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  2. Remainder - Wikipedia

    en.wikipedia.org/wiki/Remainder

    In this case, s is called the least absolute remainder. [3] As with the quotient and remainder, k and s are uniquely determined, except in the case where d = 2n and s = ± n. For this exception, we have: a = k⋅d + n = (k + 1)d − n. A unique remainder can be obtained in this case by some convention—such as always taking the positive value ...

  3. Synthetic division - Wikipedia

    en.wikipedia.org/wiki/Synthetic_division

    In algebra, synthetic division is ... is the remainder while the rest correspond to the coefficients of the quotient. ... x**2 + 3*x + 5 will be represented as [1, 3 ...

  4. Polynomial remainder theorem - Wikipedia

    en.wikipedia.org/wiki/Polynomial_remainder_theorem

    In algebra, the polynomial remainder theorem or little Bézout's theorem (named after Étienne Bézout) [1] is an application of Euclidean division of polynomials.It states that, for every number , any polynomial is the sum of () and the product by of a polynomial in of degree less than the degree of .

  5. Polynomial long division - Wikipedia

    en.wikipedia.org/wiki/Polynomial_long_division

    x 3 has been divided leaving no remainder, and can therefore be marked as used by crossing it out. The result x 2 is then multiplied by the second term in the divisor −3 = −3x 2. Determine the partial remainder by subtracting −2x 2 − (−3x 2) = x 2. Mark −2x 2 as used and place the new remainder x 2 above it.

  6. Division (mathematics) - Wikipedia

    en.wikipedia.org/wiki/Division_(mathematics)

    Sometimes this remainder is added to the quotient as a fractional part, so 10 / 3 is equal to ⁠3 + 1 / 3 ⁠ or 3.33..., but in the context of integer division, where numbers have no fractional part, the remainder is kept separately (or exceptionally, discarded or rounded). [5] When the remainder is kept as a fraction, it leads to a rational ...

  7. Chinese remainder theorem - Wikipedia

    en.wikipedia.org/wiki/Chinese_remainder_theorem

    Then one can proceed by adding 20 = 5 × 4 at each step, and computing only the remainders by 3. This gives 4 mod 4 → 0. Continue 4 + 5 = 9 mod 4 →1. Continue 9 + 5 = 14 mod 4 → 2. Continue 14 + 5 = 19 mod 4 → 3. OK, continue by considering remainders modulo 3 and adding 5 × 4 = 20 each time 19 mod 31. Continue 19 + 20 = 39 mod 3 ...