Overview
Polynomial division chooses each quotient term so its product with the divisor cancels the current highest-degree term. Subtract that whole product and continue to lower degrees. A remainder records what cannot be divided further; a missing degree should be represented by a zero coefficient.
Consider this example: For x ≠ 2, simplify (x² + x − 6)/(x − 2). Factor the numerator as (x − 2)(x + 3). Canceling the nonzero x − 2 leaves x + 3. The original expression remains undefined at 2.
The method also handles a different case: Divide 2x² + 5x − 3 by x + 3. What is the quotient? The leading quotient term 2x contributes 2x² + 6x; subtract to leave −x − 3. The next term −1 contributes −x − 3 and leaves remainder 0. The quotient is 2x − 1.
For a further application, consider: Divide x³ − 8 by x − 2. What is the quotient? Treat the dividend as x³ + 0x² + 0x − 8. Division gives x², then 2x, then 4 with remainder 0. Multiplying (x − 2)(x² + 2x + 4) restores x³ − 8.
Subtracting only the leading term of the product corrupts the next stage. A nonzero remainder cannot be dropped just because the quotient looks plausible. Check a new division with dividend = divisor × quotient + remainder. The remainder must have lower degree than the divisor, and any rational form still excludes zeros of its divisor.
When division leaves a remainder, it cannot be discarded. Divide 2x³ − x² + 4x + 1 by x + 1. The first quotient term is 2x²; subtract its product 2x³ + 2x² to leave −3x² + 4x + 1. Next use −3x; subtract −3x² − 3x to leave 7x + 1. Finally use 7; subtract 7x + 7 to leave −6. For x ≠ −1, the result is 2x² − 3x + 7 − 6/(x + 1). The negative remainder is essential: multiplying back restores the constant 1 rather than 7 or 13. The identity P(x) = (x + 1)Q(x) + r also explains a shortcut for a remainder alone. For P(x) = x² + 2x + 5, substitute the divisor root −1. The product term vanishes, so r = P(−1) = 1 − 2 + 5 = 4. Substituting +1 would use the wrong root; solve x + 1 = 0 first.