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The polynomial remainder theorem may be used to evaluate () by calculating the remainder, . Although polynomial long division is more difficult than evaluating the function itself, synthetic division is computationally easier. Thus, the function may be more "cheaply" evaluated using synthetic division and the polynomial remainder theorem.
The computation of the quotient and the remainder from the dividend and the divisor is called division, or in case of ambiguity, Euclidean division. The theorem is frequently referred to as the division algorithm (although it is a theorem and not an algorithm), because its proof as given below lends itself to a simple division algorithm for ...
Long division is the standard algorithm used for pen-and-paper division of multi-digit numbers expressed in decimal notation. It shifts gradually from the left to the right end of the dividend, subtracting the largest possible multiple of the divisor (at the digit level) at each stage; the multiples then become the digits of the quotient, and the final difference is then the remainder.
The above form of synthetic division is useful in the context of the polynomial remainder theorem for evaluating univariate polynomials. To summarize, the value of p ( x ) {\displaystyle p(x)} at a {\displaystyle a} is equal to the remainder of the division of p ( x ) {\displaystyle p(x)} by x − a . {\displaystyle x-a.}
Ruffini's rule can be used when one needs the quotient of a polynomial P by a binomial of the form . (When one needs only the remainder, the polynomial remainder theorem provides a simpler method.) A typical example, where one needs the quotient, is the factorization of a polynomial p ( x ) {\displaystyle p(x)} for which one knows a root r :
The rings for which such a theorem exists are called Euclidean domains, but in this generality, uniqueness of the quotient and remainder is not guaranteed. [8] Polynomial division leads to a result known as the polynomial remainder theorem: If a polynomial f(x) is divided by x − k, the remainder is the constant r = f(k). [9] [10]
Divide the highest term of the remainder by the highest term of the divisor (3x ÷ x = 3). Place the result (+3) below the bar. 3x has been divided leaving no remainder, and can therefore be marked as used. The result 3 is then multiplied by the second term in the divisor −3 = −9. Determine the partial remainder by subtracting −4 − (− ...
The greatest common divisor is the last non zero entry, 2 in the column "remainder". The computation stops at row 6, because the remainder in it is 0. Bézout coefficients appear in the last two columns of the second-to-last row. In fact, it is easy to verify that −9 × 240 + 47 × 46 = 2.