Problem 46 - Algebra and Trigonometry

Question

Determine \(A, B, C,\) and \(D\) in terms of \(a\) and \(b\) $$ \frac{a x^{3}+b x^{2}}{\left(x^{2}+1\right)^{2}}=\frac{A x+B}{x^{2}+1}+\frac{C x+D}{\left(x^{2}+1\right)^{2}} $$

Step-by-Step Solution

Answer

\( A = a, \ B = b, \ C = -a, \ D = -b \)

Step 1: Set Up Partial Fraction Decomposition

The expression \( \frac{a x^{3}+b x^{2}}{(x^{2}+1)^{2}} \) can be decomposed into partial fractions as \( \frac{A x+B}{x^{2}+1} + \frac{C x+D}{(x^{2}+1)^{2}} \). This step involves equating both sides of the given equation.

Step 2: Clear the Denominators

Multiply both sides of the equation by \( (x^2 + 1)^2 \) to clear the denominators: \[ a x^3 + b x^2 = (A x + B)(x^2 + 1) + (C x + D) \]

Step 3: Expand the Right Side

Expand the expression on the right:- \( (A x + B)(x^2 + 1) \) results in: \[ A x^3 + A x + B x^2 + B \] - Adding \( C x + D \), the full expression becomes: \[ A x^3 + B x^2 + A x + B + C x + D \] which simplifies further to: \[ A x^3 + B x^2 + (A+C)x + (B+D) \]

Step 4: Equate Coefficients

Align the coefficients from both sides:- For \( x^3 \): \( A = a \)- For \( x^2 \): \( B = b \)- For \( x \): \( A + C = 0 \) - Constant term: \( B + D = 0 \)

Step 5: Solve the System of Equations

Using the equations from the previous step:1. \( A = a \)2. \( B = b \)3. \( A + C = 0 \) implies \( C = -A = -a \)4. \( B + D = 0 \) implies \( D = -B = -b \)Therefore, the values are:\[ A = a, \, B = b, \, C = -a, \, D = -b \]

Step 6: Final Solution of Constants

Thus, the solution is:\[ A = a, \, B = b, \, C = -a, \, D = -b \]

Key Concepts

Polynomial Expressions

Polynomial expressions are mathematical expressions involving a sum of powers of a variable, each multiplied by a coefficient. In our problem, we have the polynomial expression \( a x^3 + b x^2 \). This expression consists of terms where the variable \( x \) is raised to different powers, the highest being 3, which is known as the degree of the polynomial.
Understanding polynomial expressions is crucial, as they often appear in algebra. Each term in a polynomial can be broken down into:

• Coefficients: The constant terms multiplying the variable powers (in this case, \( a \) and \( b \)).
• Variable: Here, \( x \) is the variable.
• Exponents: Indicate the power to which the variable is raised (3 and 2).

Factorization and simplification are common operations performed on polynomials, which frequently involve combining like terms, expanding expressions, or factoring out common factors. The manipulation of these expressions is a fundamental algebraic skill.

Rational Functions

Rational functions are quotients of two polynomials. In our exercise, the expression \( \frac{a x^3 + b x^2}{(x^2 + 1)^2} \) is a rational function, where the numerator and denominator are both polynomials. These functions are defined wherever the denominator is not zero because division by zero is undefined.
Rational functions often need to be manipulated or decomposed for simplification, easier integration, or solving algebraic equations. One common method is partial fraction decomposition, which breaks down a complex rational function into simpler, more manageable pieces. This makes it easier to integrate or apply further algebraic operations. When dealing with rational functions, students should be comfortable manipulating both the numerator and the denominator and understanding their roles in defining the domain of the function.

Algebraic Equation Solving

Algebraic equation solving involves finding the values of the variables that make the equation true. For the partial fraction decomposition performed in the exercise, we began with an equation and used algebraic manipulation to solve for unknown coefficients \( A, B, C, \) and \( D \).
Here's a quick summary of the solving process:

• Start by setting up the equation based on your decomposition setup.
• Clear denominators by multiplying through, if necessary, to avoid complex fractions.
• Expand both sides of the equation, ensuring each term is correctly sorted by power of \( x \).

After these steps, you'll typically equate coefficients for corresponding terms of the same degree, setting up a system of equations. Solving this system gives the desired values of the unknowns, ultimately solving the problem.

Coefficient Comparison

Coefficient comparison is a technique used when manipulating algebraic expressions, especially in partial fraction decomposition. Once a rational expression has been written as a sum of simpler fractions, expanding them gives a polynomial on the right-hand side.
The main idea is to compare coefficients of like terms on both sides of the equality. For example:

• If the left-hand side has a term \( a x^3 \), then the right-hand side must also have an \( x^3 \) term with the coefficient \( A \) satisfying \( A = a \).
• This comparison process is repeated for each term (e.g., \( x^2, x, \) and constant terms).

The resulting system of equations for the coefficients is solved to determine the unknowns. This systematic approach simplifies complex algebraic expressions and turns them into a solvable set of linear equations, making it indispensable in calculus and algebra.