Problem 43 - Precalculus: Mathematical for Calculus

Question

Determine \(A\) and \(B\) in terms of \(a\) and \(b\) : $$ \frac{a x+b}{x^{2}-1}=\frac{A}{x-1}+\frac{B}{x+1} $$

Step-by-Step Solution

Answer

\(A = \frac{a+b}{2}\), \(B = \frac{a-b}{2}\)

Step 1: Understand the Problem

We need to express \( \frac{a x+b}{x^{2}-1} \) as a sum of two fractions with denominators \( x-1 \) and \( x+1 \). This involves partial fraction decomposition.

Step 2: Factor the Denominator

The denominator \( x^2 - 1 \) can be factored using the difference of squares formula as \( (x-1)(x+1) \).

Step 3: Set Up the Partial Fraction Decomposition

Express \( \frac{a x + b}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} \). The goal is to find \( A \) and \( B \).

Step 4: Clear the Denominators

Multiply both sides by \( (x-1)(x+1) \) to eliminate the denominators: \[ a x + b = A(x+1) + B(x-1) \]

Step 5: Expand and Combine Terms

Expand the right side to get: \[ A(x+1) + B(x-1) = Ax + A + Bx - B \] Combine terms to get: \[ (A+B)x + (A-B) \].

Step 6: Compare Coefficients

On comparing the coefficients with \( ax + b \), we get the equations: \[ A + B = a \] \[ A - B = b \]

Step 7: Solve the System of Equations

Solve the equations simultaneously:- Add the two equations: \[ A + B + A - B = a + b \] \( 2A = a + b \) \( A = \frac{a+b}{2} \)- Subtract the second equation from the first: \[ A + B - (A - B) = a - b \] \( 2B = a - b \) \( B = \frac{a-b}{2} \)

Step 8: Finalize Solution

We've determined the values: \( A = \frac{a+b}{2} \) and \( B = \frac{a-b}{2} \).

Key Concepts

Factoring Polynomials

Factoring polynomials is a fundamental skill in algebra, and it plays a significant role in partial fraction decomposition, among other areas. To factor a polynomial is essentially to "break it down" into a product of simpler polynomials, which helps in solving equations or simplifying expressions. Given the polynomial expression \( x^2 - 1 \), this can be seen as a special case of the difference of squares because it can be rewritten using the identity \( a^2 - b^2 = (a-b)(a+b) \). Hence, \( x^2 - 1 \) can be factored into \((x-1)(x+1)\). This factorization is crucial for solving the original problem using partial fractions because it prepares the expression with distinct linear factors suitable for decomposition. Knowing how to factor polynomials is extremely useful:

• It simplifies complex expressions.
• Makes it easier to find the roots of the polynomial.
• Prepares expressions for partial fraction decomposition, as in this problem.

Solving Linear Equations

Solving linear equations is a basic, yet essential skill. It allows us to find specific variable values that satisfy given mathematical expressions. In the context of partial fraction decomposition, solving linear equations helps in finding the constants of the decomposed expression.After factoring the denominator in our original problem, you rephrase:\[ ax + b = A(x+1) + B(x-1) \]To find the unknowns \( A \) and \( B \), you first expand and simplify the right side of the equation:\[ Ax + A + Bx - B = (A+B)x + (A-B) \]By comparing coefficients with the original numerator \( ax + b \), two linear equations can be formulated:\[ A + B = a \]\[ A - B = b \]You can solve these equations using various methods such as substitution, elimination, or matrix operations. In this case, the method used was adding and subtracting the two equations to isolate \( A \) and \( B \).Knowing how to solve linear equations is crucial because:

• It allows you to quickly find values for variables.
• Helps establish the relations between different constants.
• Facilitates solving systems of equations, common in algebra.

Difference of Squares

The difference of squares is a specific algebraic pattern that emerges when two squares are subtracted from one another. This pattern is uniquely useful for factoring polynomials of the form \( a^2 - b^2 \).In our original exercise, the polynomial \( x^2 - 1 \) represents a difference of squares where:\[ x^2 - 1 = (x)^2 - (1)^2 \]Using the identity:\[ a^2 - b^2 = (a-b)(a+b) \]we can factor \( x^2 - 1 \) into \( (x-1)(x+1) \). This is critical in breaking down complex expressions into solvable parts. For partial fraction decomposition, identifying the difference of squares allows the expression to be separated into fractions with linear denominators.Benefits of mastering difference of squares include:

• Simplifies factorization of specific polynomials.
• It allows for easier integration or simplification of mathematical expressions.
• Essentials for understanding deeper algebraic concepts.