Mathematics
Synthetic division calculator
Enter the coefficients of the polynomial and a value for a: you get the table for dividing by (x − a), the quotient, the remainder, and the full factorisation with every rational root and its multiplicity.
What synthetic division is for
Synthetic division — Ruffini's rule — is a quick way to divide a polynomial by a binomial of the form (x − a): bring the first coefficient down, multiply it by a, add it to the next coefficient, and carry on to the end. The numbers on the bottom row are the coefficients of the quotient, and the last one is the remainder. It takes less room than long division because only the coefficients are written, without rewriting the powers of x each time.
That last number is more than a remainder: by the remainder theorem it is the value of the polynomial at a. If it comes to zero, a is a root, (x − a) is a factor, and the quotient is what is left to factor — one degree lower, and so easier. Repeating the process on the quotient gets to the complete factorisation, which is why division and factoring are taught together.
What remains is knowing which a to try, and that is the rational root theorem: if the polynomial has integer coefficients, every rational root p/q has p dividing the constant term and q dividing the leading one. The candidates are therefore finite, and all of them can be tried. Anything that fails is not a rational root, and this calculator says so rather than handing back an approximation: a polynomial like x² − 2 has real roots, but neither is rational, and synthetic division will not factor it.
Common mistakes
- Getting the sign of a wrong. The division is by (x − a), so to divide by (x + 3) the value to enter is −3. It is the commonest slip, and a large remainder where zero was expected is how it shows.
- Leaving out the zero coefficients. In x³ − 4x the coefficient of x² is zero and has to be entered: skipping it shifts every other coefficient along by one and the answer has nothing to do with the polynomial you started from.
- Trying whole numbers only. With a leading coefficient other than 1 the rational roots can be fractions: 2x² − 3x + 1 vanishes at 1/2, which no search through the integers alone would find.
Frequently asked questions
How does synthetic division work, step by step?
Write the coefficients from the highest power down to the constant term, zeros included. Bring the first one down, multiply it by a and add to the second; multiply that result by a and add to the third, and so on. All the numbers but the last are the coefficients of the quotient; the last is the remainder.
When is a polynomial divisible by (x − a)?
When the remainder is zero, that is, when P(a) = 0. That is the factor theorem, a direct consequence of the remainder theorem: dividing by (x − a) leaves the value of the polynomial at a as the remainder.
How do you know which roots to try?
By the rational root theorem: with integer coefficients, every rational root is p/q where p divides the constant term and q divides the leading coefficient. Every candidate is tried, with both signs. This calculator does that for you.
What does it mean for a root to have multiplicity two?
That the factor (x − a) appears twice in the factorisation — after dividing it out once, the quotient still vanishes at a. For instance x³ − 6x² + 12x − 8 is (x − 2)³: the root 2 has multiplicity three.
Can synthetic division divide by (2x − 1)?
Not directly: the rule is for monic divisors, of the form (x − a). But you can take the 2 out, writing 2x − 1 = 2(x − 1/2), divide by (x − 1/2) and then divide the quotient by 2. This calculator accepts fractional values of a, so you can enter 0.5 straight away.
How this calculation works
Synthetic division by (x − a): given coefficients aₙ, aₙ₋₁, …, a₀, set bₙ₋₁ = aₙ and then b₍ₖ₋₁₎ = aₖ + a·bₖ working down to the constant term. The b are the coefficients of the quotient and the last value computed is the remainder R, with P(x) = (x − a)·Q(x) + R. Remainder theorem: R = P(a). Factor theorem: (x − a) divides P exactly when P(a) = 0. Rational root theorem: if P has integer coefficients, every rational root p/q in lowest terms has p dividing a₀ and q dividing aₙ; the search tries all those candidates with both signs and repeats the division on each while the remainder stays zero, which gives the multiplicity. Zero is not among those candidates, since it divides no constant term, and is handled separately: it is a root when a₀ = 0. All the arithmetic runs on fractions of integers rather than floating point, because the remainder has to be compared with zero exactly.
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