Partial fraction structure

Start with a proper rational expression

Partial fractions rewrite a rational expression as simpler fractions with the same value on its original allowed inputs. First compare degrees. A proper rational expression has numerator degree less than denominator degree. If it is not proper, divide first to separate a polynomial part and a proper remainder.

For example, the denominator (x + 1)(x + 2) expands to x² + 3x + 2, so

The constant 1 is necessary: fractions with constant numerators over x + 1 and x + 2 alone can reconstruct only a linear numerator over that common quadratic denominator. They cannot supply the original x² term.

Distinct linear factors give constant numerators

For distinct linear factors,

Seek A and B making the polynomial identity 3x + 5 = A(x + 2) + B(x + 1). At x = −1 it requires A = 2; at x = −2 it requires B = 1. Expansion then verifies 2(x + 2) + (x + 1) = 3x + 5, so the decomposition holds wherever the original rational expression is defined, x ≠ −1, −2.

The substitutions at −1 and −2 determine a polynomial identity, which has no denominator. They do not assign values to the original fractions at those excluded inputs.

Apply the same method to the proper remainder x + 3 above. Its identity gives A = 2 and B = −1, hence

Recombining the right side gives numerator (x + 1)(x + 2) + 2(x + 2) − (x + 1) = x² + 4x + 5. This checks the polynomial part and fractions together.

Repeated factors need every power

For denominator (x + 1)(x + 2)², include A/(x + 1) + B/(x + 2) + C/(x + 2)². The repeated factor needs both its first and second powers. With numerator 2x + 3, require

At x = −1, 1 = A. At x = −2, −1 = −C, so C = 1. The x² coefficient on the left is zero, giving A + B = 0, hence B = −1. This produces

Check the entire numerator:

If the B term had been omitted, the x² coefficient would force A = 0, contradicting A = 1 from x = −1. This is why including only the highest repeated power is not a general decomposition method.

An irreducible quadratic needs a linear numerator

For (x + 1)(x² + 1), use A/(x + 1) + (Bx + C)/(x² + 1). A numerator of degree below two may be linear, so assuming a constant would discard a needed coefficient. For numerator 3x² + 5x + 4, require

Comparing coefficients gives A + B = 3, B + C = 5 and A + C = 4. The first and third equations minus the second give 2A = 2. Thus A = 1, B = 2 and C = 3, producing

Recombining gives (x² + 1) + (2x + 3)(x + 1) = 3x² + 5x + 4. The quadratic x² + 1 has no real zeros, so it cannot be split into real linear factors. A constant numerator may happen in a special example, but the general linear numerator is what lets the identity determine that rather than assume it.

Factor the actual denominator before choosing its form. More generally, for real c ≠ 0, x² + c² has no real zeros; if c = 0, it becomes the repeated linear factor x² and needs that structure instead. Combine coincident linear factors before counting their powers.

Verify an identity, not a few isolated values

After solving for coefficients, put the complete expression over its common denominator and compare every numerator coefficient. A spot check can find an error; use the reconstructed polynomial identity for the exact verification. Retain original exclusions even if a later cancellation makes the simplified expression appear defined at more inputs.