While tutoring Algebra, division of polynomials, I ran across those problems with remainders. As an example: (x^2 + 2x + 1)/(x-7) the result is x + 9 with a remainder of 64/(x-7). I then assumed that the remainder was a positive integer and set x-7 equal to the divisors of 64 and solved for x.(adsbygoogle = window.adsbygoogle || []).push({});

x - 7 = 1, x = 8 then the original polynomial became 64+16+1, or 81 of course is divisible by 1.

x - 7 = 2, x = 9, 81+18+1 = 100, 2(50)

x - 7 = 4, x = 11, 121+22+1 = 144, 4(36)

x - 7 = 8, x = 15, 225+30+1 = 256, 8(32)

x - 7 = 16, x = 23, 529+46+1 = 576, 16(36)

x - 7 = 32, x = 39, 1521+78+1 = 1600, 32(50)

x - 7 = 64, x = 71, 5041+142+1 = 5184, 64(81)

I thought this was interesting, however, I could not think of a problem where I could apply this solution. It may work for solving for different bases, etc.

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