# Math Help - Factoring a large Diff of Squares

1. ## Factoring a large Diff of Squares

y = x^1000 - 1

I would like to factor out an x-1 from this but I don't remember which rules apply.

1) Do I apply something similar to a difference of squares and just keep going until I get (x-1)(....)?

(x^500)^2 - (1)^2 = (x^500+1)(x^500-1) = (x^500+1)(x^250+1)(x^250-1) = (x^500+1)(x^250+1)(x^125+1)(x^125-1) = now I'm stuck and I couldn't even get it to x-1 because I ran into an odd exponent ... well I'll try difference of cubes in that last factor...: = (x^500+1)(x^250+1)(x^125+1)(x^5-1)(x^10+x^5+1)... Okay, now I'm stuck, unless I remember how to factor (x^10+x^5+1)...

2) Or do I use some kind of Binomial Theorem.

Is it true that
x^1000 - 1 = (x-1)(x^999+x^998+x^997......+1) ?

3) Do I have to remember Pascal's Triangle at all?

For example, if number 2 above is true, then do I have to put in pascal's coefficients? ie.
(x-1)(1x^999 + 999x^998 + ...... + 1)?

2. Hello, krystaline86!

This is an ugly problem . . .

Factor: . $y \:= \:x^{1000} - 1$

1) Do I apply something similar to a difference of squares
and just keep going until I get: $(x-1)(\cdots)$?

$(x^{500})^2 - (1)^2 \;= \;(x^{500}+1)(x^{500}-1)$

. . . $=\;(x^{500}+1)(x^{250}+1)(x^{250}-1)$

. . . $= \;(x^{500}+1)(x^{250}+1)(x^{125}+1)(x^{125}-1)$ . . . . Good!

now I'm stuck and I couldn't even get it to x-1 because I ran into an odd exponent.

But odd exponents are usually "a good thing".
From what you wrote, I assume you are familiar with these formulas:

. . $\begin{array}{ccc}a^5 - 1 &=&(a-1)(a^4+a^3+a^2+a+1)\\ \\[-4mm]
a^5 + 1 &=&(a+1)(a^4-a^3+a^2-a+1) \end{array}$

So: . $\begin{array}{ccccccc}x^{125}-1 &=& \left(x^{25}\right)^5 - 1 &=& \left(x^{25} - 1\right)\left(x^{100} + x^{75} + x^{50} + x^{25} + 1\right) \\
x^{125} + 1 &=& \left(x^{25}\right)^5 + 1 &=& \left(x^{25} + 1\right)\left(x^{100} - x^{75} + x^{50} - x^{25} + 1\right) \end{array}$

Then: . $\begin{array}{ccccccc}x^{25}-1 &=& \left(x^5\right)^5-1 &=& (x^5-1)(x^{20} + x^{15} + x^{10} + x^5+1) \\
x^{25} + 1 &=&\left(x^5\right)^5+1 &=& (x^5+1)(x^{20} - x^{15} + x^{10} - x^5 + 1) \end{array}$

And: . $\begin{array}{ccccc}x^5 - 1 &=&(x-1)(x^4 + x^3 + x^2 + x+1) \\ x^5+1 &=& (x+1)(x^4 - x^3 + x^2 - x + 1) \end{array}$

. . See?

3. Excellent. So my idea about using the Binomial Theorem was right then?