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Comprehensive Notes on Partial Fraction Decomposition

1. Fundamental Definitions & Classifications

A rational function (or rational fraction) is an expression of the form: $\(\frac{f(x)}{g(x)}\)$ where \(f(x)\) and \(g(x)\) are polynomials, with \(g(x) \neq 0\).

A. Proper Fractions

A rational fraction \(\frac{f(x)}{g(x)}\) is called a proper fraction if the degree of the numerator \(f(x)\) is strictly less than the degree of the denominator \(g(x)\): $\(\deg(f(x)) < \deg(g(x))\)$

  • Examples:
  • \(\frac{x}{x^2 + 4x + 4}\)
  • \(\frac{x}{x^2 + 1}\)

B. Improper Fractions

A rational fraction \(\frac{f(x)}{g(x)}\) is called an improper fraction if the degree of the numerator \(f(x)\) is greater than or equal to the degree of the denominator \(g(x)\): $\(\deg(f(x)) \ge \deg(g(x))\)$

  • Examples:
  • \(\frac{x^2 + 1}{x^2 + 3x + 2}\)
  • \(\frac{x^3 - 7x + 1}{x^3 + 2x^2 + 9}\)

Handling Improper Fractions:
Prior to decomposing an improper fraction into partial fractions, perform polynomial long division to rewrite it as: $\(\frac{f(x)}{g(x)} = Q(x) + \frac{R(x)}{g(x)}\)$ where \(Q(x)\) is the quotient polynomial and \(\frac{R(x)}{g(x)}\) is a proper fraction (i.e., \(\deg(R(x)) < \deg(g(x))\)). Apply partial fraction decomposition to \(\frac{R(x)}{g(x)}\).

2. Classification of Types & Solved Examples

Type I: Non-Repeated Linear Factors

Form:

\[\frac{f(x)}{(ax + b)(cx + d)(ex + f)} = \frac{A}{ax + b} + \frac{B}{cx + d} + \frac{C}{ex + f}\]

Example 1: Two Distinct Linear Factors

Resolve \(\frac{x}{(x+1)(x-2)}\) into partial fractions.

Solution: 1. Set up the partial fraction equation: $\(\frac{x}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2} \quad \text{--- (1)}\)$

  1. Combine terms over a common denominator: $\(\frac{x}{(x+1)(x-2)} = \frac{A(x-2) + B(x+1)}{(x+1)(x-2)}\)$

  2. Equate the numerators: $\(A(x-2) + B(x+1) = x \quad \text{--- (2)}\)$

  3. Solve for constants by substitution:

  4. Set \(x = 2\) in (2): $\(A(0) + B(2+1) = 2 \implies 3B = 2 \implies B = \frac{2}{3}\)$

  5. Set \(x = -1\) in (2): $\(A(-1-2) + B(0) = -1 \implies -3A = -1 \implies A = \frac{1}{3}\)$

  6. Substitute constants back into (1): $\(\frac{x}{(x+1)(x-2)} = \frac{1/3}{x+1} + \frac{2/3}{x-2}\)$

Example 2: Three Distinct Linear Factors

Resolve \(\frac{x^2}{(x+1)(x+2)(x-3)}\) into partial fractions.

Solution: 1. Set up the equation: $\(\frac{x^2}{(x+1)(x+2)(x-3)} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x-3} \quad \text{--- (1)}\)$

  1. Equate numerators: $\(A(x+2)(x-3) + B(x+1)(x-3) + C(x+1)(x+2) = x^2 \quad \text{--- (2)}\)$

  2. Solve for constants:

  3. Set \(x = -1\):
    $\(A(-1+2)(-1-3) = (-1)^2 \implies A(1)(-4) = 1 \implies -4A = 1 \implies A = -\frac{1}{4}\)$
  4. Set \(x = -2\):
    $\(B(-2+1)(-2-3) = (-2)^2 \implies B(-1)(-5) = 4 \implies 5B = 4 \implies B = \frac{4}{5}\)$

  5. Set \(x = 3\):
    $\(C(3+1)(3+2) = (3)^2 \implies C(4)(5) = 9 \implies 20C = 9 \implies C = \frac{9}{20}\)$

  6. Final decomposition: $\(\frac{x^2}{(x+1)(x+2)(x-3)} = \frac{-1/4}{x+1} + \frac{4/5}{x+2} + \frac{9/20}{x-3}\)$

Example 3: Three Factors including a Monomial Factor

Resolve \(\frac{x^2 + 1}{x(x+1)(x-2)}\) into partial fractions.

