Problem solved in full
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Dividing 2x³ − 3x² + 4x − 5 by x − 2 5 steps
Dividing 2x³ − 3x² + 4x − 5 by x − 2 leaves a remainder of 7. Get there by synthetic division, and then obtain the same 7 without dividing at all.
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Synthetic division is long division with the bookkeeping removed. Only the coefficients matter, and dividing by x − r means the multiplier is r.
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Bring down the 2, multiply by 2 and add to −3 to get 1, multiply by 2 and add to 4 to get 6, multiply by 2 and add to −5 to get 7. The first three are the quotient; the last is the remainder.
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Written out, that is the division algorithm for polynomials — and the identity holds for every x, not just convenient ones.
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So substitute x = 2. The (x − 2) factor annihilates the entire quotient, whatever it happens to be, and the identity collapses to P(2) = the remainder.
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Evaluating P(2) directly gives 7, matching the division without any division having been done.
Answer
The tool prints a quotient of 2x² + x + 6 and a remainder of 7. The remainder theorem is what step 4 proves, and it is worth more than the shortcut it provides: because the remainder is P(r), asking whether (x − r) divides a polynomial is the same as asking whether P(r) = 0. Here it is 7, so it does not. That equivalence turns factor-hunting into evaluation — try divisor values until the remainder reads 0, and you have found a root and factored the polynomial in the same step.
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Learning path
Beyond the quadratic
References (1)
- Insight block 3 — the CRC as polynomial division over GF(2): W. W. Peterson and D. T. Brown, "Cyclic Codes for Error Detection." Proceedings of the IRE 49(1), 228–235, 1961.