Problem 30 - College Algebra with Corequisite Support

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

The partial fraction decomposition is \( \frac{-1}{x} + \frac{4}{x^2} + \frac{2}{x+6} - \frac{9}{(x+6)^2} \).

Step 1: Identify the Factorization

First, consider the factorization of the denominator. We see that the denominator is given as \(x^2(x^2 + 12x + 36)\). Notice that \(x^2 + 12x + 36\) can be factored further as \((x + 6)^2\). Therefore, the full factorization of the denominator is \(x^2(x + 6)^2\).

Step 2: Set Up the Partial Fraction Decomposition

The function can be decomposed into partial fractions as the equation:\[ \frac{x^3 - 5x^2 + 12x + 144}{x^2(x + 6)^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x + 6} + \frac{D}{(x + 6)^2} \]Where \(A\), \(B\), \(C\), and \(D\) are constants to be determined.

Step 3: Clear the Denominator

Multiply both sides of the equation by the common denominator \(x^2(x + 6)^2\) to eliminate the fractions:\[ x^3 - 5x^2 + 12x + 144 = A(x)(x + 6)^2 + B(x + 6)^2 + C(x^2)(x + 6) + D(x^2) \].

Step 4: Expand and Simplify Each Term

Expand each term on the right side:- \(A(x)(x + 6)^2 = A(x)(x^2 + 12x + 36) = Ax^3 + 12Ax^2 + 36Ax\)- \(B(x + 6)^2 = B(x^2 + 12x + 36) = Bx^2 + 12Bx + 36B\)- \(C(x^2)(x + 6) = C(x^3 + 6x^2) = Cx^3 + 6Cx^2\)- \(D(x^2) = Dx^2\).

Step 5: Combine Like Terms

Combine terms to form an expanded polynomial on the right side:\[ Ax^3 + Cx^3 + (12A + B + 6C + D)x^2 + (36A + 12B)x + 36B = x^3 - 5x^2 + 12x + 144 \].

Step 6: Equate Coefficients

Now, equate the coefficients of \(x^3\), \(x^2\), \(x\), and the constant term from both sides.- Coefficient of \(x^3\): \(A + C = 1\)- Coefficient of \(x^2\): \(12A + B + 6C + D = -5\)- Coefficient of \(x\): \(36A + 12B = 12\)- Constant term: \(36B = 144\).

Step 7: Solve for B

From the equation \(36B = 144\), we solve for \(B\):\[ B = \frac{144}{36} = 4 \].

Step 8: Solve for A

Substitute \(B = 4\) into \(36A + 12B = 12\):\[ 36A + 48 = 12 \]\[ 36A = 12 - 48 = -36 \]\[ A = \frac{-36}{36} = -1 \].

Step 9: Solve for C

Substitute \(A = -1\) and \(B = 4\) into \(A + C = 1\):\[ -1 + C = 1 \]\[ C = 2 \].

Step 10: Solve for D

Finally, substitute \(A = -1\), \(B = 4\), and \(C = 2\) into \(12A + B + 6C + D = -5\):\[ 12(-1) + 4 + 12 + D = -5 \]\[ -12 + 4 + 12 + D = -5 \]\[ 4 + D = -5 \]\[ D = -5 - 4 = -9 \].

Step 11: Conclusion: Write the Decomposition

The partial fraction decomposition is as follows:\[ \frac{x^3 - 5x^2 + 12x + 144}{x^2(x + 6)^2} = \frac{-1}{x} + \frac{4}{x^2} + \frac{2}{x+6} + \frac{-9}{(x+6)^2} \].

Key Concepts

Repeating Linear Factors

In partial fraction decomposition, identifying **repeating linear factors** is a crucial step.
These are factors of the denominator that repeat and are linear, meaning they have a degree of one. In this exercise, the denominator is \( x^2(x + 6)^2 \), where both \( x^2 \) and \( (x + 6)^2 \) are repeating linear factors.

Handling these factors requires a specific setup:

• Each unique factor gets its term in the decomposition.
• For each repeated exponent, additional terms are included for each lesser degree down to one.

For example, because \( x^2 \) is a factor, two terms \( \frac{A}{x} \) and \( \frac{B}{x^2} \) are introduced in the decomposition. The same logic applies to \( (x + 6)^2 \), resulting in terms \( \frac{C}{x + 6} \) and \( \frac{D}{(x + 6)^2} \).
This approach ensures every aspect of the repeating linear factors is considered in the partial fraction setup.

Polynomial Factorization

**Polynomial factorization** is the process of breaking down a polynomial into its simplest components or factors. In this exercise, we start with the denominator \( x^2(x^2 + 12x + 36) \).
The expression \( x^2 + 12x + 36 \) can be further factorized by recognizing it as a perfect square:
\( x^2 + 12x + 36 = (x + 6)(x + 6) = (x + 6)^2 \).

This gives the fully factored form: \( x^2(x + 6)^2 \). Being able to identify and perform this factorization step:

• Simplifies the process of setting up partial fractions.
• Makes solving for unknown constants straightforward.

Thus, polynomial factorization sets the stage for an easier decomposition process in partial fraction scenarios.

Equating Coefficients

**Equating coefficients** is a technique used to find the constants in the partial fraction decomposition.
Once we have cleared the denominator by multiplying through, we arrive at an expanded polynomial equality.
Through this, we can directly compare coefficients of like terms across both sides of the equation.

In this exercise, from the equation:

• \( (A + C)x^3 + (12A + B + 6C + D)x^2 + (36A + 12B)x + 36B = x^3 - 5x^2 + 12x + 144 \)

we equate corresponding coefficients for each term:

• The coefficient of \( x^3 \): \( A + C = 1 \)
• The coefficient of \( x^2 \): \( 12A + B + 6C + D = -5 \)
• The coefficient of \( x \): \( 36A + 12B = 12 \)
• The constant term: \( 36B = 144 \).

This systematic approach allows us to derive each unknown constant value by solving a system of linear equations.

Algebraic Expressions

An **algebraic expression** is a mathematical phrase that can contain numbers, variables, and operational symbols.
In partial fraction decomposition, algebraic expressions form the numerator and denominator of rational functions to be decomposed.
For our exercise: the expression

\( \frac{x^3 - 5x^2 + 12x + 144}{x^2(x^2 + 12x + 36)} \)
is initially complex. Through factorization and decomposition, it becomes a simpler algebraic expression:

• \( \frac{-1}{x} + \frac{4}{x^2} + \frac{2}{x+6} + \frac{-9}{(x+6)^2} \).

Each piece of the new expression is easier to manage, understand, and integrate, or differentiate if necessary.
Utilizing partial fraction decomposition turns a complex algebraic expression into more digestible parts, which lays the foundation for further algebraic or calculus operations.