Solution: 1. Set up the decomposition: $\(\frac{x^2 + 1}{x(x+1)(x-2)} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x-2} \quad \text{--- (1)}\)$

  1. Equate numerators: $\(A(x+1)(x-2) + Bx(x-2) + Cx(x+1) = x^2 + 1 \quad \text{--- (2)}\)$

  2. Solve for constants:

  3. Set \(x = 0\):
    $\(A(1)(-2) = 0^2 + 1 \implies -2A = 1 \implies A = -\frac{1}{2}\)$
  4. Set \(x = -1\):
    $\(B(-1)(-1-2) = (-1)^2 + 1 \implies 3B = 2 \implies B = \frac{2}{3}\)$

  5. Set \(x = 2\):
    $\(C(2)(2+1) = 2^2 + 1 \implies 6C = 5 \implies C = \frac{5}{6}\)$

  6. Final decomposition: $\(\frac{x^2 + 1}{x(x+1)(x-2)} = \frac{-1/2}{x} + \frac{2/3}{x+1} + \frac{5/6}{x-2}\)$

Example 4: General Symbolic Formula

Resolve \(\frac{1}{(x-a)(x-b)(x-c)}\) into partial fractions.

Solution: 1. Set up the expansion: $\(\frac{1}{(x-a)(x-b)(x-c)} = \frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{x-c}\)$

  1. Equate numerators: $\(A(x-b)(x-c) + B(x-a)(x-c) + C(x-a)(x-b) = 1\)$

  2. Evaluate at roots:

  3. Set \(x = a\): \(A(a-b)(a-c) = 1 \implies A = \frac{1}{(a-b)(a-c)}\)
  4. Set \(x = b\): \(B(b-a)(b-c) = 1 \implies B = \frac{1}{(b-a)(b-c)}\)

  5. Set \(x = c\): \(C(c-a)(c-b) = 1 \implies C = \frac{1}{(c-a)(c-b)}\)

  6. Final Result: $\(\frac{1}{(x-a)(x-b)(x-c)} = \frac{1}{(a-b)(a-c)(x-a)} + \frac{1}{(b-a)(b-c)(x-b)} + \frac{1}{(c-a)(c-b)(x-c)}\)$

Type II: Improper Fractions with Non-Repeated Linear Factors

Method:

Divide the numerator by the denominator using long division first to separate the polynomial part, then perform partial fraction decomposition on the remainder fraction.

Example 1:

Resolve \(\frac{4x^2 + 1}{(2x+1)(x+2)}\) into partial fractions.

Solution: 1. Expand the denominator: $\((2x+1)(x+2) = 2x^2 + 5x + 2\)$

  1. Perform Long Division: Divide \((4x^2 + 1)\) by \((2x^2 + 5x + 2)\): $\(\frac{4x^2 + 1}{2x^2 + 5x + 2} = 2 + \frac{-10x - 3}{2x^2 + 5x + 2} \quad \text{--- (1)}\)$

  2. Decompose the remainder fraction: $\(\frac{-10x - 3}{(2x+1)(x+2)} = \frac{A}{2x+1} + \frac{B}{x+2}\)$ $\(A(x+2) + B(2x+1) = -10x - 3 \quad \text{--- (2)}\)$

  3. Solve for constants:

  4. Set \(x = -2\):
    $\(B(2(-2) + 1) = -10(-2) - 3 \implies -3B = 17 \implies B = -\frac{17}{3}\)$

  5. Set \(x = -\frac{1}{2}\):
    $\(A\left(-\frac{1}{2} + 2\right) = -10\left(-\frac{1}{2}\right) - 3 \implies \frac{3}{2}A = 2 \implies A = \frac{4}{3}\)$

  6. Substitute back into (1): $\(\frac{4x^2 + 1}{(2x+1)(x+2)} = 2 + \frac{4/3}{2x+1} - \frac{17/3}{x+2}\)$

Example 2:

Resolve \(\frac{x^3 + 7x + 1}{x^2 + 5x + 6}\) into partial fractions.

