Problem 48 - College Algebra

Question

For the following exercises, find the decomposition of the partial fraction for the irreducible repeating quadratic factor. $$\frac{x^{3}+2 x^{2}+4 x}{\left(x^{2}+2 x+9\right)^{2}}$$

Step-by-Step Solution

Answer

\( \frac{x}{x^2+2x+9} - \frac{5x}{(x^2+2x+9)^2} \)

Step 1: Identify the Denominator Structure

The given expression is \( \frac{x^{3}+2x^{2}+4x}{(x^{2}+2x+9)^{2}} \). Here, the denominator has an irreducible quadratic factor \((x^{2}+2x+9)\) which is repeated twice.

Step 2: Setup the Partial Fraction Decomposition

For a repeated irreducible quadratic factor like \((x^{2}+2x+9)^{2}\), the partial fraction decomposition takes the form: \(\frac{Ax+B}{x^2+2x+9} + \frac{Cx+D}{(x^2+2x+9)^2}.\)

Step 3: Multiply Both Sides by the Denominator

Eliminate the denominators by multiplying both sides of the equation by \((x^2+2x+9)^2\). This gives:\[ x^{3}+2x^{2}+4x = (Ax+B)(x^2+2x+9) + (Cx+D).\]

Step 4: Expand the Expressions

Expand the left-hand side:\[ (Ax+B)(x^2+2x+9) + (Cx+D) = Ax^3 + (2A+B)x^2 + (9A+2B)x + 9B +Cx + D. \] Equate this to the right-hand side to get:\[ x^3+2x^2+4x. \]

Step 5: Compare Coefficients

Compare coefficients of like terms from both sides:- For \(x^3\): \(A = 1\)- For \(x^2\): \(2A + B = 2\)- For \(x\): \(9A + 2B + C = 4\)- Constant: \(9B + D = 0\)

Step 6: Solve the System of Equations

Solving the equations:1. \(A = 1\)2. \(2(1) + B = 2 \Rightarrow B = 0\)3. \(9(1) + 2(0) + C = 4 \Rightarrow C = -5\)4. \(9(0) + D = 0 \Rightarrow D = 0\) Thus, \(A = 1, B = 0, C = -5, D = 0.\)

Step 7: Write the Partial Fraction Decomposition

Substitute the values into the partial fraction decomposition:\[ \frac{x^{3}+2x^{2}+4x}{(x^{2}+2x+9)^{2}} = \frac{1 \cdot x + 0}{x^2+2x+9} - \frac{5 \cdot x + 0}{(x^2+2x+9)^2}.\] Thus:\[ = \frac{x}{x^2+2x+9} - \frac{5x}{(x^2+2x+9)^2}.\]

Key Concepts

Irreducible Quadratic Factor

In mathematics, an irreducible quadratic factor is a quadratic expression that cannot be factored into real linear factors. This means that it cannot be broken down further into simpler polynomial terms. Instead, it has complex roots, and in exercises involving polynomial fractions, it's often the trickiest part of the process.
In the original exercise, we see the denominator \( (x^2+2x+9)^2 \). This quadratic factor \( x^2+2x+9 \) is irreducible because it does not have real numbers that satisfy the equation when set to zero. This lack of further factorization makes it crucial in forming the terms of the partial fraction decomposition.

Step-by-Step Solution

Let’s go through the process of partial fraction decomposition in a step-by-step manner. When confronted with a fraction where the denominator includes an irreducible quadratic factor, especially a repeated one like \((x^2+2x+9)^2\), each step carefully breaks down the complex expression into simpler components.
Unlike decomposing fractions with only linear factors, an irreducible quadratic requires setting up the decomposition in a specific form. The correct setup involves terms \( \frac{Ax+B}{x^2+2x+9} \) and \( \frac{Cx+D}{(x^2+2x+9)^2} \) to account for one and two appearances of the quadratic factor, respectively. This step is essential, as correct setup ensures proper simplification later in the exercise.

Coefficient Comparison

Once the partial fraction decomposition is set up, the next step involves coefficient comparison. This technique simplifies both the mathematical expression and the solution process.
After expanding the numerator terms, you're left with an expression on both sides of the equation. This necessitates matching the coefficients from each corresponding term with the given polynomial. Specifically, if the expression is equated to \(x^{3}+2x^{2}+4x\), as in the exercise, coefficients of powers of \(x\) must match, resulting in equations like:

• \(x^3: A = 1\),
• \(x^2: 2A + B = 2\),
• \(x: 9A + 2B + C = 4\),
• Constant term: \(9B + D = 0\).

These comparisons allow us to find specific values for \(A, B, C,\) and \(D\) when solving a set of algebraic equations.

System of Equations

Once the coefficients are compared, the problem transforms into a system of equations. A system of equations is a set of two or more equations that need to be solved together to find the values of variables involved.
Referring back to the set of equations in the exercise, solving involves substituting known values and solving algebraically:

• From \( A = 1 \), substitution in the equation \( 2A + B = 2 \) gives \( B = 0 \).
• Using \( B = 0 \) and \( A = 1 \), substituting in \( 9A + 2B + C = 4 \) yields \( C = -5 \).
• Lastly, \( B = 0 \) is plugged into \( 9B + D = 0 \), giving \( D = 0 \).

Using these methods efficiently leads us to the correct decomposition and exposes a common yet vital technique in solving complex algebraic expressions.