A worked example: x³ − 6x² + 11x − 6 at x = 4
Enter coefficients "1, -6, 11, -6" (representing x³ − 6x² + 11x − 6) with x = 4 into the Polynomial Calculator, and it returns result = 6 with degree = 3. That matches direct substitution: 4³ − 6(4²) + 11(4) − 6 = 64 − 96 + 44 − 6 = 6.
The formula: Horner's method step by step
Rather than computing each power of x separately, the calculator evaluates the polynomial with Horner's method, which repeatedly multiplies the running total by x and adds the next coefficient. For coefficients "2, 0, -8" (representing 2x² − 8) at x = 3, the steps are: start at 0, then 0 × 3 + 2 = 2, then 2 × 3 + 0 = 6, then 6 × 3 + (-8) = 10. The calculator's own output confirms this: result = 10, degree = 2.
Another example: x² − 3x + 2 at x = 10
For coefficients "1, -3, 2" (representing x² − 3x + 2) evaluated at x = 10, the calculator returns result = 72 with degree = 2. Checking by direct substitution: 10² − 3(10) + 2 = 100 − 30 + 2 = 72, which matches exactly.
Why entering coefficients in the right order matters
The calculator reads coefficients from the highest-degree term down to the constant, so a missing term still needs an explicit 0 placeholder — otherwise every coefficient after the gap shifts to the wrong power of x and the whole evaluation comes out wrong, even though no error is thrown. Getting the order and the placeholders right matters more for the final answer than the evaluation method itself.