Whether this statement is true or false

[an_mui]

'x - 1' is a factor of a polynomial in the form (x^n - 1) where 'n' is a positive integer.

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.

 

[toam]

(x^n - 1) = (x^(n/2) -1) (x^(n/2) - 1)
= (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.

 

[Tide]

Your logic is faulty. Think about what your statement means when n is an odd integer and what is meant by factoring a polynomial.

 

[an_mui]

oops i think i know now

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

 

[Tide]

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

 

[shmoe]

To help with the long division, you might want to try it in the special cases where n=2, 3, 4, ... or as many as needed before you see a pattern. Then try to prove this pattern works for a general n.

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

Or you can do both and be even more convinced.

 

[HallsofIvy]

More simply, when x= 1, xⁿ- 1 becomes 1-1= 0. Therefore, x- 1 is a factor (I just noticed that shmoe referred to the "remainder theorem"- that's what this is).