Polynomial long division: (x³ − 2x² + 4x − 8) ÷ (x − 2)
The systematic recipe for dividing one polynomial by another, written out the same way you divided whole numbers in primary school.
What this shows
Polynomial long division divides one polynomial (the dividend) by another (the divisor) to produce a quotient polynomial and a remainder polynomial — the same shape of result as integer long division. The algorithm is mechanical:
1. Divide the leading term of the dividend by the leading
term of the divisor. That gives the next quotient term.
2. Multiply the whole divisor by that quotient term and
subtract from the dividend.
3. Repeat with the new (smaller-degree) dividend until the
remaining polynomial has lower degree than the divisor.
Working through (x³ − 2x² + 4x − 8) ÷ (x − 2):
Step 1: x³ / x = x². Multiply: x²·(x − 2) = x³ − 2x².
Subtract from dividend:
(x³ − 2x² + 4x − 8) − (x³ − 2x²) = 4x − 8.
Step 2: 4x / x = 4. Multiply: 4·(x − 2) = 4x − 8.
Subtract: (4x − 8) − (4x − 8) = 0.So x³ − 2x² + 4x − 8 = (x − 2)(x² + 4), exactly — the remainder is zero, which means x = 2 is a root of the dividend. This is the converse of the factor theorem: if (x − r) divides a polynomial cleanly, then r is a root, and vice versa.
Where it shows up
Long division is the building block of root-finding for higher- degree polynomials. Once you've found one root r by guessing or by the rational-root theorem, dividing by (x − r) reduces a cubic to a quadratic, a quartic to a cubic, and so on — until the residual polynomial is small enough that the quadratic formula finishes the job.
The same algorithm shows up in computer algebra systems' polynomial-GCD routines and in error-correcting codes, where cyclic redundancy checks (CRCs) and Reed-Solomon decoders both rely on polynomial long division over finite fields.
Frequently asked questions
What if the remainder isn't zero?
Then the divisor isn't a factor of the dividend. The general result is dividend = (divisor · quotient) + remainder, where the remainder polynomial has lower degree than the divisor. For divisor (x − r), the remainder is just the constant value of the dividend evaluated at r — that's the remainder theorem.
Can I always divide a polynomial by (x − r)?
Yes — for any constant r, polynomial division by the linear factor (x − r) always succeeds and produces a quotient of degree one less than the dividend, plus a constant remainder equal to dividend(r).
How does this compare to synthetic division?
Synthetic division is a shortcut for the specific case where the divisor is (x − r). It tracks only the coefficients, in a tabular layout that's faster by hand. Long division generalises to any divisor degree; synthetic division does not.
Why was the remainder zero in the example?
Because x = 2 happens to be a root of x³ − 2x² + 4x − 8. Substitute: 8 − 8 + 8 − 8 = 0. The factor theorem then says (x − 2) must divide the polynomial exactly.