Solution: 1. Perform Long Division: $\(\frac{x^3 + 7x + 1}{x^2 + 5x + 6} = x - 5 + \frac{26x + 31}{x^2 + 5x + 6} \quad \text{--- (1)}\)$

  1. Factor denominator and set up partial fractions for the remainder: $\(\frac{26x + 31}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3}\)$ $\(A(x+3) + B(x+2) = 26x + 31 \quad \text{--- (2)}\)$

  2. Solve for constants:

  3. Set \(x = -2\):
    $\(A(-2+3) = 26(-2) + 31 \implies A = -21\)$

  4. Set \(x = -3\):
    $\(B(-3+2) = 26(-3) + 31 \implies -B = -47 \implies B = 47\)$

  5. Final Result: $\(\frac{x^3 + 7x + 1}{x^2 + 5x + 6} = x - 5 - \frac{21}{x+2} + \frac{47}{x+3}\)$

Example 3:

Resolve \(\frac{x^2}{(3x+2)(2x+3)}\) into partial fractions.

Solution: 1. Expand denominator: $\((3x+2)(2x+3) = 6x^2 + 13x + 6\)$

  1. Perform Long Division: $\(\frac{x^2}{6x^2 + 13x + 6} = \frac{1}{6} + \frac{-\frac{13}{6}x - 1}{6x^2 + 13x + 6} \quad \text{--- (1)}\)$

  2. Decompose the remainder fraction: $\(\frac{-\frac{13}{6}x - 1}{(3x+2)(2x+3)} = \frac{A}{3x+2} + \frac{B}{2x+3}\)$ $\(A(2x+3) + B(3x+2) = -\frac{13}{6}x - 1 \quad \text{--- (2)}\)$

  3. Solve for constants:

  4. Set \(x = -\frac{2}{3}\):
    $\(A\left(2\left(-\frac{2}{3}\right)+3\right) = -\frac{13}{6}\left(-\frac{2}{3}\right)-1 \implies \frac{5}{3}A = \frac{4}{9} \implies A = \frac{1}{3}\)$

  5. Set \(x = -\frac{3}{2}\):
    $\(B\left(3\left(-\frac{3}{2}\right)+2\right) = -\frac{13}{6}\left(-\frac{3}{2}\right)-1 \implies -\frac{5}{2}B = \frac{9}{4} \implies B = -\frac{9}{10}\)$

  6. Final Result: $\(\frac{x^2}{(3x+2)(2x+3)} = \frac{1}{6} + \frac{1/3}{3x+2} - \frac{9/10}{2x+3}\)$

Type III: Repeated Linear Factors

Form:

If \(g(x) = (ax + b)(cx + d)^n\), assume: $\(\frac{f(x)}{g(x)} = \frac{A}{ax + b} + \frac{B}{cx + d} + \frac{C}{(cx + d)^2} + \dots + \frac{Z}{(cx + d)^n}\)$

Note: Direct substitution alone is usually insufficient to determine all constants in repeated factor cases. Combine substitution with equating coefficients of like powers of \(x\).

Example 1:

Resolve \(\frac{x^2}{(x+1)(x-1)^2}\) into partial fractions.

Solution: 1. Set up the equation: $\(\frac{x^2}{(x+1)(x-1)^2} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{(x-1)^2} \quad \text{--- (1)}\)$

  1. Equate numerators: $\(A(x-1)^2 + B(x+1)(x-1) + C(x+1) = x^2 \quad \text{--- (2)}\)$

  2. Find constants:

  3. Set \(x = 1\) in (2):
    $\(C(1+1) = 1^2 \implies 2C = 1 \implies C = \frac{1}{2}\)$

  4. Set \(x = -1\) in (2):
    $\(A(-1-1)^2 = (-1)^2 \implies 4A = 1 \implies A = \frac{1}{4}\)$

  5. Expand terms in (2) to solve for \(B\): $\(A(x^2 - 2x + 1) + B(x^2 - 1) + C(x + 1) = x^2\)$ $\((A + B)x^2 + (-2A + C)x + (A - B + C) = x^2\)$

  6. Equate coefficients of \(x^2\): $\(A + B = 1 \implies \frac{1}{4} + B = 1 \implies B = \frac{3}{4}\)$

  7. Final Result: $\(\frac{x^2}{(x+1)(x-1)^2} = \frac{1/4}{x+1} + \frac{3/4}{x-1} + \frac{1/2}{(x-1)^2}\)$

Example 2:

Resolve \(\frac{x^2 + 4x + 1}{(1-x)^4}\) into partial fractions.

