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Use partial fractions to expand rational functions: distinct and repeated factors, improper fractions, validity, cancellations and numerical approximations with original worked practice.
Before you startPartial fractions, algebraic division and generalised binomial expansion
01 / Split the fraction, then expand each part
Expanding two reciprocal factors and multiplying works, but partial fractions often gives simpler arithmetic. Decompose the rational function first, expand each part and add coefficients of matching powers.
The model compares distinct factors, a repeated factor and an improper fraction. Select a power to see the separate signed contributions and their exact total.
(5 − 2x)/[(1 − x)(2 + x)]
A = 1/(1 − x); B = 3/(2 + x); C = 0.
A contributes 1; B contributes 3/8; C contributes 0.
Coefficient of x²: 1 + 3/8 + 0 = 11/8.
Guaranteed for |x| < 1.
Bars above zero are positive; bars below are negative. The vertical scale updates with the chosen degree. C = 0 means no third part is needed.
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02 / A reliable order of work
Do not expand farther than needed, unless a request for a number of nonzero terms requires you to pass a cancellation.
03 / Two distinct linear factors
Write A/(1 − x) + B/(2 + x)
Both factors are distinct and linear.
5 − 2x = A(2 + x) + B(1 − x)
Clear the denominators.
At x = 1: 3 = 3A. At x = −2: 9 = 3B
These substitutions solve the polynomial identity, giving A = 1 and B = 3.
1/(1 − x) = 1 + x + x² + x³ + …
Guaranteed for |x| < 1.
3/(2 + x) = (3/2)(1 + x/2)⁻¹ = 3/2 − 3x/4 + 3x²/8 − 3x³/16 + …
Guaranteed for |x| < 2.
Total: 5/2 + x/4 + 11x²/8 + 13x³/16 + …
Add each column; the common interval is |x| < 1.
Substituting x = 1 or −2 into the cleared identity does not make the original fraction defined there. Its domain still excludes those denominator zeros.
04 / A repeated factor needs every power
Use A/(1 − x) + B/(1 − x)² + C/(1 + x)
Do not omit A/(1 − x).
1 = A(1 − x²) + B(1 + x) + C(1 − 2x + x²)
Clear the denominator.
At x = 1: B = ½. At x = −1: C = ¼. Then x = 0 gives A = ¼
Verify the resulting identity.
A part: ¼(1 + x + x² + x³ + …)
The geometric coefficients are all 1 before scaling.
B part: ½(1 + 2x + 3x² + 4x³ + …)
Use exponent −2.
C part: ¼(1 − x + x² − x³ + …)
The plus sign inside gives alternating signs.
Total: 1 + x + 2x² + 2x³ + …, for |x| < 1
All three expansions have the same guaranteed interval.
The coefficient of xⁿ is ¼ + (n + 1)/2 + (−1)ⁿ/4 for n ≥ 0. This also lets you request a higher coefficient without writing every preceding term.
05 / Divide an improper fraction first
(1 − x)(2 + x) = 2 − x − x²
Use a consistent denominator sign.
x² + 1 = −(2 − x − x²) + (3 − x)
So the quotient is −1 and the proper remainder is (3 − x)/[(1 − x)(2 + x)].
(3 − x)/[(1 − x)(2 + x)] = (2/3)/(1 − x) + (5/3)/(2 + x)
Check: (2/3)(2 + x) + (5/3)(1 − x) = 3 − x.
Expand: −1 + (2/3)(1 + x + x² + x³) + (5/6)(1 − x/2 + x²/4 − x³/8) + …
Keep the quotient −1.
= ½ + x/4 + 7x²/8 + 9x³/16 + …, for |x| < 1
The quotient changes the constant; its coefficients at positive degrees are zero.
A higher-degree polynomial quotient can contribute at more than one degree. Expand that finite polynomial exactly, then add it to the fractional series.
06 / Normalise constants in repeated brackets
(2 − x)⁻² = ¼(1 − x/2)⁻²
Factoring out 2 contributes 2⁻² = ¼.
= ¼[1 + x + 3x²/4 + x³/2 + …]
The coefficient of the cubic input is −4, and the input is −x/2.
= ¼ + x/4 + 3x²/16 + x³/8 + …
Multiply every term by ¼.
Guaranteed for |x| < 2
Use |−x/2| < 1.
In a decomposition with several brackets, perform this normalisation separately for each one. A common mistake is to divide by 2 instead of 4 for a squared bracket.
07 / Conditions, cancellation and holes
The original domain excludes x = ½ and x = 1
Record both before cancelling.
