HomePartial Fractions: Decompose Rational Expressions Step by Step Free original lessons
Partial Fractions: Decompose Rational Expressions Step by Step
Learn to break rational expressions into simpler partial fractions using linear and repeated factors. Includes worked example, practice, and rubric.
Free lesson
Four exercises
Worked answers
Algebra and Function Practice
Partial fraction decomposition rewrites a single rational expression as a sum of simpler fractions. This skill is useful for integration and for understanding rational functions. The method applies when the denominator is factored into linear factors (possibly repeated) and the numerator degree is less than the denominator degree. You will set up unknown numerators, clear denominators, and solve for constants. The convention for handwriting is that x-height is measured from baseline to midline; letter styles may vary, so focus on correct algebraic structure rather than exact glyph shapes. All numerical examples are hypothetical and chosen for easy arithmetic.
Practice materials: Printable downloads are not available for this lesson yet. Use the instructions and four exercises below.
Setting Up the Decomposition
First ensure the rational expression is proper: the degree of the numerator must be less than the degree of the denominator. If not, perform polynomial long division first. Then factor the denominator completely into linear factors over the real numbers. For each distinct linear factor (ax+b), assign a constant numerator A. For each repeated linear factor (ax+b)^k, assign k constants: A1, A2, ..., Ak, with denominators (ax+b), (ax+b)^2, ..., (ax+b)^k. Write the decomposition as a sum of these fractions. Do not simplify the original fraction before decomposing; keep the factored denominator.
For example, if the denominator is (x-1)(x+2)^2, the setup is A/(x-1) + B/(x+2) + C/(x+2)^2. The constants A, B, C are unknown. Multiply both sides by the original denominator to clear fractions, yielding a polynomial equation. Then solve for the constants by substituting convenient values of x (roots of the factors) or by equating coefficients. Always check your final decomposition by combining the partial fractions back into a single fraction and comparing to the original.
- Check that the rational expression is proper; if not, divide first.
- Factor the denominator completely into linear factors.
- Write the partial fraction setup with unknown constants for each factor and power.
- Multiply both sides by the factored denominator to clear fractions.
- Solve for constants using substitution or coefficient matching.
- Verify by recombining the partial fractions.
Solving for Constants: Substitution and Coefficient Matching
After clearing denominators, you obtain an identity that holds for all x. Substitution is fastest when the factors are linear: plug in the roots of each factor to isolate constants. For repeated factors, substitution gives one constant per distinct root; the remaining constants require equating coefficients of like powers of x or choosing additional x-values. Coefficient matching means expanding both sides and setting the coefficients of x^0, x^1, etc., equal. Both methods are valid; choose the one that yields simpler arithmetic.
Be careful with signs and with the degree of the numerator. If the original numerator has missing powers, include zero coefficients when matching. After finding constants, substitute them back into the setup. A common error is forgetting the repeated-factor terms or assigning a linear numerator to a linear denominator. Since the denominator factors are linear, numerators are constants. Always perform a final check by combining the partial fractions and simplifying to see if you recover the original expression.
- Substitute roots of distinct linear factors to find corresponding constants.
- For repeated factors, use the root to find the highest-power constant, then match coefficients for the rest.
- Equate coefficients of like powers of x if substitution is insufficient.
- Solve the resulting linear system for remaining constants.
- Write the final decomposition with numerical constants.
- Combine and simplify to verify equivalence.
Handling Repeated Factors and Verification
Repeated factors require multiple terms with increasing powers. For (x-r)^2, use A/(x-r) + B/(x-r)^2. For (x-r)^3, use A/(x-r) + B/(x-r)^2 + C/(x-r)^3. When clearing denominators, the equation will involve these constants. Substituting x=r gives the constant for the highest power directly. For example, with B/(x-r)^2, multiplying by (x-r)^2 and setting x=r isolates B. The other constants are found by equating coefficients or substituting other x-values. Keep track of which constant belongs to which power.
Verification is essential. Combine the partial fractions by finding a common denominator, expand the numerator, and compare with the original numerator. If they match, the decomposition is correct. If not, recheck the setup and arithmetic. Remember that the decomposition is unique for a given factored denominator. Handwriting conventions: x-height is baseline to midline; letter styles vary, so ensure algebraic symbols are unambiguous. Numerical values in examples are hypothetical and chosen for simplicity.
- For each repeated factor (x-r)^k, include terms A1/(x-r) + ... + Ak/(x-r)^k.
- Substitute x=r to find Ak directly.
- Use coefficient matching or additional substitutions to find A1 through A(k-1).
- Write the final sum with all constants.
- Combine the partial fractions over a common denominator.
- Compare the simplified numerator to the original numerator.
Worked example
Decompose (3x+5)/((x-1)(x+2)) into partial fractions.
Worked answer
(8/3)/(x-1) + (1/3)/(x+2)
Why: Set up A/(x-1) + B/(x+2). Multiply both sides by (x-1)(x+2): 3x+5 = A(x+2) + B(x-1). Substitute x=1: 3(1)+5=8 and A(1+2)=3A, so 3A=8 and A=8/3. Substitute x=-2: 3(-2)+5=-1 and B(-2-1)=-3B, so -3B=-1 and B=1/3. Thus the decomposition is (8/3)/(x-1) + (1/3)/(x+2). Check by combining: (8/3)/(x-1) + (1/3)/(x+2) = [8(x+2)+1(x-1)]/[3(x-1)(x+2)] = (8x+16+x-1)/[3(x-1)(x+2)] = (9x+15)/[3(x-1)(x+2)] = (3x+5)/[(x-1)(x+2)], which matches the original.
