Question

Find the partial fraction decomposition of each rational expression. $$ \frac{x^{2}+9}{x^{4}-2 x^{2}-8} $$

Short answer

The partial fraction decomposition is:\(\frac{5}{16}\frac{1}{x-2} + \frac{1}{16}\frac{1}{x+2} + \frac{-x+2}{x^{2}+2}\).

Step-by-step solution

Factor the Denominator

Factor the denominator of the rational expression. Given the denominator is \(x^{4}-2x^{2}-8\), we first make a substitution to simplify the factoring process. Let \(u = x^{2}\), making the denominator \(u^{2}-2u-8\). Now factor \(u^{2}-2u-8\) to get \((u-4)(u+2)\). Re-substitute \(u = x^{2}\) back into the expression, providing \((x^{2}-4)(x^{2}+2)\).

Factor Further

Factor \(x^{2}-4\) as a difference of squares: \(x^{2}-4 = (x-2)(x+2)\). Now, the full factorization of the denominator is \((x-2)(x+2)(x^{2}+2)\).

Set up the Partial Fractions

Express the original fraction \(\frac{x^{2}+9}{(x-2)(x+2)(x^{2}+2)}\) as a sum of partial fractions. Assume the decomposition takes the form: \(\frac{x^{2}+9}{(x-2)(x+2)(x^{2}+2)} = \frac{A}{x-2} + \frac{B}{x+2} + \frac{Cx+D}{x^{2}+2}\).

Clear the Denominator

Multiply both sides by the common denominator \((x-2)(x+2)(x^{2}+2)\). This gives: \(x^{2}+9 = A(x+2)(x^{2}+2) + B(x-2)(x^{2}+2) + (Cx+D)(x-2)(x+2)\).

Expand and Simplify

Expand and simplify the right side of the equation: \(x^{2}+9 = A(x^{3} + 2x + 2x + 4) + B(x^{3} - 2x - 2x + 4) + (Cx+D)(x^{2}-4)\). This further simplifies to: \(x^{2}+9 = Ax^{3} + 4Ax + 4A + Bx^{3} - 4Bx + 4B + Cx^{3}x^{2} - 4Cx + Dx^{2} - 4D\).

Compare Coefficients

Group and compare the coefficients of like terms on both sides of the equation: - Coefficient of \(x^{3}\): \(A + B + C = 0\) - Coefficient of \(x^{2}\): \(A + B + D = 1\) - Coefficient of \(x\): \(4A - 4B + 1= 0\) - Constant term: \(4A + 4B - 4D = 9\)Solving these equations will provide the values of \(A\), \(B\), \(C\), and \(D\).

Solve the System of Equations

Solve the system of linear equations: - \(C = -1\) - \(4A - 4B - 1 = 0\), implies \(4A - 4B = 1\) or \(A - B = \frac{1}{4}\) - Further steps yield \(A = \frac{5}{16}\), \(B = \frac{1}{16}\), and \(D = 2\)With these values, the partial fraction decomposition is finalized.

Write the Final Decomposition

Substitute the values back into the partial fraction decomposition setup to get: \(\frac{x^{2}+9}{(x-2)(x+2)(x^{2}+2)} = \frac{\frac{5}{16}}{x-2} + \frac{\frac{1}{16}}{x+2} + \frac{-x+2}{x^{2}+2}\).

Key concepts

Factoring Polynomials

Understanding how to factor polynomials is essential for solving partial fraction decomposition problems. Factoring simplifies the polynomial into a product of lower-degree polynomials. Let's break down the process used in our example:

• First, we recognize that the polynomial in our denominator, \(x^{4}-2x^{2}-8\), is complex. To simplify, we substitute \(u = x^{2}\). This changes our polynomial to \(u^{2}-2u-8\).
• Next, we factor \(u^{2}-2u-8\) into \((u-4)(u+2)\). By re-substituting \(u = x^{2}\), we get \((x^{2}-4)(x^{2}+2)\).
• In the final step, we further factor \(x^{2}-4\) into \( (x-2)(x+2)\), giving us the fully factored form \((x-2)(x+2)(x^{2}+2)\).

Factoring helps reduce complexity, making it easier to solve the rational expression in partial fraction decomposition.

Rational Expressions

Rational expressions are fractions where the numerator and denominator are polynomials. To solve these, especially in partial fractions, we need to manipulate and simplify these expressions.

• An important first step is to ensure the polynomial in the denominator is factored completely, as highlighted in the previous section.
• For example, the rational expression \(\frac{x^{2}+9}{(x-2)(x+2)(x^{2}+2)}\) involves polynomials in both the numerator \(x^{2}+9\) and the denominator \((x-2)(x+2)(x^{2}+2)\).
• The goal is to express this rational expression as a sum of simpler fractions, making it easier to work with and solve.

Simplifying rational expressions through partial fraction decomposition helps solve systems of equations more efficiently.

System of Equations

In partial fraction decomposition, setting up and solving a system of equations is crucial. Here's how to tackle it:

• Once the denominator is factored, we set up an equation to express the rational expression as a sum of simpler fractions. For instance, \(\frac{x^{2}+9}{(x-2)(x+2)(x^{2}+2)}\) becomes \( \frac{A}{x-2} + \frac{B}{x+2} + \frac{Cx+D}{x^{2}+2} \).
• We then multiply both sides by the common denominator to eliminate the fractions, simplifying them into polynomial form.
• Next, we equate the coefficients of like terms from both sides of the equation to set up our system of equations. This yields equations like \(A + B + C = 0\), \(A + B + D = 1\), etc.
• Solving this system helps find the values of constants A, B, C, and D.

This systematic approach is essential to break down the problem and find the solution to partial fractions.

Comparing Coefficients

Comparing coefficients is a technique used to solve for unknowns in partial fraction decomposition. Here's how it works:

• Once the denominators are cleared, we expand and simplify both sides. For our example, this means expanding \(A(x+2)(x^{2}+2) + B(x-2)(x^{2}+2) + (Cx+D)(x-2)(x+2)\).
• After expansion, we group like terms (terms with the same degree of x) from both sides of the equation.
• Next, we set the coefficients of these like terms equal to each other, forming our system of equations. E.g., equate coefficients of \(x^{3}\), \(x^{2}\), x, and constants separately.
• Solving the system provides the values for A, B, C, and D, like solving \(A + B + C = 0\), etc.

Comparing coefficients ensures we accurately solve for each variable, leading us to the correct partial fraction decomposition.