Thursday

March 5, 2015

March 5, 2015

Posted by **Donielle** on Friday, December 16, 2011 at 11:10am.

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

- Algebra 2 -
**Steve**, Friday, December 16, 2011 at 1:15pmJust 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

**Answer this Question**

**Related Questions**

Algebra 2 - Look at the following polynomials and their factorizations: x^2-1=(x...

math - each of the natural numbers 2 through 100, inclusive, is factored in its ...

Maths - 1) Factorise x2 -6x +8 (the 2 is squared) 2) Hence solve this equation: ...

Algebra I - Multiplying Polynomials...... Do you have a question? Use the ...

math - Factor 2x+13x+40 I thought with this sort of problem you are supposed to...

algebra 2 - Factor completely with respect to the integers. 1. 9x^2 - 4 2. x^3...

algebra - Consider the following four polynomials, labeled A - D: A. ( 2x2 - 4x...

math - describe general rule for which fraction that have decimal forms that ...

factoring - can this equation be factored further? y= x^4+2x^3+4x^2+8x+16 Not in...

college-Linear Algebra - Let V={f(x)=c0 + c1x + c2x2 : 01 f(x)dx=1}. In other ...