The dividend is the number you are dividing so you know it is 2 divided by something. Now to get a remainder of 2, you need the divisor to be larger than than 2, so:
2/3 works as does 2/4, 2/5,2/6,2/7 and so on
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Divide the divisor into the dividend which will result as a quotient and sometimes having a remainder
that it can go in and not having trouble in others words there r no left over
Fill in the missing powers of x in the dividend by having a coefficient of zero (0); then use long division: _________________ 6x³ - 3x² - 2x + 1 ________--------------------------------- 2x + 1 | 12x⁴ + 0x³ - 7x² + 0x + 1 _________ 12x⁴ + 6x³ _________ ------------- _______________ - 6x³ - 7x² _______________ - 6x³ - 3x² _______________ ------------- ____________________ - 4x² + 0x ____________________ - 4x² - 2x ____________________ ------------ ____________________________ 2x + 1 ____________________________ 2x + 1 ____________________________ -------- __________________________________0 ____________________________ ===== → (12x⁴ - 7x² + 1) ÷ (2x + 1) = 6x³ - 3x² - 2x + 1 with no remainder.
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how to find the missing numbers in a division math with only having the answer