Finding integer values by matching polynomial remainders to divisors

[coolul007]

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.
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.

 

[cool_jessica]

This is tricky method to solve, yes definitely this method is going to use to find the value of X, this is sort cut method to solve this problem of finding the value of x

 

[ramsey2879]

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.
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.

You would get similar results if you chose x - 2^n + 1 in lieu of x - 7. Or try other negative prime powers plus 1.