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Partial Fraction Decomposition

chatgpt image jun 14, 2026, 04 45 06 pm

Partial Fraction Decomposition: Simplifying Complex Rational Functions

Partial Fraction Decomposition is a fundamental algebraic technique used to rewrite complicated rational expressions as the sum of simpler fractions. This method is widely applied in calculus, differential equations, Laplace transforms, and various engineering fields because it makes difficult mathematical operations much easier to perform.

In this article, we’ll discuss what Partial Fraction Decomposition is, when to use it, its different cases, and solved examples for each case.


What Is Partial Fraction Decomposition?

Partial Fraction Decomposition is the process of expressing a rational function as a sum of simpler rational functions called partial fractions.

A rational function is any expression of the form:P(x)Q(x)\frac{P(x)}{Q(x)}

where P(x)P(x) and Q(x)Q(x) are polynomials and Q(x)0Q(x) \neq 0.

By decomposing a complicated fraction into simpler pieces, many mathematical operations become easier to solve.


Learning Objectives

After studying Partial Fraction Decomposition, students should be able to:

  • Define Partial Fraction Decomposition.
  • Distinguish between proper and improper rational functions.
  • Recognize Cases 0–4 of Partial Fraction Decomposition.
  • Set up the correct decomposition for each case.
  • Solve rational expressions using Partial Fraction Decomposition.
  • Apply the technique in calculus and engineering problems.

Case 0: Improper Rational Functions

A rational function is improper when the degree of the numerator is greater than or equal to the degree of the denominator.

Before decomposing, perform polynomial long division.

Solved Example

Simplify:x3+2x2+1x2+1\frac{x^3+2x^2+1}{x^2+1}

Solution

Perform polynomial long division.

Divide:x3÷x2=xx^3 \div x^2=x

Multiply:x(x2+1)=x3+xx(x^2+1)=x^3+x

Subtract:2x2x+12x^2-x+1

Next,2x2÷x2=22x^2 \div x^2=2

Multiply:2(x2+1)=2x2+22(x^2+1)=2x^2+2

Subtract:x1-x-1

Therefore,x3+2x2+1x2+1=x+2x+1x2+1\boxed{ \frac{x^3+2x^2+1}{x^2+1} = x+2-\frac{x+1}{x^2+1} }


Case 1: Distinct Linear Factors

When the denominator consists of different linear factors with no repetition:(xa)(xb)(x-a)(x-b)

the decomposition takes the formAxa+Bxb\frac{A}{x-a}+\frac{B}{x-b}

Solved Example

Decompose5x+1(x1)(x+2)\frac{5x+1}{(x-1)(x+2)}

Step 1

Assume5x+1(x1)(x+2)=Ax1+Bx+2\frac{5x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}

Step 2

Multiply both sides by the denominator.5x+1=A(x+2)+B(x1)5x+1=A(x+2)+B(x-1)

Expand:5x+1=(A+B)x+(2AB)5x+1=(A+B)x+(2A-B)

Equate coefficients:A+B=5A+B=52AB=12A-B=1

Solve:A=2,B=3A=2,\qquad B=3

Final Answer

5x+1(x1)(x+2)=2x1+3x+2\boxed{ \frac{5x+1}{(x-1)(x+2)} = \frac{2}{x-1} + \frac{3}{x+2} }


Case 2: Repeated Linear Factors

When a factor repeats,(xa)n(x-a)^n

include every power in the decomposition.

General Form

A1xa+A2(xa)2++An(xa)n\frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} +\cdots+ \frac{A_n}{(x-a)^n}

Solved Example

Decompose3x+5(x2)2\frac{3x+5}{(x-2)^2}

AssumeAx2+B(x2)2\frac{A}{x-2} + \frac{B}{(x-2)^2}

Multiply both sides:3x+5=A(x2)+B3x+5=A(x-2)+B

Expand:3x+5=Ax2A+B3x+5=Ax-2A+B

Compare coefficients:A=3A=36+B=5-6+B=5B=11B=11

Final Answer

3x+5(x2)2=3x2+11(x2)2\boxed{ \frac{3x+5}{(x-2)^2} = \frac{3}{x-2} + \frac{11}{(x-2)^2} }


Case 3: Irreducible Quadratic Factors

When the denominator contains quadratic factors that cannot be factored over the real numbers, use a linear numerator.

General Form

Ax+Bx2+a\frac{Ax+B}{x^2+a}

Solved Example

Decompose2x2+3(x2+1)(x2+4)\frac{2x^2+3}{(x^2+1)(x^2+4)}

AssumeAx+Bx2+1+Cx+Dx2+4\frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+4}

Multiply through:2x2+3=(Ax+B)(x2+4)+(Cx+D)(x2+1)2x^2+3 = (Ax+B)(x^2+4) + (Cx+D)(x^2+1)

Expanding and comparing coefficients givesA=0,B=13,C=0,D=53A=0,\qquad B=\frac13,\qquad C=0,\qquad D=\frac53

Final Answer

2x2+3(x2+1)(x2+4)=13(x2+1)+53(x2+4)\boxed{ \frac{2x^2+3}{(x^2+1)(x^2+4)} = \frac{1}{3(x^2+1)} + \frac{5}{3(x^2+4)} }


Case 4: Repeated Irreducible Quadratic Factors

If an irreducible quadratic factor repeats, every power receives its own linear numerator.

General Form

Ax+Bx2+1+Cx+D(x2+1)2\frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2}

Solved Example

Decomposex3+2x+1(x2+1)2\frac{x^3+2x+1}{(x^2+1)^2}

AssumeAx+Bx2+1+Cx+D(x2+1)2\frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2}

Multiply through:x3+2x+1=(Ax+B)(x2+1)+(Cx+D)x^3+2x+1 = (Ax+B)(x^2+1) + (Cx+D)

Expanding and equating coefficients yieldsA=1,B=0,C=1,D=1A=1,\qquad B=0,\qquad C=1,\qquad D=1

Final Answer

x3+2x+1(x2+1)2=xx2+1+x+1(x2+1)2\boxed{ \frac{x^3+2x+1}{(x^2+1)^2} = \frac{x}{x^2+1} + \frac{x+1}{(x^2+1)^2} }


Applications of Partial Fraction Decomposition

Partial Fraction Decomposition is widely used in:

  • Calculus for evaluating rational integrals
  • Laplace transforms
  • Differential equations
  • Electrical and electronic engineering
  • Control systems
  • Signal processing
  • Mathematical modeling

Conclusion

Partial Fraction Decomposition is a powerful mathematical tool that converts complex rational expressions into simpler fractions. By understanding the five major cases improper rational functions, distinct linear factors, repeated linear factors, irreducible quadratic factors, and repeated irreducible quadratic factors students can solve a wide variety of algebraic and calculus problems with confidence.

Mastering this technique provides a strong foundation for advanced studies in mathematics, engineering, physics, and other technical disciplines.

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evallokarlstephen

This is an excellent post. Thanks Mr. Corpuz and the group.

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