Solution: 1. Set up the equation: $\(\frac{x^2 + 4x + 1}{(1-x)^4} = \frac{A}{1-x} + \frac{B}{(1-x)^2} + \frac{C}{(1-x)^3} + \frac{D}{(1-x)^4} \quad \text{--- (1)}\)$

  1. Equate numerators: $\(A(1-x)^3 + B(1-x)^2 + C(1-x) + D = x^2 + 4x + 1 \quad \text{--- (2)}\)$

  2. Solve for \(D\):

  3. Set \(x = 1\):
    $\(D = 1^2 + 4(1) + 1 = 6\)$

  4. Expand polynomial terms to find \(A, B, C\): $\(A(1 - 3x + 3x^2 - x^3) + B(1 - 2x + x^2) + C(1 - x) + D = x^2 + 4x + 1\)$ $\((-A)x^3 + (3A + B)x^2 + (-3A - 2B - C)x + (A + B + C + D) = x^2 + 4x + 1\)$

  5. Equate coefficients of like powers:

  6. Coefficient of \(x^3\):
    $\(-A = 0 \implies A = 0\)$
  7. Coefficient of \(x^2\):
    $\(3A + B = 1 \implies 3(0) + B = 1 \implies B = 1\)$

  8. Coefficient of \(x\):
    $\(-3A - 2B - C = 4 \implies -3(0) - 2(1) - C = 4 \implies -2 - C = 4 \implies C = -6\)$

  9. Final Result: $\(\frac{x^2 + 4x + 1}{(1-x)^4} = \frac{1}{(1-x)^2} - \frac{6}{(1-x)^3} + \frac{6}{(1-x)^4}\)$

Example 3:

Resolve \(\frac{x^2 + 2x + 3}{(x+1)^3}\) into partial fractions.

Solution: 1. Set up the expansion: $\(\frac{x^2 + 2x + 3}{(x+1)^3} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{(x+1)^3} \quad \text{--- (1)}\)$

  1. Equate numerators: $\(A(x+1)^2 + B(x+1) + C = x^2 + 2x + 3 \quad \text{--- (2)}\)$

  2. Find constants:

  3. Set \(x = -1\):
    $\(C = (-1)^2 + 2(-1) + 3 = 1 - 2 + 3 = 2\)$
  4. Expand (2):
    $\(A(x^2 + 2x + 1) + B(x + 1) + C = x^2 + 2x + 3\)$ $\(Ax^2 + (2A + B)x + (A + B + C) = x^2 + 2x + 3\)$
  5. Equate coefficient of \(x^2\):
    $\(A = 1\)$

  6. Equate coefficient of \(x\):
    $\(2A + B = 2 \implies 2(1) + B = 2 \implies B = 0\)$

  7. Verification: $\(\text{RHS} = \frac{1}{x+1} + \frac{2}{(x+1)^3} = \frac{(x+1)^2 + 2}{(x+1)^3} = \frac{x^2 + 2x + 1 + 2}{(x+1)^3} = \frac{x^2 + 2x + 3}{(x+1)^3} \quad \text{(Matches LHS)}\)$

  8. Final Result: $\(\frac{x^2 + 2x + 3}{(x+1)^3} = \frac{1}{x+1} + \frac{2}{(x+1)^3}\)$

Type IV: Non-Repeated Quadratic Factors

Form:

If \(g(x)\) contains an irreducible quadratic factor \(bx^2 + cx + d\) (i.e. cannot be factored into real linear factors), assign a linear numerator \(Bx + C\): $\(\frac{f(x)}{(x+a)(bx^2 + cx + d)} = \frac{A}{x+a} + \frac{Bx + C}{bx^2 + cx + d}\)$

3. General Step-by-Step Problem Solving Strategy

  1. Check Fraction Type:
    Compare \(\deg(f(x))\) and \(\deg(g(x))\). If \(\deg(f(x)) \ge \deg(g(x))\), perform long division first.
  2. Factor Denominator:
    Factor \(g(x)\) completely into linear and irreducible quadratic factors.
  3. Set Up Partial Fraction Expression:
  4. Linear factor \((ax+b) \implies \frac{A}{ax+b}\)
  5. Repeated linear factor \((ax+b)^n \implies \frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \dots + \frac{A_n}{(ax+b)^n}\)
  6. Irreducible quadratic factor \((ax^2+bx+c) \implies \frac{Ax+B}{ax^2+bx+c}\)
  7. Form Algebraic Equation:
    Multiply through by \(g(x)\) to clear all denominators.
  8. Solve for Unknown Constants:
  9. Substitute roots of linear factors to isolate constants directly.
  10. Expand terms and equate coefficients of corresponding powers of \(x\) to solve remaining linear systems.