Factoring Perfect Squares: x^8 - 1

[boxxer03]

I have a math problem that I can't figure out. We're covering Perfect Square Trinomials and Factoring Perfect Squares.

The problem is x^8-1=? or x to the 8th power minus 1 equals what?

I'm thinking it's Prime because it can't be broken down any further and 8 isn't a square number. Any other thoughts would be greatly appreciated.

 

[pka]

This what you want.
\(\displaystyle x^8 - 1 = \left( {x^4 + 1} \right)\left( {x^2 + 1} \right)\left( {x + 1} \right)\left( {x - 1} \right)\)
Nothing on the right can be factored any more.

 

[boxxer03]

Thanks a lot for your help!

 

[stapel]

Thanks a lot for your help!

Do you understand how that result was achieved? :?:

If this is being asked in the homework, you should expect to see this sort of thing on the test, so you need to understand the "how"! :wink:

Eliz.

 

[boxxer03]

Yeah, I think so. 4 + 4 = 8, so you keep the first 4 but you break the second down into 2 and x and x which equals x^4 because its division and since its a negative 1 its a difference of two squares which means in the last two results it will be the same thing but one will be a minus and one will be positive. Which is how you get (x+1)(x-1). I was just thinking it was prime because 8 isn't a square number however the implied 1 in front of the x is square and the following 1 is also square. I was just assuming that the exponent needed to be square also to be able to be broken down. Thanks you again.

 

[stapel]

Yeah, I think so. 4 + 4 = 8, so you keep the first 4 but you break the second down into 2 and....

Sorry; no.

The difference of squares formula is as follows:

. . . . .a² - b² = (a - b)(a + b)

In your case, since "1" is equal to any power of 1, you have:

. . . . .x⁸ - 1 = (x⁴)² - 1²

Apply the formula to the last form above. Then apply the formula again, to the one (new) factor to which the formula applies.

Then memorize this formula!! :wink:

Eliz.