01 · Equal degrees
Decompose (2x² + 3x + 4)/[(x − 1)(x + 2)].
Hint
Subtract 2(x² + x − 2).
Worked solution
Q = 2 and R = x + 8. Write x + 8 ≡ A(x + 2) + B(x − 1). At 1, A = 3; at −2, B = −2. Answer: 2 + 3/(x − 1) − 2/(x + 2), x ≠ 1, −2.
Understand · explore · practise
Improper partial fractions
Learn improper partial fractions by dividing first, then decomposing the proper remainder. Equal-degree, higher-degree and repeated-factor examples with worked practice.
Before you startAlgebraic division and partial fractions with distinct or repeated factors
01 / Check the degrees
An improper fraction needs a polynomial part.
A rational fraction is proper when its numerator degree is smaller than its denominator degree. If the numerator has equal or greater degree, divide first. A list of simple fractions alone will generally be missing part of the answer.
N/D = Q + R/D
Find Q and R by polynomial division, with degree R < degree D or R = 0. Then decompose only R/D.
Keep the original denominator restrictions throughout. A zero remainder means division has finished the algebra, but it does not restore excluded inputs from the original fraction.
Why does a polynomial part matter? Each proper simple fraction tends to zero as |x| becomes large. A fraction with numerator and denominator of equal degree tends to a nonzero constant; a sum of those simple fractions cannot supply that constant.
02 / Equal degrees
Separate the constant before finding partial fractions.
Decompose (3x² + 5x + 1)/(x² + x − 2). Its numerator and denominator both have degree 2, so division gives a constant quotient.
A constant plus a proper fractionWorked example
3x² + 5x + 1
≡ 3(x² + x − 2) + (2x + 7)
The leading coefficients give quotient 3; subtraction leaves 2x + 7.
N/D = 3 + (2x + 7)/[(x − 1)(x + 2)]
Factor the denominator. Keep x ≠ 1, −2.
2x + 7 ≡ A(x + 2) + B(x − 1)
At x = 1: A = 3. At x = −2: B = −1.
Answer: 3 + 3/(x − 1) − 1/(x + 2)
Check the remainder numerator: 3(x + 2) − (x − 1) = 2x + 7.
03 / Choose the polynomial part
A true identity can still be an unfinished division.
You can subtract any multiple k of the denominator and write the remaining numerator as R. That always produces an identity N = kD + R. Only k = 3 cancels the quadratic term in this example.
Use the control to test different constants. Do not confuse “the two sides are equal” with “the division has finished”. Both equality and the degree condition are required.
Which polynomial part makes the remainder proper?Explore
(3x² + 5x + 1)/(x² + x − 2)
R(x) = (3 − k)x²
+ (5 − k)x + (1 + 2k)
k = 0: remainder = 3x² + 5x + 1.
The remainder still has degree 2. Division is not finished.
N ≡ kD + R is true for every setting. The completed division also requires degree R < degree D. Original restrictions: x ≠ 1, −2.
Watch division prepare the partial fractions
Pause, replay or seek freely. The notes explain the same idea and stay in view.
04 / Higher numerator degree
Keep the whole quotient, including its constant.
Decompose (x³ + 2x² + 4x − 1)/[(x − 1)(x + 2)]. Expand the denominator for division, then use its factors for the partial fractions.
A linear polynomial partWorked example
D = x² + x − 2
First quotient term: x. Subtract xD to leave x² + 6x − 1.
Next quotient term: 1
Subtract D to leave 5x + 1. Thus Q = x + 1.
N/D = x + 1
+ (5x + 1)/[(x − 1)(x + 2)]
The remaining fraction is now proper.
5x + 1 ≡ A(x + 2) + B(x − 1)
At 1, A = 2. At −2, B = 3.
Answer: x + 1
+ 2/(x − 1) + 3/(x + 2)
Retain x ≠ 1, −2. Check the complete numerator: (x + 1)(x² + x − 2) + 2(x + 2) + 3(x − 1) = x³ + 2x² + 4x − 1.
05 / Repeated factors
Division does not change the repeated-factor rule.
Decompose (2x³ + x² + x + 5)/[(x − 1)²(x + 2)]. The denominator expands to x³ − 3x + 2. The equal degrees give quotient 2 and remainder x² + 7x + 1.
N/D = 2
+ (x² + 7x + 1)/[(x − 1)²(x + 2)]
Now use A/(x − 1) + B/(x − 1)² + C/(x + 2) for the proper fraction. Clearing denominators gives x² + 7x + 1 ≡ A(x − 1)(x + 2) + B(x + 2) + C(x − 1)².
At x = 1, B = 3. At x = −2, C = −1. Comparing x² gives A + C = 1, so A = 2.
