You need to decompose (3x + 1) / ((x + 2)(x² + 9)) into partial fractions. What is the correct form to set up?
AA/(x + 2) + B/(x² + 9)
BA/(x + 2) + (Bx + C)/(x² + 9)
CA/(x + 2) + B/(x² + 9) + C/(x² + 9)²
DA/(x + 2) + B/x + C/9
An irreducible quadratic factor (x² + 9) requires a linear numerator (Bx + C), not a constant. This is because there are two degrees of freedom — the factor is degree 2, so a constant numerator cannot match both the coefficient and derivative constraints. Option A is the most common error: using A/(x² + 9) with a constant numerator. Options C and D are structurally wrong — C treats the quadratic as a repeated factor, D factors the constant incorrectly.
Question 2 Multiple Choice
Before decomposing (x³ + 5x) / (x² − 4) into partial fractions, what must you do first — and why?
AFactor the denominator as (x − 2)(x + 2) and immediately set up A/(x − 2) + B/(x + 2)
BPerform polynomial long division, because the numerator degree (3) is not less than the denominator degree (2)
CCancel common factors between numerator and denominator
DSet the denominator equal to zero to find the roots
Partial fraction decomposition only works on *proper* rational functions — those where the numerator degree is strictly less than the denominator degree. Here, degree 3 ≥ degree 2, so the fraction is improper. Long division produces a polynomial quotient plus a proper remainder fraction; you then decompose only the remainder. Skipping this step and jumping to A/(x−2) + B/(x+2) is the most common error and produces an incorrect decomposition.
Question 3 True / False
The factor (x − 3)² in the denominator requires only one partial fraction term: B/(x − 3)².
TTrue
FFalse
Answer: False
A repeated linear factor (x − 3)² requires *two* terms: A/(x − 3) + B/(x − 3)². One term for each power up to the multiplicity. Using only B/(x − 3)² cannot account for the full structure of the original numerator — you'd be missing a degree of freedom. The general rule: (x − r)^k contributes k separate terms, from power 1 up through power k.
Question 4 True / False
Partial fraction decomposition can only be applied after the rational expression has been converted into a proper fraction (numerator degree < denominator degree).
TTrue
FFalse
Answer: True
This is a prerequisite that is often overlooked. Partial fractions require the fraction to be proper because the decomposition is based on the denominator's factors — and those factors only account for a polynomial of the same degree as the denominator. If the numerator is larger, there is a polynomial portion that no sum of proper partial fractions can represent. Long division extracts that polynomial first, leaving a proper remainder to decompose.
Question 5 Short Answer
Why does an irreducible quadratic factor like (x² + 4) require a linear numerator (Ax + B) in its partial fraction term, rather than a constant A?
Think about your answer, then reveal below.
Model answer: Because a degree-2 denominator factor introduces two independent degrees of freedom that the numerator must match. A constant numerator has only one free parameter, which is insufficient to satisfy the polynomial identity that results from clearing denominators. A linear numerator Ax + B provides two parameters (A and B), giving enough flexibility to correctly match both the even and odd coefficient constraints across the entire identity.
Think of it this way: when you clear the denominator and equate coefficients, a quadratic factor generates two equations (one for each power of x up to degree 1 in the numerator). One free parameter (constant) can only satisfy one equation; two free parameters (linear) can satisfy two. This is why the technique is consistent: the number of unknowns always matches the number of equations when you set up the form correctly.