How this instrument works
Synthetic division is a coefficient-only shortcut for dividing a polynomial by a linear factor of the specific form (x − r). Instead of writing out full terms and subtracting like standard long division, you carry the coefficients a, b, c across a short chain of multiply-and-add steps: bring a straight down as the quotient's leading coefficient, multiply by r and add to b to get the quotient's constant term, then multiply that by r and add to c to get the remainder. Two steps replace a page of algebra, which is why the method survives in every algebra course despite calculators existing.
The shortcut works because of a coefficient-matching argument you can run in one line. Write ax² + bx + c as (x − r)(Ax + B) + R and expand the right side: Ax² + (B − Ar)x + (R − Br). Matching coefficients term by term forces A = a, then B = b + a·r, then R = c + r·B — exactly the three formulas this calculator evaluates. Nothing is approximated; the identity is exact algebra, not a numerical estimate.
The remainder carries more information than it looks like it should. By the Remainder Theorem, dividing any polynomial P(x) by (x − r) always leaves R = P(r) — so this calculator's Remainder field is silently evaluating the dividend at x = r without you ever expanding a square. When that remainder lands on exactly zero, the Factor Theorem takes over: r is a root of the dividend, and (x − r) is confirmed as one of its genuine linear factors, not merely a close divisor.
- Enter the dividend's coefficients into Dividend: x² coefficient, a, Dividend: x coefficient, b, and Dividend: constant, c, reading them straight off ax² + bx + c.
- If the divisor is written as (x + k), remember the root is −k, not k: type that signed value into Divisor's root, r (dividing by x − r).
- Read Quotient: x coefficient (= a) and Quotient: constant term together as the quotient line Ax + B.
- Check Remainder: zero means (x − r) divides the dividend exactly; any other value is the dividend evaluated at r.
Worked example — factoring x² + 5x + 6 by synthetic division
Take the dividend x² + 5x + 6, so a = 1, b = 5, c = 6, and divide it by (x + 2). Written the required way that divisor is (x − (−2)), so the root is r = −2 — enter −2, not 2, into Divisor's root, r. The quotient's x coefficient is simply a, so quotientA = 1, and its constant term is b + a·r = 5 + 1 × (−2) = 3, so quotientB = 3.
The remainder is c + r·(b + a·r) = 6 + (−2) × 3 = 6 − 6 = 0. A zero remainder means (x + 2) divides x² + 5x + 6 exactly, with quotient x + 3, and it reproduces the familiar factoring x² + 5x + 6 = (x + 2)(x + 3) — the same two factors you would find by inspection, reached here by three multiplications and two additions instead of trial and error.
Questions
What is synthetic division and why is it faster than long division?
Synthetic division is a coefficient-only shortcut for dividing a polynomial by a linear factor (x − r). It drops the x terms and subtraction steps of standard long division in favor of one multiply-and-add per coefficient: bring down a, multiply by r, add to the next coefficient, repeat. For a quadratic that is exactly two multiply-add steps, B = b + a·r and R = c + r·B, instead of a full page of column arithmetic.
How do I get the sign of r right when dividing by (x + 2)?
Synthetic division always divides by (x − r), so a divisor written as (x + 2) means r = −2, not 2 — rewrite (x + 2) as (x − (−2)) before typing it in. Flipping this sign is the most common mistake with the method: every multiply-and-add step still runs, but the wrong r silently changes both the quotient and the remainder.
What does a nonzero remainder actually tell you?
By the Remainder Theorem, dividing a polynomial P(x) by (x − r) always leaves a remainder equal to P(r), so the Remainder field is quietly evaluating the dividend at x = r without expanding any powers. A remainder of 3 means P(r) = 3; only an exact zero confirms that r is a root and (x − r) a genuine factor of the dividend.
Does synthetic division work for any divisor, not just (x − r)?
No — the coefficient shortcut only applies to a linear divisor with leading coefficient 1, that is (x − r). Dividing by a quadratic or higher-degree factor, or by a linear term like (2x − 3), needs full polynomial long division instead; synthetic division is a special case chosen for speed, not a general-purpose replacement for it.
How does the Factor Theorem connect to this calculator's remainder?
The Factor Theorem states that (x − r) is a factor of P(x) exactly when P(r) equals zero, which here means the Remainder field reads zero. That is the golden case above: dividing x² + 5x + 6 by (x + 2) leaves remainder 0, confirming x = −2 as a root and (x + 2) as one of the two factors in x² + 5x + 6 = (x + 2)(x + 3).