Problem 38 - Algebra and Trigonometry

Question

Find the partial fraction decomposition of the rational function. \(\frac{x^{2}+x+1}{2 x^{4}+3 x^{2}+1}\)

Step-by-Step Solution

Answer

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

Step 1: Factoring the Denominator

The denominator of the function is \(2x^4 + 3x^2 + 1\). Notice it can be rewritten using substitution \(y = x^2\), yielding \(2y^2 + 3y + 1\). This quadractic can be factored: \((2y + 1)(y + 1)\). Substituting back gives the factorization: \((2x^2 + 1)(x^2 + 1)\).

Step 2: Set Up the Partial Fractions

Set up the partial fraction decomposition based on the factors of the denominator: \(\frac{x^2 + x + 1}{(2x^2 + 1)(x^2 + 1)} = \frac{Ax + B}{2x^2 + 1} + \frac{Cx + D}{x^2 + 1}\).

Step 3: Clear the Denominator

Multiply both sides by \((2x^2 + 1)(x^2 + 1)\) to clear the fraction: \(x^2 + x + 1 = (Ax + B)(x^2 + 1) + (Cx + D)(2x^2 + 1)\).

Step 4: Distribute and Combine Like Terms

Expand the expressions on the right:\((Ax + B)(x^2 + 1) = Ax^3 + Ax + Bx^2 + B\)\((Cx + D)(2x^2 + 1) = 2Cx^3 + Cx + 2Dx^2 + D\)Combine these:\((A+2C)x^3 + (B+2D)x^2 + (A+C)x + (B+D)\).

Step 5: Equate Coefficients

Equate the coefficients from both sides of the equation:1. \(A + 2C = 0\) for \(x^3\)2. \(B + 2D = 0\) for \(x^2\)3. \(A + C = 1\) for \(x\)4. \(B + D = 1\) for the constant term.

Step 6: Solve the System of Equations

Solve the system of equations:1. From \(A + 2C = 0\) and \(A + C = 1\), substitute \(A = -2C\) into \(A + C = 1\) to find \(C = -1\) and consequently \(A = 2\).2. From \(B + 2D = 0\) and \(B + D = 1\), substitute \(B = -2D\) into \(B + D = 1\) to find \(D = -1\) and consequently \(B = 2\).

Step 7: Write the Decomposition

Substitute \(A = 2\), \(B = 2\), \(C = -1\), and \(D = -1\) back into the partial fractions:\(\frac{2x + 2}{2x^2 + 1} - \frac{x + 1}{x^2 + 1}\).Thus, the partial fraction decomposition is:\(\frac{x^{2}+x+1}{2 x^{4}+3 x^{2}+1} = \frac{2x + 2}{2x^2 + 1} - \frac{x + 1}{x^2 + 1}\).

Key Concepts

Rational Function

A rational function is a type of mathematical function represented by the ratio of two polynomials. Think of it similar to a fraction, but instead of integers, it involves polynomials. For example, in the function \(\frac{x^{2}+x+1}{2x^{4}+3x^{2}+1}\), the numerator \(x^2 + x + 1\) and the denominator \(2x^4 + 3x^2 + 1\) are both polynomials.
This rational function can often be simplified or expressed in different forms, such as through partial fraction decomposition. This process helps in breaking down complex fractions into simpler components, making them easier to integrate or differentiate.
Understanding rational functions is key in calculus and many other areas of mathematics, as it applies in real-world scenarios like engineering and physics.

Factoring Polynomial

Factoring a polynomial involves writing it as a product of simpler polynomials. It's a crucial step in partial fraction decomposition because we need to break down the denominator into its factors.
In our example, the denominator is \(2x^4 + 3x^2 + 1\). By substituting \(y = x^2\), it gets simplified to \(2y^2 + 3y + 1\), which is a quadratic polynomial. Using techniques like factoring by grouping or the quadratic formula, this can be factored into \((2y + 1)(y + 1)\).

Finally, substitute back \(y = x^2\) to get \((2x^2 + 1)(x^2 + 1)\). Factoring helps to identify the terms involved in the partial fraction decomposition accurately, simplifying further steps in solving the equation.

System of Equations

A system of equations is a set of multiple equations that are all simultaneously considered. In partial fraction decomposition, setting up and solving a system of equations is mandatory to find the coefficients accompanying each fraction.

Once the polynomial is factored, we express the original function as a sum of simpler fractions. Each fraction's coefficients are unknowns, and we need to establish relationships between them. For our exercise, we had two fractions yielding equations for coefficients \(A, B, C, D\). This forms a system of equations:

• \(A + 2C = 0\)
• \(B + 2D = 0\)
• \(A + C = 1\)
• \(B + D = 1\)

Solving such systems involves techniques like substitution or elimination to pinpoint these unknowns. They are essential in obtaining the correct values for the partial fraction decomposition.

Equating Coefficients

The equating coefficients method is a mathematical technique used to find unknown coefficients in a polynomial expression. This process involves expanding and simplifying both sides of an equation and matching the coefficients of like terms from both sides.

For our partial fraction decomposition exercise, after clearing the denominator by multiplying through, we expand:

• \((Ax + B)(x^2 + 1)\)
• \((Cx + D)(2x^2 + 1)\)

which collectively give terms equating to the original polynomial in the numerator \(x^2 + x + 1\).
As we group and combine similar terms, we form equations equated to their counterparts in the polynomial. Then, by solving them like in a typical system of equations, we identify the coefficients. This step is critical to ensure that each part of the decomposed fraction represents a true part of the original rational function.