Four practice exercises
Exercise 1
Decompose (5x+1)/((x-2)(x+3)) into partial fractions.
Show answer and check
(11/5)/(x-2) + (14/5)/(x+3)
Why: Set up A/(x-2) + B/(x+3). Multiply by (x-2)(x+3): 5x+1 = A(x+3) + B(x-2). Substitute x=2: 5(2)+1=11 and A(2+3)=5A, so A=11/5. Substitute x=-3: 5(-3)+1=-14 and B(-3-2)=-5B, so B=14/5. Thus the decomposition is (11/5)/(x-2) + (14/5)/(x+3). Check by combining: [11(x+3)+14(x-2)]/[5(x-2)(x+3)] = (11x+33+14x-28)/[5(x-2)(x+3)] = (25x+5)/[5(x-2)(x+3)] = (5x+1)/[(x-2)(x+3)], which matches the original.
Check: Combine (11/5)/(x-2) + (14/5)/(x+3) over the common denominator 5(x-2)(x+3); the numerator simplifies to 25x+5, which divided by 5 gives 5x+1, recovering the original expression.
Exercise 2
Decompose (2x+7)/((x+1)(x-4)) into partial fractions.
Show answer and check
-1/(x+1) + 3/(x-4)
Why: Set up A/(x+1) + B/(x-4). Multiply by (x+1)(x-4): 2x+7 = A(x-4) + B(x+1). Substitute x=-1: 2(-1)+7=5 and A(-1-4)=-5A, so -5A=5 and A=-1. Substitute x=4: 2(4)+7=15 and B(4+1)=5B, so B=3. Thus the decomposition is -1/(x+1) + 3/(x-4). Check by combining: [-1(x-4)+3(x+1)]/[(x+1)(x-4)] = (-x+4+3x+3)/[(x+1)(x-4)] = (2x+7)/[(x+1)(x-4)], which matches the original.
Check: Combine -1/(x+1) + 3/(x-4) over the common denominator (x+1)(x-4); the numerator simplifies to 2x+7, recovering the original expression.
Exercise 3
Decompose (4x+3)/((x-1)^2) into partial fractions.
Show answer and check
4/(x-1) + 7/(x-1)^2
Why: Set A/(x-1)+B/(x-1)^2. Multiply: 4x+3=A(x-1)+B. x=1: 7=B. Then 4x+3=A(x-1)+7 => 4x-4=A(x-1) => A=4. Check: 4/(x-1)+7/(x-1)^2 = (4(x-1)+7)/(x-1)^2 = (4x-4+7)/(x-1)^2 = (4x+3)/(x-1)^2.
Check: Combine 4/(x-1)+7/(x-1)^2 to get (4x+3)/(x-1)^2.
Exercise 4
Decompose (x+5)/((x+2)(x-1)^2) into partial fractions.
Show answer and check
1/3/(x+2) - 1/3/(x-1) + 2/(x-1)^2
Why: Set A/(x+2)+B/(x-1)+C/(x-1)^2. Multiply: x+5=A(x-1)^2+B(x+2)(x-1)+C(x+2). x=-2: 3=9A => A=1/3. x=1: 6=3C => C=2. Match x^2: 0=A+B => B=-1/3. Check: (1/3)/(x+2)+(-1/3)/(x-1)+2/(x-1)^2. Combine: common denom 3(x+2)(x-1)^2. Numerator: (x-1)^2 - (x+2)(x-1) + 6(x+2) = (x^2-2x+1) - (x^2+x-2) +6x+12 = x^2-2x+1 -x^2-x+2+6x+12 =3x+15=3(x+5). Divide by 3 gives (x+5)/den.
Check: Combine the three fractions and simplify to (x+5)/((x+2)(x-1)^2).
Common mistakes
- Setting up a linear numerator for a linear denominator, such as (Ax+B)/(x-1). For linear factors, numerators are constants. Use A/(x-1), not (Ax+B)/(x-1).
- Omitting terms for repeated factors, e.g., using only A/(x+2) for (x+2)^2. Include one term per power: A/(x+2) + B/(x+2)^2.
- Forgetting to divide first when the rational expression is improper. Check degrees; if numerator degree ≥ denominator degree, perform polynomial long division before decomposing.
- Making sign errors when substituting negative roots. Substitute carefully and double-check arithmetic; use parentheses to avoid sign mistakes.
Self-check rubric
- Correctly identifies proper rational expression and factors denominator.
- Sets up partial fractions with correct number and form of terms, including repeated factors.
- Clears denominators correctly and forms a valid polynomial identity.
- Solves for constants accurately using substitution or coefficient matching.
- Writes final decomposition with constants in simplest form.
- Verifies by recombining and simplifying to original expression.
- Handwriting: x-height from baseline to midline; letter styles may vary but algebraic symbols are clear and unambiguous.
Questions and answers
When can I use partial fractions?
When the rational expression is proper and the denominator factors into linear factors (possibly repeated). If the denominator has irreducible quadratic factors, different numerators are needed, but that is beyond this brief.
Do I always need to factor the denominator completely?
Yes, the setup depends on the factored form. Factor completely over the real numbers into linear factors for this method.
How do I check my answer?
Combine the partial fractions into a single fraction and simplify. The result should equal the original expression.
What if the numerator has missing powers?
Include zero coefficients when matching. For example, treat x^2+1 as 1x^2+0x+1.