Determine The Partial Fraction Expansion For The Rational Function Below

Partial Fraction Expansion: A Step‑by‑Step Guide to Decomposing Rational Functions

Partial fraction expansion is a powerful algebraic technique that breaks down a complex rational function into a sum of simpler fractions. Because of that, this decomposition is invaluable in calculus, differential equations, and engineering, where integrating or solving expressions becomes far easier once the original function is split into manageable pieces. In this article we will explore the theory behind partial fractions, outline a systematic procedure for determining the expansion, and walk through a detailed example that illustrates each stage of the process.

Introduction to Partial Fraction Expansion

A rational function is defined as the ratio of two polynomials, (R(x)=\frac{P(x)}{Q(x)}), where the degree of (P) is less than the degree of (Q). When the denominator (Q(x)) can be factored into linear or irreducible quadratic factors, the function can be expressed as a sum of simpler fractions—each having a denominator that is one of those factors. This representation is called the partial fraction expansion.

The primary benefits of this method include:

• Simpler integration: Each term can often be integrated using basic logarithmic or arctangent formulas.
• Easier inverse Laplace transforms: Many engineering problems rely on converting functions back to the time domain.
• ** clearer analysis**: The behavior of the function near poles (roots of the denominator) becomes transparent.

Understanding how to determine this expansion is therefore a foundational skill for students and professionals working with algebraic manipulations.

The General Procedure

1. Verify Proper Rational Form

First, see to it that the rational function is proper (degree of numerator < degree of denominator). If it is improper, perform polynomial long division to rewrite it as a polynomial plus a proper fraction before proceeding.

2. Factor the Denominator Completely

Factor (Q(x)) into linear factors ((x-a)) and irreducible quadratic factors ((x^2+bx+c)) over the real numbers. Repeated factors must be noted, because they affect the form of the expansion But it adds up..

3. Set Up the Partial Fraction Template

Based on the factor types, write the appropriate generic expression:

• Distinct linear factors:
  [ \frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b} ]

• Repeated linear factors (e.g., ((x-a)^2)):
  [ \frac{P(x)}{(x-a)^2} = \frac{A}{x-a} + \frac{B}{(x-a)^2} ]

• Irreducible quadratic factors:
  [ \frac{P(x)}{x^2+bx+c} = \frac{Cx+D}{x^2+bx+c} ]

• Mixed cases combine the above patterns Worth knowing..

4. Clear Denominators and Equate Coefficients

Multiply the entire equation by the original denominator to eliminate fractions. Then expand the right‑hand side and collect like terms. By matching coefficients of like powers of (x) on both sides, you obtain a system of linear equations for the unknown constants (A, B, C, D,) etc Easy to understand, harder to ignore. No workaround needed..

5. Solve the System

Use substitution, elimination, or matrix methods to find the constants. In many cases, the cover‑up method can quickly give the values for simple linear factors by evaluating the numerator at the factor’s root The details matter here..

6. Verify the Expansion

Plug the obtained constants back into the partial fraction form and combine the terms over a common denominator. The result should simplify exactly to the original rational function, confirming correctness.

Detailed Example

Let us determine the partial fraction expansion for the rational function

[ R(x)=\frac{3x^{2}+5x+2}{(x-1)(x+2)^{2}(x^{2}+4)}. ]

Step 1: Proper Form

The numerator degree is 2, denominator degree is 5, so the function is already proper It's one of those things that adds up..

Step 2: Factor the Denominator

The denominator is already factored as:

• Linear factor: ((x-1))
• Repeated linear factor: ((x+2)^{2})
• Irreducible quadratic factor: ((x^{2}+4)) (since its discriminant (b^{2}-4ac = 0-16 < 0))

Step 3: Set Up the Template

Following the rules for repeated linear and irreducible quadratic factors, we write:

[ \frac{3x^{2}+5x+2}{(x-1)(x+2)^{2}(x^{2}+4)}= \frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^{2}}+\frac{Dx+E}{x^{2}+4}. ]

Step 4: Clear Denominators

Multiply both sides by ((x-1)(x+2)^{2}(x^{2}+4)):

[ 3x^{2}+5x+2 = A(x+2)^{2}(x^{2}+4) + B(x-1)(x+2)(x^{2}+4) + C(x-1)(x^{2}+4) + (Dx+E)(x-1)(x+2)^{2}. ]

Step 5: Expand and Collect Terms

We expand each term (the algebra is lengthy but systematic). After expansion, we gather coefficients for (x^{4}, x^{3}, x^{2}, x^{1}, x^{0}). This yields a system of five equations:

[ \begin{cases} \text{Coeff. of }x^{4}: & A + B + D = 0 \ \text{Coeff. of }x^{3}: & 4A + 2B + 2D + E = 0 \ \text{Coeff. of }x^{2}: & 8A + 4B + C + 2E = 3 \ \text{Coeff. of }x^{1}: & 8A + 4B + 2C + 2D + 4E = 5 \ \text{Coeff.

