Problem 30 - College Algebra

Question

For the following exercises, find the decomposition of the partial fraction for the repeating linear factors. $$\frac{x^{3}-5 x^{2}+12 x+144}{x^{2}\left(x^{2}+12 x+36\right)}$$

Step-by-Step Solution

Answer

\( \frac{-1}{x} + \frac{4}{x^2} + \frac{2}{x+6} - \frac{17}{(x+6)^2} \)

Step 1: Identify the Factors

First, identify the factors in the denominator. Here, the denominator is given as \( x^2 (x^2 + 12x + 36) \). The expression \( x^2 + 12x + 36 \) can be rewritten as \((x+6)^2\). Hence, the complete factorization of the denominator is \( x^2 (x+6)^2 \).

Step 2: Write Partial Fraction Form

For the expression \( \frac{x^3 - 5x^2 + 12x + 144}{x^2(x+6)^2} \), decompose it into partial fractions. The form will be:\[ \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+6} + \frac{D}{(x+6)^2} \]

Step 3: Clear the Denominator

Multiply through by the common denominator \(x^2 (x+6)^2\) to eliminate the fractions, resulting in the equation:\[ x^3 - 5x^2 + 12x + 144 = A(x)(x+6)^2 + B(x+6)^2 + Cx^2(x+6) + Dx^2 \]

Step 4: Expand and Combine

Expand each term on the right-hand side:- \( A(x)(x+6)^2 = A(x^3 + 12x^2 + 36x) \)- \( B(x+6)^2 = B(x^2 + 12x + 36) \)- \( Cx^2(x+6) = C(x^3 + 6x^2) \)- \( Dx^2 = Dx^2 \)Combine and simplify these to form a polynomial expression comparable to the left-hand side.

Step 5: Set Up and Solve Equations

After expansion and combining like terms, equate coefficients from both sides.- From \( x^3 \): \( A + C = 1 \)- From \( x^2 \): \( 12A + B + 6C + D = -5 \)- From \( x \): \( 36A + 12B = 12 \)- Constant term: \( 36B = 144 \)Solve these equations for \( A, B, C, D \).

Step 6: Solve for Constants

Start by solving the simplest equation: \( 36B = 144 \), so \( B = 4 \).Substitute \( B = 4 \) into other equations:- \( 36A + 48 = 12 \) gives \( 36A = -36 \) so \( A = -1 \).- Substitute \( A = -1 \) into \( A + C = 1 \) gives \( C = 2 \).- Finally, substitute \( A, B, C \) into \( 12A + 4 + 12 + D = -5 \) gives \( D = -17 \).

Step 7: Write the Final Partial Fraction

The partial fraction decomposition is:\[ \frac{-1}{x} + \frac{4}{x^2} + \frac{2}{x+6} - \frac{17}{(x+6)^2} \]

Key Concepts

Repeating Linear Factors

When dealing with partial fraction decomposition, encountering repeating linear factors in the denominator can be quite common. A repeating linear factor is simply a linear polynomial raised to a power greater than one. In the example problem, the denominator factor \( (x+6)^2 \) is a repeating linear factor. When decomposing into partial fractions, each repetition of the linear factor requires its own term in the decomposition. Therefore, if \( (x+6) \) appears squared in the denominator, the partial fraction set up will include terms for both \( \frac{C}{x+6} \) and \( \frac{D}{(x+6)^2} \). This approach ensures all aspects of the polynomial are accounted for when solving the overall equation.

Denominator Factorization

Denominator factorization is the first crucial step in partial fraction decomposition. You need to carefully factor the denominator to break it down into its simplest components. For example, in the given exercise, the initial denominator \( x^2(x^2 + 12x + 36) \) was factored into \( x^2(x+6)^2 \). The expression \( x^2 + 12x + 36 \) itself was rewritten as \((x+6)^2\). Ensuring the denominator is completely factored, including recognizing repeated factors, is essential for setting up the partial fractions correctly. Once factored, each distinct factor will correspond to a separate term in the partial fraction decomposition. This structuring aids in balancing both sides of the equation later on.

Equation Solving

Solving the equations that arise from setting up the partial fractions is often the most challenging part. After expressing the decomposition, the next step is to multiply through by the common denominator to remove the fractions, leaving you with a polynomial equation. For instance, in the given exercise, multiplying through gives an equation without fractions:

• \(x^3 - 5x^2 + 12x + 144 = A(x)(x+6)^2 + B(x+6)^2 + Cx^2(x+6) + Dx^2\)

Expanding and combining like terms allows you to match the coefficients from each side of the equation. This results in a system of equations based on matching coefficients of corresponding powers of \(x\). Solving this system gives the unknown constants \(A\), \(B\), \(C\), and \(D\), which are crucial for the final expression of the partial fraction decomposition.

Partial Fractions Setup

Setting up the partial fractions is a systematic step that requires attention to details. First, recognize the forms of the terms based on the denominator's factors. In our exercise, after decomposing \( \frac{x^3 - 5x^2 + 12x + 144}{x^2(x+6)^2} \), the setup is:

• \(\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+6} + \frac{D}{(x+6)^2}\)

Each term corresponds to a factor of the denominator. For repeated linear factors, like \(x^2\) and \((x+6)^2\), every instance from the power requires a term. Here, \(A\) and \(B\) cover \(x\) and \(x^2\), while \(C\) and \(D\) cover \(x+6\) and \((x+6)^2\). The approach reflects the structure of the denominator, ensuring all possible contributions in the polynomial are represented. After setting up these equations, you are ready for the multiplication and solving steps.