Problem 2 - Thomas Calculus

Question

Expand the quotients in Exercises \(1-8\) by partial fractions. $$\frac{5 x-7}{x^{2}-3 x+2}$$

Step-by-Step Solution

Answer

The partial fraction decomposition is \(\frac{2}{x-1} + \frac{3}{x-2}\).

Step 1: Factor the Denominator

First, factor the quadratic denominator \(x^2 - 3x + 2\). This quadratic can be factored into \((x - 1)(x - 2)\).

Step 2: Express as Partial Fractions

Express the function \(\frac{5x - 7}{(x - 1)(x - 2)}\) in terms of partial fractions. Set up the equation: \[ \frac{5x - 7}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2} \]

Step 3: Clear the Denominator

Multiply through by the common denominator \((x-1)(x-2)\) to eliminate the fractions: \[ 5x - 7 = A(x - 2) + B(x - 1) \].

Step 4: Expand and Simplify

Expand the right-hand side: \(A(x - 2) + B(x - 1) = Ax - 2A + Bx - B = (A + B)x - (2A + B)\). Simplify the expression: \[(A + B)x - (2A + B)\].

Step 5: Equate Coefficients

Compare coefficients from both sides of the equation \(5x - 7 = (A + B)x - (2A + B)\). This gives us two equations: 1. \(A + B = 5\)2. \(-2A - B = -7\).

Step 6: Solve the System of Equations

Solve the system of equations:1. \(A + B = 5\)2. \(-2A - B = -7\).From equation 1: \(B = 5 - A\). Substitute into equation 2: \(-2A - (5 - A) = -7\). Simplify: \(-2A - 5 + A = -7\), which gives \(-A = -2\) or \(A = 2\). Substitute \(A = 2\) back into \(B = 5 - A\) to find \(B = 3\).

Step 7: Final Expression of Partial Fractions

Substitute \(A\) and \(B\) back into the partial fractions expression: \[ \frac{5x - 7}{(x-1)(x-2)} = \frac{2}{x-1} + \frac{3}{x-2} \].

Key Concepts

Rational Functions

Rational functions are fractions that consist of a numerator and a denominator, both of which are polynomials. A function like \(\frac{5x - 7}{x^2 - 3x + 2}\) is a rational function because it has a polynomial on the top and the bottom. Rational functions are interesting and useful because they can represent a wide range of behaviors and are often involved in phenomena like asymptotes and intercepts, which are crucial in calculus and real-world modeling.

Working with rational functions typically involves factoring polynomials, simplifying expressions, and sometimes finding partial fractions, which makes them more manageable. If the degree of the numerator is less than the degree of the denominator, the function is called "proper." If not, it's "improper" and might need polynomial long division before proceeding further. Rational functions are used to model situations where outputs decrease as inputs increase, which commonly occurs in many scientific and engineering contexts.

System of Equations

A system of equations is a set where two or more equations are solved simultaneously. In solving rational functions by partial fractions, we often use systems of equations to find unknown constants. For example, when we decompose \(\frac{5x - 7}{(x-1)(x-2)}\) into partial fractions, we find constants \(A\) and \(B\).
This process involves:

• Setting up equations like \(A(x - 2) + B(x - 1) = 5x - 7\).
• Expanding and simplifying the expression to identify like terms.
• Equating the coefficients of similar terms to form simultaneous equations, such as \(A + B = 5\) and \(-2A - B = -7\).

Solving these equations provides the values for \(A\) and \(B\). These solutions help construct the partial fraction decomposition, aiding in various calculus and algebra applications including integration and solving differential equations.

Factoring Quadratics

Factoring quadratics is a fundamental skill in algebra that helps simplify expressions and solve equations. In the given step-by-step solution, the quadratic \(x^2 - 3x + 2\) is factored into \((x - 1)(x - 2)\). This is a classic example of factoring a trinomial where the leading coefficient is 1.

• Start by identifying two numbers that multiply to the constant term (2) and add up to the linear coefficient (-3).
• In this instance, the numbers \(-1\) and \(-2\) satisfy both conditions.
• The expression splits into \((x - 1)(x - 2)\), indicating the points where the function's numerator will have discontinuities if not properly accounted for.

Understanding factoring is crucial because it forms the basis for dividing rational functions into more manageable pieces, leading to easier simplifications and solving strategies. When factoring, always look for patterns or use the quadratic formula if the expression doesn't seem to factor easily.