(The exact coefficients arise from multiplying out each bracket; the reader can verify them by expanding.)

Step 6: Solve for Constants

We can solve this linear system using substitution or matrix methods. Here is a quick solution using the cover‑up method for the simple linear factors:

• Find (A): Set (x=1) (zeroes the terms containing (x-1)):
  [ A = \frac{3(1)^{2}+5(1)+2}{(1+2)^{2}(1^{2}+4)} = \frac{

Here's a thinking process:

1. Analyze User Input:

• User wants me to continue an article about partial fraction decomposition.
• Specific instruction: "Continue the article without friction. Do not repeat previous text. Finish with a proper conclusion."
• They provided a partial text that ends at: A = \frac{3(1)^{2}+5(1)+2}{(1+2)^{2}(1^{2}+4)} = \frac
• The text has sections 1-6, a detailed example, Step 1-6, and Step 6 is partially written (solving for constants, starting with cover-up method for A).
• I need to continue from where it left off, fill in the missing math, complete the solution for A, then likely B, C, D, E, verify, and conclude properly.

2. Identify the Gap:

• The text ends abruptly at A = \frac{3(1)^{2}+5(1)+2}{(1+2)^{2}(1^{2}+4)} = \frac
• I need to compute that fraction, finish the step for A, then proceed to find B, C, D, E.
• The system of equations is already set up in Step 6 (the 5 equations).
• I should use the cover-up method or substitution/elimination to solve for all constants.
• The user said "Do not repeat previous text." So I shouldn't reprint the whole system or steps verbatim unless necessary, but I need to continue the narrative.
• I need to finish with a "proper conclusion."

3. Plan the Continuation:

• Finish calculating A: 3(1)^2 + 5(1) + 2 = 3+5+2 = 10. Denominator: (1+2)^2 = 9, (1^2+4) = 5, so 9*5 = 45. Thus A = 10/45 = 2/9.
• Then I can find B using cover-up for the repeated linear factor? Actually, cover-up for (x+2)^2 isn't direct because it's repeated. Usually, we can find C by plugging x = -2, since terms with (x+2) and (x+2)^2 vanish except the C term. Let's check: If we set x = -2, the equation after clearing denominators becomes: 3(-2)^2 + 5(-2) + 2 = C(-2-1)((-2)^2+4) => 12 -10 +2 = C(-3)(8) => 4 = -24C => C = -1/4. That's standard.
• For B, since it's a repeated linear factor, cover-up doesn't directly give B, but we can use other values or solve the system. Alternatively, we can differentiate and plug in, but easier to just plug in other convenient x values or use the system. I'll probably use x = 0, x = -1, etc., or just solve the 5x5 system. Given the system is already set up, I can reference it and solve it, or use the method of equating coefficients.
• Actually, the text already has the system of 5 equations. I could just solve it, but the user might expect me to continue the example logically.
• Let me structure the continuation:
  • Finish A = 2/9.
  • Use x = -2 to find C = -1/4.
  • Then perhaps use other x values (like 0, 1 is already used, -1, 2) to get equations for B, D, E.
  • Or just solve the linear system using matrix methods as mentioned in section 5.
  • I'll do a mix: find C via cover-up/direct substitution, then maybe find B by plugging x = 0 or using the system, and D, E.
  • Actually, to keep it seamless and not repeat, I'll just compute A, then find C by plugging x = -2, then maybe plug x = 0 to get one equation, x = -1 another, x = 2 another, and solve the remaining. But the system is already there, so I could just say "solving the system yields..." and give results.
  • Wait, the user said "Do not repeat previous text." The previous text already has the system and says "(The exact coefficients arise from multiplying out each bracket; the reader can verify them by expanding.)" So I should probably not just copy the system again. I need to continue the narrative as if writing the article.
  • I'll continue from where it left off: finish A's calculation, then find C, then perhaps use the system or substitution to find B, D, E. I'll actually solve it properly.
  • Let's compute A = 10/45 = 2/9.
  • Find C: set x = -2. LHS: 3(4) + 5(-2) + 2 = 12 -10 +2 = 4. RHS: C(-2-1)((-2)^2+4) = C(-3)(8) = -24C. So 4 = -24C => C = -1/4.
  • Now we have A and C. To find B, D, E, we can plug in other x values or use the coefficient system. Since the article already set up the system, I might just solve it eleg