- #1

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my guess:

This statement is always true because (x^n - 1) is a difference of square. When factored even more, (x^n - 1) = (x^n/2 - 1)(x^n/2 + 1). Therefore, (x-1) can be factor of (x^n - 1) and is always true.

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- Thread starter an_mui
- Start date

- #1

- 47

- 0

my guess:

This statement is always true because (x^n - 1) is a difference of square. When factored even more, (x^n - 1) = (x^n/2 - 1)(x^n/2 + 1). Therefore, (x-1) can be factor of (x^n - 1) and is always true.

- #2

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= (x^(n/4) - 1)(x^(n/4) + 1)(x^(n/2) + 1)

and so on, where the power of x is n/(2^k).

Consider the case where n is odd, dividing n by 2^k will never equal 1.

- #3

Tide

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- #4

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(x-1)(x^(n-1) + 1) =(x^n - 1), where n is a positive integer greater than or equal to 2

- #5

Tide

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Not quite! I suggest trying long division! If there is no remainder then x-1 is a factor. :)

- #6

shmoe

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Or you can avoid long division by using the remainder theorem.

Or you can do both and be even more convinced.

- #7

HallsofIvy

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