What is the main goal of resolving a rational function into partial fractions?
A
To multiply the rational function
B
To divide the rational function into simpler fractions
C
To differentiate the rational function
D
To find the inverse of the rational function
Analysis & Theory
Partial fraction decomposition breaks a rational function into simpler fractions that are easier to integrate or manipulate.
If the degree of the numerator is greater than or equal to the denominator, what should be done first?
A
Apply partial fractions directly
C
Multiply numerator and denominator by same factor
D
Take reciprocal of the function
Analysis & Theory
When numerator's degree ≥ denominator's degree, polynomial long division must be performed first before partial fraction decomposition.
Resolve 1/(x(x+1)) into partial fractions.
Analysis & Theory
1/(x(x+1)) = A/x + B/(x+1). Solving gives A=1, B=-1. So, 1/x - 1/(x+1).
Resolve (2x+3)/(x^2+x) into partial fractions.
Analysis & Theory
(2x+3)/(x(x+1)) = A/x + B/(x+1). Solving gives A=2, B=1. So, 2/x + 1/(x+1).
For repeated linear factors, such as 1/(x(x-1)^2), the partial fraction form is:
B
A/x + B/(x-1) + C/(x-1)^2
Analysis & Theory
For repeated factors, we include terms for each power: A/x + B/(x-1) + C/(x-1)^2.
For irreducible quadratic factors, the numerator in partial fractions is of the form:
Analysis & Theory
If the denominator contains an irreducible quadratic (e.g., x^2+1), numerator is linear (Ax + B).
Resolve (x+5)/((x+2)(x+3)) into partial fractions.
Analysis & Theory
(x+5)/((x+2)(x+3)) = A/(x+2) + B/(x+3). Solving gives A=2, B=-1.
Resolve (2x^2+3x+1)/((x+1)(x^2+1)) into partial fractions.
A
1/(x+1) + (2x+1)/(x^2+1)
B
2/(x+1) + (x+1)/(x^2+1)
C
3/(x+1) + (x+2)/(x^2+1)
Analysis & Theory
Decompose as A/(x+1) + (Bx+C)/(x^2+1). Solving gives A=1, B=2, C=1.
Which of the following is NOT a valid case for partial fraction decomposition?
A
Distinct linear factors
B
Repeated linear factors
C
Irreducible quadratic factors
D
Prime number in numerator
Analysis & Theory
Partial fraction decomposition depends only on the denominator factorization, not on numerator being prime.
Partial fraction decomposition is most useful in which mathematical operation?
D
Factorial simplification
Analysis & Theory
Partial fractions simplify integration of rational functions into manageable terms.