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Home/ Questions/Q 7933319
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Editorial Team
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Editorial Team
Asked: June 3, 20262026-06-03T21:17:43+00:00 2026-06-03T21:17:43+00:00

I remember reading somewhere (maybe someone can help remember where), that there is a

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I remember reading somewhere (maybe someone can help remember where), that there is a method that is the fastest for evaluating a polynomial. Something reminds me that it had something to do with Vietta’s formula, or the fact that the 0-power coefficient is the product of the 0-power coefficients of any factors of the polynomial.

I know wikipedia says it’s Horner’s scheme for evaluating fastest. But I recall that you actually did not have to evaluate in that way at all – it had something with the roots?

All I know for sure is that there was a method for evaluating a polynomial that has gives you a “oh that is clever” kind of feeling when you see it, but it’s not too difficult and is kind of obvious.

Anyone kind or smart enough to help me out?

It is something along the lines of “you can evaluate P at x by … ” and then there is a really simple little thing that actually avoid having to do any real additions and multiplications on the order of the polynomial degree.

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  1. Editorial Team
    Editorial Team
    2026-06-03T21:17:45+00:00Added an answer on June 3, 2026 at 9:17 pm

    Are you evaluating the polynomial more than once? Is the polynomial particularly simple? Consider the following polynomial:

    f(a) = a^(14)
    

    If we want to reduce the number of multiplications required to evaluate f(a) we can compute the minimal addition chain from addition-chain exponentiation:

    ((a × a→b) × b→d) × d × d × b
    

    Which shows we can compute the f(a) using only 5 multiplications. For a fixed polynomial with small coefficients this can be signifigant savings. Wikipeida notes:

    In practice … shortest addition-chain exponentiation is primarily used for small fixed exponents for which a shortest chain can be precomputed and is not too large.

    For many real-world cases where f(a) can vary another method may be appropriate, but it’s worth noting alternate solutions!

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