2 + 2/(x − 1)
+ 3/(x − 1)² − 1/(x + 2)
Keep x ≠ 1, −2. The initial 2 is essential.
06 / Factor theorem link
Find the denominator factors, then organise the work.
Consider (x³ + x² + x − 6)/(x³ − x² − 4x + 4). To factor the denominator D, notice D(1) = 0. Thus x − 1 is a factor; division leaves x² − 4 = (x − 2)(x + 2).
A combined algebra problemWorked example
D = (x − 1)(x − 2)(x + 2)
Restrictions: x ≠ 1, 2, −2.
N − D = 2x² + 5x − 10
The quotient is 1 and this remainder is proper.
2x² + 5x − 10
≡ A(x − 2)(x + 2)
+ B(x − 1)(x + 2)
+ C(x − 1)(x − 2)
Give each distinct linear factor one constant numerator.
x = 1: −3 = −3A
x = 2: 8 = 4B
x = −2: −12 = 12C
Hence A = 1, B = 2 and C = −1.
Answer: 1 + 1/(x − 1)
+ 2/(x − 2) − 1/(x + 2)
Recombination restores the full cubic numerator, including the polynomial part 1.
07 / Your turn
Divide first, decompose second, check everything.
State the original exclusions with each answer. A complete check reconstructs the polynomial part as well as the proper remainder.
02 · Linear quotient
Decompose (x³ + 3x² + 2x + 5)/[x(x + 2)].
Hint
Division gives quotient x + 1.
Worked solution
Subtract x(x² + 2x), leaving x² + 2x + 5; then subtract x² + 2x, leaving 5. For 5/[x(x + 2)], 5 ≡ A(x + 2) + Bx gives A = 5/2 and B = −5/2. Answer: x + 1 + 5/(2x) − 5/[2(x + 2)], x ≠ 0, −2.
03 · Repeated denominator
Decompose (x² + 4x + 1)/(x − 1)².
Hint
Subtract the denominator first.
Worked solution
Q = 1 and R = 6x. Then 6x ≡ A(x − 1) + B gives A = 6, B = 6. Answer: 1 + 6/(x − 1) + 6/(x − 1)², x ≠ 1.
04 · Zero remainder, restricted domain
Simplify (x² − 1)/(x − 1). Is the answer the same function as x + 1 on all real numbers?
Hint
Division may have zero remainder, but check the original denominator.
Worked solution
The expression equals x + 1 for x ≠ 1, with Q = x + 1 and R = 0. The original is undefined at 1; the unrestricted function x + 1 is defined there. They agree on the original domain, not on all real numbers.
05 · Non-monic denominator
Decompose (4x² + 5x + 1)/[(2x − 1)(x + 1)].
Hint
The denominator expands to 2x² + x − 1.
Worked solution
Q = 2 and R = 3x + 3. Use 3x + 3 ≡ A(x + 1) + B(2x − 1). At 1/2, A = 3; at −1, B = 0. Answer: 2 + 3/(2x − 1), with both original restrictions x ≠ 1/2, −1.
06 · A missing polynomial part
A proposed answer to (3x² + 5x + 1)/(x² + x − 2) is 3/(x − 1) − 1/(x + 2). Explain two ways to detect the error.
Hint
Recombine the answer and consider large |x|.
Worked solution
The proposed numerator is 2x + 7, missing 3(x² + x − 2); the answer needs +3. Also, the proposed expression tends to 0 for large |x| while the original tends to 3. Either check detects the missing constant.
07 · Quadratic polynomial part
Decompose (x⁴ − x² + x + 1)/(x² − 1).
Hint
The leading division leaves quotient x² and remainder x + 1.
Worked solution
x⁴ − x² + x + 1 = x²(x² − 1) + (x + 1). The proper part (x + 1)/[(x − 1)(x + 1)] equals 1/(x − 1) on the original domain. Answer: x² + 1/(x − 1), with x ≠ −1, 1. The zero coefficient of 1/(x + 1) does not restore x = −1.
08 · Design an improper fraction
Write 2 + 1/(x − 2) − 3/(x + 1) as one fraction and state its exclusions.
Hint
Put the constant 2 over the full common denominator too.
Worked solution
The denominator is (x − 2)(x + 1) = x² − x − 2. The numerator is 2(x² − x − 2) + (x + 1) − 3(x − 2) = 2x² − 4x + 3. Answer: (2x² − 4x + 3)/[(x − 2)(x + 1)], x ≠ 2, −1. Dividing this result recovers quotient 2 and remainder −2x + 7.
08 / Recap
One complete answer includes three pieces.
- Polynomial part from division.
- Partial fractions for the proper remainder.
- All original denominator restrictions.
Check by bringing the entire answer to a common denominator. Use a large-|x| check to catch a missing polynomial part, but use exact recombination to verify every coefficient.
Section 1 of 8 · Check the degrees