For permitted x, the expression equals 1/(1 − x)
Cancel the common factor exactly.
Use 1 + x + x² + …, with |x| < 1
There is no need to expand a separate (1 − 2x)⁻¹ factor.
For the original expression also exclude x = ½
The series itself converges there to 2, but that value belongs to the filled-in extension, not the original expression.
Without exact cancellation, intersect all separate series conditions. For instance factors (1 − 3x) and (2 + x) give |x| < ⅓ and |x| < 2, so the common guarantee is |x| < ⅓.
08 / Use the expansion numerically
Use 5/2 + x/4 + 11x²/8 + 13x³/16
The condition |x| < 1 holds.
2.5 + 0.005 + 0.00055 + 0.0000065 = 2.5055565
Keep exact coefficients during substitution.
The reference value is 2.505556678…
The absolute error is approximately 0.000000178.
Evaluate the original expression as a separate check when a calculator is available. The agreement tests this numerical value; the decomposition and coefficient calculations justify the expansion.
09 / Your turn
Decompose (7 + x)/[(1 − x)(3 + x)] and expand through x².
Use A/(1 − x) + B/(3 + x).
7 + x = A(3 + x) + B(1 − x). At x = 1, A = 2; at x = −3, B = 1. Thus 2/(1 − x) + 1/(3 + x) = 7/3 + (17/9)x + (55/27)x² + …, guaranteed for |x| < 1.
Expand (3 − 2x)/(1 − x)² through x³.
Try A/(1 − x) + B/(1 − x)².
3 − 2x = A(1 − x) + B gives A = 2 and B = 1. Add 2(1 + x + x² + x³) and (1 + 2x + 3x² + 4x³): 3 + 4x + 5x² + 6x³ + …, for |x| < 1.
Expand 1/[(1 − 2x)(1 + x)] through x³ using partial fractions.
The constants are 2/3 and 1/3.
1 = (2/3)(1 + x) + (1/3)(1 − 2x). Expand (2/3)(1 − 2x)⁻¹ + (1/3)(1 + x)⁻¹. The sum is 1 + x + 3x² + 5x³ + …, guaranteed for |x| < ½.
Find the first three nonzero terms of (x² + 2)/[(1 − x)(1 + x)].
Divide to obtain −1 + 3/(1 − x²), then decompose if needed.
The decomposition is −1 + (3/2)/(1 − x) + (3/2)/(1 + x). Odd powers cancel: 2 + 3x² + 3x⁴ + …, for |x| < 1.
Find the coefficient of x⁴ in (2 − x)⁻².
Use ¼(1 − x/2)⁻².
The coefficient is ¼ × 5 × (½)⁴ = 5/64. The guaranteed interval is |x| < 2.
Find the first four nonzero terms of (1 + 3x)/[(1 − x)(1 + 2x)].
The partial fractions are (4/3)/(1 − x) − (1/3)/(1 + 2x).
The coefficients are 4/3 − (−2)ⁿ/3 at degree n. They give 1, 2, 0, 4, −4 for n = 0,…,4. Hence 1 + 2x + 4x³ − 4x⁴ + …, guaranteed for |x| < ½.
A decomposition has nonzero terms A/(1 − 3x) and B/(2 + x). What common open interval justifies their separate binomial expansions?
Intersect the two conditions.
|3x| < 1 and |x/2| < 1 give |x| < ⅓ and |x| < 2. The common guarantee is −⅓ < x < ⅓.
For (1 − 2x)/[(1 − 2x)(1 − x)], may the resulting geometric expansion represent the original expression at x = ½?
Check the original denominator before cancellation.
No. The original expression is undefined at x = ½. The simplified formula and its series have value 2 there, but this fills the hole. For the original, use |x| < 1 with x ≠ ½.
Find the coefficient of x⁷ in 1/[(1 − x)²(1 + x)].
Use ¼ + (n + 1)/2 + (−1)ⁿ/4.
At n = 7, the coefficient is ¼ + 4 − ¼ = 4. The expansion is guaranteed for |x| < 1.
Use the cubic expansion from the distinct-factor example to approximate (5 − 2x)/[(1 − x)(2 + x)] at x = 0.02.
Use the same x in every term.
5/2 + 0.02/4 + 11(0.02)²/8 + 13(0.02)³/16 = 2.5055565. Since |0.02| < 1, the expansion’s open-interval condition holds; the exact reference is about 2.505556678.
10 / Recap
Section 1 of 10 · Split the fraction, then expand each part