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Learn partial fractions with repeated linear factors: include every power, isolate coefficients and use an extra equation. Original worked examples, a manual model and practice.
Before you startPartial fractions with distinct linear factors and coefficient comparison
01 / The template
A repeated denominator factor needs a term for each power. For a squared factor, include both the first and second powers, even when the first-power factor is not written separately in the original denominator.
N(x)/[(x − a)²(x − b)]
= A/(x − a)
+ B/(x − a)² + C/(x − b)
For this template, a ≠ b and the fraction is proper. Keep x ≠ a, b.
For a cubed factor, include 1/(x − a), 1/(x − a)² and 1/(x − a)³, each with its own constant numerator. A coefficient may turn out to be zero; find that from the identity, rather than omitting the term at the start.
02 / Why all powers?
Writing A/(x − 1) + B/(x − 1) just gives (A + B)/(x − 1). You still only have a first-power denominator; this cannot represent every fraction with denominator (x − 1)².
Even writing only B/(x − 1)² can be too restrictive. For instance, (3x + 5)/(x − 1)² has a changing numerator, so a single constant B is not enough.
(3x + 5)/(x − 1)²
= A/(x − 1) + B/(x − 1)²
x ≠ 1.
3x + 5 ≡ A(x − 1) + B
Clear denominators. Compare x coefficients: A = 3.
5 = −A + B ⇒ B = 8
Answer: 3/(x − 1) + 8/(x − 1)², for x ≠ 1.
03 / Make the identity
Decompose (x² + 7x + 1)/[(x − 1)²(x + 2)]. The original restrictions are x ≠ 1, −2. Begin with A/(x − 1) + B/(x − 1)² + C/(x + 2).
x² + 7x + 1
≡ A(x − 1)(x + 2)
+ B(x + 2) + C(x − 1)²
The A term loses one copy of x − 1, leaving the other copy. The B term loses both copies. The C term loses x + 2 and keeps the square. Write these products carefully before substituting.
x² + 7x + 1
≡ A(x − 1)(x + 2)
+ B(x + 2) + C(x − 1)²
x = 1: 9 = 0A + 3B + 0C.
Only B survives, so B = 3. A still needs another equation.
Try −2 to isolate C, then 0 to obtain an equation involving A. These are substitutions into a polynomial identity, not into the original fractions.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
04 / Find the last coefficient
Substituting x = 1 gives 9 = 3B, so B = 3. Substituting x = −2 gives −9 = 9C, so C = −1. Substituting either root again repeats the same information: it cannot determine A.
Use x = 0:
1 = −2A + 2B + C
Substitute the known B = 3 and C = −1.
1 = −2A + 6 − 1
⇒ A = 2
Any convenient extra input can work.
Alternatively, compare x²:
1 = A + C ⇒ A = 2
The B term has no x² coefficient.
Answer: 2/(x − 1)
+ 3/(x − 1)² − 1/(x + 2)
Retain x ≠ 1, −2. Recombination gives 2(x² + x − 2) + 3(x + 2) − (x² − 2x + 1) = x² + 7x + 1.
05 / Non-monic square
If the repeated factor is 2x + 1, include A/(2x + 1) and B/(2x + 1)². The root is −1/2. Remember that (2x + 1)² = 4x² + 4x + 1.
A/(2x + 1)
+ B/(2x + 1)² + C/(x − 2)
x ≠ −1/2, 2.
14x² + 7x + 5
≡ A(2x + 1)(x − 2)
+ B(x − 2) + C(2x + 1)²
Clear denominators using whole factors.
x = −1/2: 5 = −5B/2
⇒ B = −2
The other two terms vanish.
x = 2: 75 = 25C ⇒ C = 3
Now compare x²: 14 = 2A + 4C, hence A = 1.
Answer: 1/(2x + 1)
− 2/(2x + 1)² + 3/(x − 2)
Keep both original exclusions. Recombining yields the original numerator.
06 / Higher powers
For (2x² − 3x + 4)/(x − 1)³, use A/(x − 1) + B/(x − 1)² + C/(x − 1)³, with x ≠ 1.
2x² − 3x + 4
≡ A(x − 1)² + B(x − 1) + C
At x = 1, C = 3. The x² coefficient gives A = 2. The x coefficient gives −2A + B = −3, so B = 1. Check the constant: A − B + C = 2 − 1 + 3 = 4.
2/(x − 1) + 1/(x − 1)²
+ 3/(x − 1)³, x ≠ 1
For multiplicity m, the template contains all powers 1 through m. This does not mean every coefficient must be nonzero.
07 / Your turn
Use root substitution where it helps, then a new input or coefficient equation for the remaining constants.
Give the partial-fraction template for a proper fraction with denominator x²(x + 4).
x is a repeated linear factor.
A/x + B/x² + C/(x + 4), with x ≠ 0, −4. Omitting A/x would restrict the numerator unnecessarily.
Decompose (5x − 1)/(x + 2)².
Write 5x − 1 = 5(x + 2) − 11.
A/(x + 2) + B/(x + 2)² gives 5x − 1 ≡ A(x + 2) + B. Therefore A = 5, B = −11. Answer: 5/(x + 2) − 11/(x + 2)², x ≠ −2.
Decompose (2x² + x + 6)/[x²(x + 3)].
Use A/x + B/x² + C/(x + 3).
2x² + x + 6 ≡ Ax(x + 3) + B(x + 3) + Cx². At 0, B = 2. At −3, 21 = 9C, so C = 7/3. From x², A + C = 2, hence A = −1/3. Answer: −1/(3x) + 2/x² + 7/[3(x + 3)], x ≠ 0, −3.
Decompose (3x² − 2x + 2)/[(x − 1)²(x + 1)].
Root substitution finds the squared-factor coefficient and the x + 1 coefficient.
The identity is 3x² − 2x + 2 ≡ A(x − 1)(x + 1) + B(x + 1) + C(x − 1)². At 1, B = 3/2; at −1, C = 7/4; x² gives A = 5/4. Answer: 5/[4(x − 1)] + 3/[2(x − 1)²] + 7/[4(x + 1)], x ≠ 1, −1.
Decompose (6x + 7)/(2x + 1)².
Match 6x + 7 ≡ A(2x + 1) + B.
2A = 6 gives A = 3. Then A + B = 7 gives B = 4. Answer: 3/(2x + 1) + 4/(2x + 1)², x ≠ −1/2.
Decompose (x² + 1)/(x + 1)³.
Use all three powers of x + 1.
x² + 1 ≡ A(x + 1)² + B(x + 1) + C. Comparing x² gives A = 1; comparing x gives 2A + B = 0, hence B = −2. The constant gives A + B + C = 1, hence C = 2. Answer: 1/(x + 1) − 2/(x + 1)² + 2/(x + 1)³, x ≠ −1.
Decompose 4/[(x − 2)²] and explain why the full template still matters.
Use A/(x − 2) + B/(x − 2)² and compare coefficients.
4 ≡ A(x − 2) + B gives A = 0 and B = 4. This particular expression needs only the squared term. The zero coefficient is a result of the calculation; a general numerator would require both template terms. x ≠ 2.
After substituting the two distinct roots into a three-coefficient repeated-factor identity, why is a third equation normally needed?
A repeated root is still only one input value.
The two substitutions usually isolate the highest-power repeated-factor coefficient and the other-factor coefficient. They leave the lower-power coefficient undetermined. Repeating either substitution supplies no new information; use another input or compare a polynomial coefficient.
08 / Recap
Section 1 of 8 · The template