Asked by Donielle
Look at the following polynomials and their factorizations:
x^2-1=(x-1)(x+1)
x^3-1=(x-1)(x^2+x+1)
x^4-1=(x-1)(x^3+x^2+x+1)
In general, how can x^n-1 be factored. Show that this factorization works by multiplying the factors
x^2-1=(x-1)(x+1)
x^3-1=(x-1)(x^2+x+1)
x^4-1=(x-1)(x^3+x^2+x+1)
In general, how can x^n-1 be factored. Show that this factorization works by multiplying the factors
Answers
Answered by
Steve
Just do a long multiplication and line up the like exponents:
You'll notice that all the exponents have a plus entry and a minus entry, except the highest and lowest:
x^n + x^n-1 + x^n-2 + ... + x^2 + x^1
000 - x^n-1 - x^n-2 - ... - x^2 - x^1 - 1
-----------------------------------------
x^n - 1
You'll notice that all the exponents have a plus entry and a minus entry, except the highest and lowest:
x^n + x^n-1 + x^n-2 + ... + x^2 + x^1
000 - x^n-1 - x^n-2 - ... - x^2 - x^1 - 1
-----------------------------------------
x^n - 1
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