How this instrument works
A cubic ax³+bx²+cx+d does not hand over its solutions the way a quadratic does; there is no simple three-term formula most people memorize. What often happens instead is that one solution is already known — handed to you, spotted by inspection, or guessed from the rational root theorem — and the remaining work is to strip that value out and solve whatever two-term problem is left. Synthetic division does the stripping: it takes the coefficients a, b, c, d and a candidate value r, and walks through a short chain of multiply-and-add steps that either confirms r belongs to the equation or exposes that it does not.
The chain runs like this: multiply a by r and add to b, multiply that result by r and add to c, multiply again by r and add to d. That last sum is the remainder. A remainder of exactly zero means r truly satisfies the cubic, and the numbers produced along the way are the coefficients of a plain quadratic — the original problem with one solution already removed. Feed those coefficients into the ordinary quadratic formula and the other two solutions fall out together, no guessing required for either of them.
This combination — one division step plus one classic formula — solves a cubic completely once a single real solution is already in hand, without reaching for Cardano's full cubic formula, which handles the general case but carries far more arithmetic. That makes this sheet a companion to a plain quadratic-formula calculator rather than a competitor to it: here, the quadratic formula is the second half of a two-step process, applied to a smaller equation manufactured by the division step rather than to the original three-term problem.
A remainder that is not zero is not a failure of the sheet — it is information. It tells you plainly that the value you tried does not actually satisfy d + (c + (b + a·r)·r)·r = 0, so it was never a genuine solution of your cubic in the first place. The two values still reported alongside a nonzero remainder come from the same quotient arithmetic, but they belong to a different, related quadratic rather than to the equation you started with — a distinction worth sitting with before trusting any answer this method returns.
- Enter the cubic's four numbers into a, b, c, and d, matching ax³+bx²+cx+d.
- Enter your candidate solution into r — a value you already suspect solves the equation, whether it was given or guessed.
- Read Remainder first: exactly 0 confirms r genuinely solves the cubic; any other value means the guess missed.
- When Remainder reads 0, read Second root and Third root for the two remaining solutions, produced by the quadratic formula.
- When Remainder is nonzero, treat Second root and Third root only as belonging to the quotient's own quadratic, not to your original cubic.
Worked example — checking that x = 1 solves x³ − 6x² + 11x − 6
Set a=1, b=−6, c=11, d=−6, and test r=1. Synthetic division proceeds: 1×1=1, added to −6 gives −5; −5×1=−5, added to 11 gives 6; 6×1=6, added to −6 gives 0. Remainder reads 0, so r=1 genuinely solves this cubic. The quotient left behind is x²−5x+6, and the quadratic formula gives [5 ± √(25−24)] ⁄ 2 = [5±1] ⁄ 2, or 3 and 2 — Second root and Third root. A quick check: 1, 3, and 2 are exactly the three solutions of x³−6x²+11x−6, since that expression equals (x−1)(x−2)(x−3).
A second cubic, x³−4x, tests r=2 and again gets remainder 0, with Second root 0 and Third root −2 — matching x³−4x = x(x−2)(x+2) exactly. But test r=2 against 2x³+3x²−8x+3 and the chain gives 4, then 7, then 12, then a final remainder of 15 — not zero, so 2 was never a genuine solution of that cubic. The −1.5 and −2.0 still shown for Second root and Third root come from the quotient 2x²+7x+6 built along the way; they solve that quotient, not the original cubic, since the starting guess of 2 was simply wrong.
Questions
What does a nonzero remainder actually tell me?
It tells you the value you tried is not a genuine solution of your cubic. Synthetic division only confirms a candidate when the final sum comes out to exactly 0; a remainder of 15, as with r=2 tested against 2x³+3x²−8x+3, means 2 fails the equation even though the arithmetic still produces two extra numbers along the way.
How does synthetic division pull a known solution out of a cubic?
It repeats one small step three times: multiply the running value by r, then add the next coefficient. Starting from a, then b, then c, then d, the chain produces the quotient's two coefficients plus a final remainder. When that remainder is 0, the chain has effectively divided the cubic by (x − r) with nothing left over.
Why is the quadratic formula needed here at all?
Because after one solution is divided out, a three-term equation still remains — a plain quadratic — and the quadratic formula is the direct route to both of its solutions at once. Testing r=1 against x³−6x²+11x−6 leaves x²−5x+6, and the formula turns that straight into 3 and 2 without further guessing.
Does this avoid needing Cardano's full cubic formula?
Yes, as long as one solution is already known or guessed correctly. Cardano's formula solves any cubic from scratch and involves considerably more arithmetic; this method skips all of that by leaning on a solution you already have, reducing the remaining work to one division and one quadratic formula application.
What should I do if my guessed root turns out to be wrong?
Try a different candidate. A wrong guess shows up immediately as a nonzero remainder, as it does when r=2 is tested against 2x³+3x²−8x+3 and returns 15 instead of 0. The two extra numbers reported alongside that remainder describe a different quadratic, not your original cubic, so they should not be trusted as solutions.
Where would a reasonable candidate solution come from in the first place?
Often from the rational root theorem, which lists a short set of possible whole-number or fraction solutions built from the cubic's leading and constant terms — a and d. Testing each candidate here with synthetic division is the fast way to check the list, since a genuine solution announces itself with a remainder of exactly 0.