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| − | Well the derivative <math>F'(x)=x^7</math>. the only factors of this are <math>x</math>,<math>x^2</math>,<math>x^3</math>,<math>x^4</math>,<math>x^5</math>,&<math>x^6</math>. | + | Well the derivative is <math>F'(x)=x^7</math>. the only factors of this are <math>x</math>,<math>x^2</math>,<math>x^3</math>,<math>x^4</math>,<math>x^5</math>,& <math>x^6</math>. |
Since F(x) has the +1, they can never share a common factor, right? | Since F(x) has the +1, they can never share a common factor, right? | ||
| − | Unless my derivative is wrong. I took it to be in <math>Z_3</math>, so the <math>21x^20</math> term disappeared. | + | Unless my derivative is wrong. I took it to be in <math>Z_3</math>, so the <math>21x^{20}</math> term disappeared. |
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| + | I got the same derivative by using mod 3. | ||
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| + | I think you just have to show that f and f' dont have any factors in common with F | ||
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| + | -If anyone cares after the fact, you are excatly right on how to solve this problem, there are no common factors between f and its derivative. That will suffice bt thms. | ||
| + | Allen | ||
Latest revision as of 17:46, 23 November 2008
I can show that -1 isn't a multiple root, but I don't think that's really answering the question. Can anyone elaborate?
I used Theorem 20.5. Let f = the polynomial in the question and take the derivative. Prove there is no common factor between f and f prime. I'm not sure if this is correct though.
right. that's what i was using, but can you tell me how you showed that they don't have common factors?
Well the derivative is $ F'(x)=x^7 $. the only factors of this are $ x $,$ x^2 $,$ x^3 $,$ x^4 $,$ x^5 $,& $ x^6 $.
Since F(x) has the +1, they can never share a common factor, right? Unless my derivative is wrong. I took it to be in $ Z_3 $, so the $ 21x^{20} $ term disappeared.
I got the same derivative by using mod 3.
I think you just have to show that f and f' dont have any factors in common with F
-If anyone cares after the fact, you are excatly right on how to solve this problem, there are no common factors between f and its derivative. That will suffice bt thms.
Allen
