Guides / Rational expressions

Partial fraction decomposition: pick the right form, then find A and B

Partial fractions runs adding fractions backwards: one fraction in, a sum of simpler ones out. The algebra for A and B is short. Points are lost earlier, when the form is set up with too few terms or the fraction needed dividing first.

Why it works

When you add 3x − 1 + 4x + 2, you get one fraction over (x − 1)(x + 2). Decomposition reverses that, so the denominators of the pieces are always the factors of the original denominator. That's why the factors fix the form and only the numbers on top are unknown.

After you multiply both sides by the denominator, the equation is true for every x, so you can pick convenient values. At a root of the denominator, every term but one becomes zero, and that term hands you one constant directly. Plugging in roots like this is often called the cover-up method. The same factoring skills from simplifying rational expressions do most of the work, and polynomial long division handles the case where the top is too big.

Example 1: two different linear factors

Decompose 7x + 2(x − 1)(x + 2).

1. 7x + 2(x − 1)(x + 2) = Ax − 1 + Bx + 2Degree 1 on top, 2 on the bottom, so no division. One constant over each linear factor.
2. 7x + 2 = A(x + 2) + B(x − 1)Multiply both sides by (x − 1)(x + 2). Each term keeps the factor its denominator lacked.
3. x = 1: 9 = 3A, A = 3At x = 1 the B term is B · 0, so it disappears.
4. x = −2: −12 = −3B, B = 4The root of x + 2 is −2, not 2. At x = −2 the A term disappears.
5. 3x − 1 + 4x + 2Check: 3(x + 2) + 4(x − 1) = 3x + 6 + 4x − 4 = 7x + 2.

Example 2: a squared factor

Decompose 4x(x − 1)(x + 1)².

1. Ax − 1 + Bx + 1 + C(x + 1)²The squared factor gets two terms: one over (x + 1) and one over (x + 1)². A cubic denominator, three constants.
2. 4x = A(x + 1)² + B(x − 1)(x + 1) + C(x − 1)Multiply both sides by (x − 1)(x + 1)².
3. x = 1: 4 = 4A, A = 1Only the A term survives.
4. x = −1: −4 = −2C, C = 2Only the C term survives.
5. x² terms: 0 = A + B, B = −1No root isolates B, so compare the x² terms. The left side has none; the right side has Ax² + Bx².
6. 1x − 1 − 1x + 1 + 2(x + 1)²Check at x = 2: the original is 8/9, and 1 − 1/3 + 2/9 = 8/9.

Example 3: a quadratic that won't factor

Decompose 3x² + x + 4x(x² + 2).

1. Ax + Bx + Cx² + 2x² + 2 has no real roots, so it stays whole. Over a quadratic, the top is Bx + C, not just a constant.
2. 3x² + x + 4 = A(x² + 2) + (Bx + C)xMultiply both sides by x(x² + 2).
3. x = 0: 4 = 2A, A = 2The only real root of the denominator.
4. x²: 3 = A + B, so B = 1. x: 1 = CMatch coefficients. The right side expands to (A + B)x² + Cx + 2A.
5. 2x + x + 1x² + 2Check at x = 1: the original is 8/3, and 2 + 2/3 = 8/3.

Common mistakes, and the exact line they happen on

1. Skipping the division

Decompose x² + 1x² − 1.

1. x² + 1(x − 1)(x + 1) = 1x − 1 − 1x + 1This is the mistake. The top and bottom are both degree 2, so this fraction must be divided first. Plugging in roots still produces numbers, which is why it feels fine. But the right side adds up to 2/(x² − 1), and at x = 2 that's 2/3 while the original is 5/3.
2. 1 + 2x² − 1Correct line: x² + 1 = 1 · (x² − 1) + 2, so the quotient is 1 and the remainder is 2.
3. 1 + 1x − 1 − 1x + 1Decompose only the remainder. Check at x = 2: 1 + 1 − 1/3 = 5/3.

2. One term for a squared factor

Decompose 4x(x − 1)(x + 1)² (Example 2).

1. Ax − 1 + C(x + 1)²This is the mistake. (x + 1)² was treated as one factor, so it got one term. Clearing gives 4x = A(x + 1)² + C(x − 1). The roots give A = 1 and C = 2, but then the x² terms say 0 = A = 1. No numbers work.

When you can't find constants that check, the form is usually the problem, not the arithmetic. A squared factor needs a term for each power: B/(x + 1) + C/(x + 1)².

3. A plain constant over a quadratic

Decompose 3x² + x + 4x(x² + 2) (Example 3).

1. Ax + Cx² + 2This is the mistake. Clearing gives 3x² + x + 4 = A(x² + 2) + Cx. The x² terms say A = 3, the x terms say C = 1, and the constants say 2A = 4, so A = 2. A can't be both. Over a quadratic, the top has to be Bx + C.

4. Covering up at the wrong root

The cover-up shortcut finds a constant in one line: cover the constant's own factor, and evaluate what's left at the value that makes that factor zero. Decompose 7x + 2(x − 1)(x + 2) (Example 1).

1. A = 7(1) + 21 + 2 = 3Right so far: cover x − 1 and evaluate (7x + 2)/(x + 2) at x = 1.
2. B = 7(2) + 22 − 1 = 16This is the mistake. The 2 was read off x + 2 without its sign. The shortcut only works where the covered factor is zero, and x + 2 is zero at x = −2. The check fails: 3(x + 2) + 16(x − 1) = 19x − 10, not 7x + 2.
3. B = 7(−2) + 2−2 − 1 = −12−3 = 4Correct line: evaluate at x = −2.

Practice

1. Decompose x + 11(x + 3)(x − 1).

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   x = −3: 8 = −4A, A = −2. x = 1: 12 = 4B, B = 3. Answer: −2/(x + 3) + 3/(x − 1). Check: −2(x − 1) + 3(x + 3) = x + 11 ✓.

2. Decompose 5x − 4x² − x − 2.

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   x² − x − 2 = (x − 2)(x + 1). x = 2: 6 = 3A, A = 2. x = −1: −9 = −3B, B = 3. Answer: 2/(x − 2) + 3/(x + 1).

3. Decompose 3x + 5(x + 1)².

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   Form A/(x + 1) + B/(x + 1)², so 3x + 5 = A(x + 1) + B. Matching: A = 3, and A + B = 5 gives B = 2. Answer: 3/(x + 1) + 2/(x + 1)².

4. Decompose x² + 2x + 3x(x² + 1).

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   Form A/x + (Bx + C)/(x² + 1). x = 0: A = 3. x²: 1 = A + B, B = −2. x: C = 2. Answer: 3/x + (−2x + 2)/(x² + 1).

5. Decompose 2x² + x − 4x² − 4.

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   Same degree, so divide first: 2x² + x − 4 = 2(x² − 4) + (x + 4). Then (x + 4)/((x − 2)(x + 2)): at x = 2, A = 6/4 = 3/2; at x = −2, B = 2/(−4) = −1/2. Answer: 2 + 32(x − 2) − 12(x + 2).