needed to perform partial fraction decompositions effectively.
  

Mastering algebraic expressions empowers students to tackle various mathematical challenges and is fundamental in both high school and college-level mathematics.