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  2. Long division - Wikipedia

    en.wikipedia.org/wiki/Long_division

    If necessary, simplify the long division problem by moving the decimals of the divisor and dividend by the same number of decimal places, to the right (or to the left), so that the decimal of the divisor is to the right of the last digit. When doing long division, keep the numbers lined up straight from top to bottom under the tableau.

  3. Division (mathematics) - Wikipedia

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

    This is denoted as 20 / 5 = 4, or ⁠ 20 / 5 ⁠ = 4. [2] In the example, 20 is the dividend, 5 is the divisor, and 4 is the quotient. Unlike the other basic operations, when dividing natural numbers there is sometimes a remainder that will not go evenly into the dividend; for example, 10 / 3 leaves a remainder of 1, as 10 is not a multiple of 3.

  4. Division algorithm - Wikipedia

    en.wikipedia.org/wiki/Division_algorithm

    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.

  5. Short division - Wikipedia

    en.wikipedia.org/wiki/Short_division

    The first number to be divided by the divisor (4) is the partial dividend (9). One writes the integer part of the result (2) above the division bar over the leftmost digit of the dividend, and one writes the remainder (1) as a small digit above and to the right of the partial dividend (9).

  6. Third grade - Wikipedia

    en.wikipedia.org/wiki/Third_grade

    Third grade (also 3rd Grade or Grade 3) is the third year of formal or compulsory education. It is the third year of primary school . Children in third grade are usually 8–9 years old.

  7. 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.

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