Problem 15 - Precalculus

Question

(a) factor the denominator of the given \(\mathrm{ra}\) tional expression; (b) determine the form of the partial fraction decomposition for the given rational expression; and (c) determine the values of the constants in the partial fraction decomposition that you gave in part (b). To help you in spotting errors, use the fact that in part (c), each of the required constants turns out to be an integer. $$\frac{2 x^{3}+5 x-4}{x^{4}+2 x^{2}+1}$$

Step-by-Step Solution

Answer

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

Step 1: Factor the Denominator

First, we need to factor the polynomial in the denominator \(x^4 + 2x^2 + 1\). Notice that this can be seen as a quadratic in terms of \(x^2\): \((x^2)^2 + 2(x^2) + 1\). This can be factored as \((x^2 + 1)^2\) since it follows the pattern \(a^2 + 2ab + b^2 = (a+b)^2\). Thus, we have:\[x^4 + 2x^2 + 1 = (x^2 + 1)^2\]

Step 2: Determine Form for Partial Fraction Decomposition

Now that we know the denominator is \((x^2 + 1)^2\), the rational expression becomes \(\frac{2x^3 + 5x - 4}{(x^2 + 1)^2}\). Since \(x^2 + 1\) is an irreducible quadratic factor raised to the second power, the partial fraction decomposition should take the form:\[\frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{(x^2 + 1)^2}\] where \(A\), \(B\), \(C\), and \(D\) are constants to be determined.

Step 3: Set Up Equation to Find Constants

Multiply both sides of the partial fraction equation by \((x^2 + 1)^2\) to clear the denominators:\[2x^3 + 5x - 4 = (Ax + B)(x^2 + 1) + (Cx + D)\]Expand the right-hand side:\[(Ax + B)(x^2 +1) = Ax^3 + Ax + Bx^2 + B\]So:\[Ax^3 + Bx^2 + Ax + B + Cx + D\]Combine the terms:\[Ax^3 + Bx^2 + (A + C)x + (B + D)\]

Step 4: Equate Coefficients to Solve for Constants

Equate the coefficients of the expanded polynomial to those on the left side (\(2x^3 + 5x - 4\)):1. \(A = 2\) (coefficient of \(x^3\))2. \(B = 0\) (coefficient of \(x^2\))3. \(A + C = 5\) (coefficient of \(x\))4. \(B + D = -4\) (constant term)Substitute \(A = 2\) and \(B = 0\) into the equations:1. \(2 + C = 5 \Rightarrow C = 3\)2. \(0 + D = -4 \Rightarrow D = -4\)

Step 5: Write Final Partial Fraction Decomposition

Now that we have the constants \(A = 2\), \(B = 0\), \(C = 3\), and \(D = -4\), substitute these into the form \(\frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{(x^2 + 1)^2}\):\[\frac{2x}{x^2 + 1} + \frac{3x - 4}{(x^2 + 1)^2}\] is the partial fraction decomposition of the original expression.

Key Concepts

Factorization of Polynomials

When tackling rational expressions, factorization plays a crucial role. In this exercise, we started by factorizing the denominator of the given rational expression: \[x^4 + 2x^2 + 1\]. Recognizing patterns in the polynomial is key. Here, the polynomial is expressed in terms of \(x^2\). Spotting that it resembles a perfect square trinomial, we rewrite it as:

• \((x^2)^2 + 2(x^2) + 1 = (x^2+1)^2\).

This highlights the importance of recognizing special algebraic identities, such as the square of a binomial: \[a^2 + 2ab + b^2 = (a+b)^2\].
This skill aids in simplifying complex polynomials, facilitating further operations like Partial Fraction Decomposition.
By breaking down the polynomial into easily manageable factors, we set a firm foundation for subsequent steps, making further calculations simpler and more intuitive.

Irreducible Quadratic Factors

In Partial Fraction Decomposition, identifying irreducible quadratic factors is crucial. This occurs when a quadratic cannot be factored further over the real numbers. In our example, we identified:

• The factor \(x^2 + 1\) as an irreducible quadratic factor.

Such factors have a distinct form in decomposition because they don't split into linear terms. When faced with \((x^2 + 1)^2\), the decomposition acknowledges.
Each power of the irreducible quadratic factor separately:

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

Understanding this concept helps in constructing the correct form of a partial fraction, ensuring all factors and their multiplicities are represented.
This step is integral in breaking down the expression into easier-to-handle sections, making finding constants straightforward.

Polynomial Long Division

While not explicitly required in solving this specific exercise, polynomial long division is often a complementary skill when working with rational expressions. Typically, if the numerator's degree is equal to or exceeds the denominator's degree, division is required.
In our case, since the degree of the numerator (3) is less than the degree of the denominator (4), division wasn't necessary. However, understanding polynomial long division remains valuable. This method involves:

• Dividing the first term in the numerator by the first term in the denominator.
• Multiplying the entire divisor by this result.
• Subtracting it from the original dividend.
• Repeating the steps with the new polynomial formed.

This process continues until the degree of the remainder is less than the degree of the divisor.
Polish this technique to bolster your toolbox for algebraic manipulations, aiding you whenever the terms necessitate it.