Problem 18 - Precalculus

Question

In Exercises 7 - 18 , find the partial fraction decomposition of the following rational expressions. $$ \frac{2 x^{2}+3 x+14}{\left(x^{2}+2 x+9\right)\left(x^{2}+x+5\right)} $$

Step-by-Step Solution

Answer

The decomposition is \( \frac{3x-9}{x^2+2x+9} + \frac{-3x+7}{x^2+x+5} \).

Step 1: Set Up the Partial Fraction Decomposition Structure

Given the rational expression \(\frac{2x^2+3x+14}{(x^2+2x+9)(x^2+x+5)}\), start by setting it up as a sum of partial fractions. Since the denominator is a product of two distinct quadratic factors, the decomposition has the form: \(\frac{Ax+B}{x^2+2x+9} + \frac{Cx+D}{x^2+x+5}\).

Step 2: Clear the Denominator

Multiply both sides of the equation by the common denominator \((x^2+2x+9)(x^2+x+5)\) to clear the fractions. This results in the equation: \[ 2x^2+3x+14 = (Ax+B)(x^2+x+5) + (Cx+D)(x^2+2x+9) \]

Step 3: Expand the Right-Hand Side

Expand the right-hand side by applying the distributive property:\[(Ax+B)(x^2+x+5) = Ax^3 + Ax^2 + 5Ax + Bx^2 + Bx + 5B\]\[(Cx+D)(x^2+2x+9) = Cx^3 + 2Cx^2 + 9Cx + Dx^2 + 2Dx + 9D\]

Step 4: Combine Like Terms

Combine all like terms on the right-hand side:\[Ax^3 + Cx^3 + (Ax^2 + Bx^2 + 2Cx^2 + Dx^2) + (5Ax + Bx + 9Cx + 2Dx) + (5B + 9D)\]which simplifies to:\[(A+C)x^3 + (A+B+2C+D)x^2 + (5A+B+9C+2D)x + (5B+9D)\]

Step 5: Equate Coefficients

Equate the coefficients of the terms on both sides of the equation. This gives us the following system:1. \(A+C = 0\) for \(x^3\)2. \(A+B+2C+D = 2\) for \(x^2\)3. \(5A+B+9C+2D = 3\) for \(x\)4. \(5B+9D = 14\) for the constant term.

Step 6: Solve the System of Equations

Solve the system of equations:- From \(A+C=0\), we have \(C=-A\).- Substitute \(C=-A\) in equations 2, 3, and 4.- Solving the equations, we have the values: \(A = 3, B = -9, C = -3, D = 7\).

Step 7: Write the Decomposed Form

Substitute the values of \(A, B, C,\) and \(D\) back into the partial fraction structure:\[ \frac{3x-9}{x^2+2x+9} + \frac{-3x+7}{x^2+x+5}\]

Key Concepts

Rational Expressions

A rational expression is simply a fraction where both the numerator and the denominator are polynomials. In the context of partial fraction decomposition, we focus on breaking down these rational expressions into simpler, more manageable components. The goal here is to represent a complex rational expression, especially one involving high-degree polynomials, as a sum of simpler fractions. This process is analogous to simplifying a regular fraction into its components.For example, given a rational expression \[\frac{2x^2+3x+14}{(x^2+2x+9)(x^2+x+5)}\], the numerator \(2x^2+3x+14\) and the denominator \((x^2+2x+9)(x^2+x+5)\) are both polynomials. Our job is to decompose this into a sum of two simpler rational expressions that are easier to work with.

Denominator

The denominator in a rational expression plays a crucial role, especially in partial fraction decomposition. It's the expression below the fraction line, indicating what the numerator is divided by. In our exercise, the denominator is a product of two distinct quadratic polynomials: \( (x^2+2x+9)(x^2+x+5) \).When we are setting up our partial fraction decomposition, we analyze these quadratic factors to determine the form of our partial fractions. Since we have two distinct factors in the denominator, each quadratic expression gives rise to a separate partial fraction in the decomposition. This structure is essential for setting up the system of equations needed to find the coefficients for the numerators of these fractions.

System of Equations

To complete the partial fraction decomposition process, we need to equate the coefficients from the expanded equations to those of the original expression. Setting these coefficients equal gives rise to a system of equations. For our current problem, after expanding and rearranging the expression, we are left with equations derived from comparing coefficients on both sides of equation.Our system for this exercise is:

• \(A+C = 0\) for the coefficient of \(x^3\)
• \(A+B+2C+D = 2\) for the coefficient of \(x^2\)
• \(5A+B+9C+2D = 3\) for the coefficient of \(x\)
• \(5B+9D = 14\) for the constant term.

By solving this system, we can determine the values of \(A, B, C,\) and \(D\). These coefficients are then substituted back into our partial fractions to express our original rational expression in its decomposed form. This reveals simpler expressions that are much easier to integrate or manipulate.