Problem 39 - Precalculus Mathematics for Calculus

Question

Find the partial fraction decomposition of the rational function. $$\frac{2 x^{3}+7 x+5}{\left(x^{2}+x+2\right)\left(x^{2}+1\right)}$$

Step-by-Step Solution

Answer

The partial fraction decomposition is \(\frac{x - 3}{x^2 + x + 2} + \frac{x + 4}{x^2 + 1}\).

Step 1: Identify the Denominator

Examine the denominator of the given rational function, which is \((x^2 + x + 2)(x^2 + 1)\). Notice that both factors are irreducible quadratic polynomials.

Step 2: Set Up the Partial Fractions

Because the denominator consists of two irreducible quadratic factors, the partial fraction decomposition will have the form \(\frac{Ax + B}{x^2 + x + 2} + \frac{Cx + D}{x^2 + 1}\).

Step 3: Multiply Through by the Denominator

Eliminate the denominator from each term by multiplying throughout by \((x^2 + x + 2)(x^2 + 1)\), resulting in:\[2x^3 + 7x + 5 = (Ax + B)(x^2 + 1) + (Cx + D)(x^2 + x + 2)\]

Step 4: Expand and Combine Like Terms

Expand the polynomial expressions on the right-hand side:\[(Ax + B)(x^2 + 1) = Ax^3 + Ax + Bx^2 + B\]\[(Cx + D)(x^2 + x + 2) = Cx^3 + Cx^2 + 2Cx + Dx^2 + Dx + 2D\]Combine these into a single expression:\[(A + C)x^3 + (B + C + D)x^2 + (A + 2C + D)x + (B + 2D)\]

Step 5: Equate Coefficients

Compare coefficients from both sides of the expanded equation. This gives us:1. \(A + C = 2\)2. \(B + C + D = 0\)3. \(A + 2C + D = 7\)4. \(B + 2D = 5\)

Step 6: Solve the System of Equations

Solve the system of equations for \(A\), \(B\), \(C\), and \(D\).- First equation implies \(C = 2 - A\).- Substitute \(C = 2 - A\) into the second and third equations to simplify and find each unknown step by step.- Solving, you get \(A = 1\), \(C = 1\), \(B = -3\), \(D = 4\).

Step 7: Write the Final Decomposition

Substitute \(A\), \(B\), \(C\), and \(D\) into the partial fraction form:\[\frac{x - 3}{x^2 + x + 2} + \frac{x + 4}{x^2 + 1}\]

Key Concepts

Understanding Irreducible Quadratic Polynomials

When faced with the task of decomposing a rational function, it's crucial to correctly identify irreducible quadratic polynomials in the denominator. These are quadratics that cannot be factored further over the real number system, meaning their roots are complex. In the given exercise, the denominator is composed of two quadratic expressions: \(x^2 + x + 2\) and \(x^2 + 1\). Both are irreducible because they do not have real roots.
For a quadratic \(ax^2 + bx + c\), calculate the discriminant using \(b^2 - 4ac\):

• If the discriminant is less than zero, the quadratic is irreducible over the reals.
• For example, for \(x^2 + x + 2\), the discriminant is \(1^2 - 4 \times 1 \times 2 = -7\), which is negative. Hence, it's irreducible.
• Similarly, the discriminant for \(x^2 + 1\) is \(0^2 - 4 \times 1 \times 1 = -4\).

Recognizing these irreducible quadratics allows us to set up the correct form for partial fraction decomposition.

Solving with a System of Equations

To perform a partial fraction decomposition of a rational function, especially when involving irreducible quadratics, we set up a system of equations to find unknown coefficients in the numerators. In our exercise, after multiplying through by the common denominator and expanding, we equate coefficients from both sides of the equation.
This leads to a set of linear equations that can be solved simultaneously. For example, in the exercise, we find:

• \(A + C = 2\)
• \(B + C + D = 0\)
• \(A + 2C + D = 7\)
• \(B + 2D = 5\)

These equations determine values of \(A, B, C,\) and \(D\). Solving them step by step involves substitution and elimination methods, which may include:

• Expressing one variable in terms of another, like \(C = 2 - A\).
• Substituting into the other equations and solving sequentially.
• These methods ensure that each coefficient matches, simplifying the right-hand side back to the original left-hand side polynomial.

This systematic approach is powerful for breaking down complex rational functions.

Rational Function Decomposition Insights

Rational function decomposition is the process of expressing the given rational function as a sum of simpler fractions. This procedure is pivotal in calculus, especially when integrating complex expressions. The original function in the exercise \(\frac{2 x^{3}+7 x+5}{\left(x^{2}+x+2\right)\left(x^{2}+1\right)}\) is decomposed into two fractions, each with irreducible quadratic denominators.
Here's how the decomposition helps:

• It simplifies integrals of complex rational functions.
• It transforms difficult algebraic expressions into easier-to-manage components.
• It aids in solving differential equations by breaking tasks into simpler parts.

To solve the decomposition, understanding the need for handling irreducible quadratics in specific forms \(\frac{Ax + B}{x^2 + x + 2} + \frac{Cx + D}{x^2 + 1}\) is crucial.
Partial fraction decomposition not only simplifies the algebra but also provides deeper insights into the behavior of rational functions, which are essential for advanced mathematical concepts and real-world applications.