# Help with factor theorem and quintic equation

• Jul 6th 2007, 05:29 PM
starswept
Help with factor theorem and quintic equation
Hi, I'm having difficulty understanding this question:
Show that x-y is a factor of x^5-y^5, then determine the other factor.

It seems obvious to me that (x-y) is a factor of (x^5-y^5), but I'm not sure how to show it as a proof with the theorem or how to determine the other factor. Any help would be greatly appreciated. Thank you in advance :)
• Jul 6th 2007, 05:54 PM
Jhevon
Quote:

Originally Posted by starswept
Hi, I'm having difficulty understanding this question:
Show that x-y is a factor of x^5-y^5, then determine the other factor.

It seems obvious to me that (x-y) is a factor of (x^5-y^5), but I'm not sure how to show it as a proof with the theorem or how to determine the other factor. Any help would be greatly appreciated. Thank you in advance :)

treating y as a constant and plugging in y into the equation we get zero. that is the easiest proof to show x - y is a factor. however, it doesn't get us anywhere after that. i'd do long division (or synthetic division, but you say you've never learned that).

do you know long division?

an even simpler way would be to use the general formula for the difference of two quintic (or whatever they're called). do you remember the formula? i know i don't
• Jul 6th 2007, 06:00 PM
starswept
Quote:

Originally Posted by Jhevon
treating y as a constant and plugging in y into the equation we get zero. that is the easiest proof to show x - y is a factor. however, it doesn't get us anywhere after that. i'd do long division (or synthetic division, but you say you've never learned that).

do you know long division?

Yes, thankfully I do know long division. My first attempt at this was doing long division with x-y as the divisor and x^5-y^5 as the dividend. That proves that (x-y) is a factor as you get 0, but it leaves a quotient of x^4, which didn't seem very useful as another factor. What should the next step be after that? And thanks again for the help :)

Oh, and sadly, I have not yet been taught the formula for the difference of two quintics. Today, my class was taught the formula for the difference between two cubes. That's all the textbook has too. So why the homework deals with quintics, I don't know ;)
• Jul 6th 2007, 06:08 PM
Jhevon
Quote:

Originally Posted by starswept
Yes, thankfully I do know long division. My first attempt at this was doing long division with x-y as the divisor and x^5-y^5 as the dividend. That proves that (x-y) is a factor as you get 0, but it leaves a quotient of x^4, which didn't seem very useful as another factor. What should the next step be after that? And thanks again for the help :)

Oh, and sadly, I have not yet been taught the formula for the difference of two quintics. Today, my class was taught the formula for the difference between two cubes. That's all the textbook has too. So why the homework deals with quintics, I don't know ;)

yeah, i'm having trouble seeing a factor for the resulting quartic, it may seem that it is never zero. thus, i would be daring and say that the new quotient (the quartic) is the other factor. it's true! if i divided the original quintic by the quartic, i would get the quotient x - y and zero as the remainder

in fact, now i am convinced that's the only possible solution. if we treat x^5 - y^5 either as a function of x or as a function of y and graphed it, the result would be the same, a straight line, meaning it will have only one real root. the resulting quartic has no real root, it is never zero
• Jul 6th 2007, 06:34 PM
starswept
Quote:

Originally Posted by Jhevon
yeah, i'm having trouble seeing a factor for the resulting quartic, it may seem that it is never zero. thus, i would be daring and say that the new quotient (the quartic) is the other factor. it's true! if i divided the original quintic by the quartic, i would get the quotient x - y and zero as the remainder

in fact, now i am convinced that's the only possible solution. if we treat x^5 - y^5 either as a function of x or as a function of y and graphed it, the result would be the same, a straight line, meaning it will have only one real root. the resulting quartic has no real root, it is never zero

Thank you, good to know I was on the right track. Strangely, the back of the book lists the other factor as (x^4+x^3y+x^2y^2+xy^3+y^4). I have no idea what they've done to the quartic there, unless they've expanded it in some way?
• Jul 6th 2007, 06:38 PM
Jhevon
Quote:

Originally Posted by starswept
Thank you, good to know I was on the right track. Strangely, the back of the book lists the other factor as (x^4+x^3y+x^2y^2+xy^3+y^4). I have no idea what they've done to the quartic there, unless they've expanded it in some way?

they did nothing, that's exactly the quartic i got. are you sure YOU did the long division correctly?
• Jul 6th 2007, 06:45 PM
starswept
Quote:

Originally Posted by Jhevon
they did nothing, that's exactly the quartic i got. are you sure YOU did the long division correctly?

Clearly I must not have :o I'm sure I've already asked you for way too much help today already, but if you don't mind, could you show me how you did the division (not that I should need that at this level, but alas)? Thanks again.
• Jul 6th 2007, 07:07 PM
Jhevon
Quote:

Originally Posted by starswept
Clearly I must not have :o I'm sure I've already asked you for way too much help today already, but if you don't mind, could you show me how you did the division (not that I should need that at this level, but alas)? Thanks again.

first, you haven't asked me too many questions. secondly, long division of polynomials is something i use to this day, and i'm past calc 3 when it comes to math, so you should know it, and never forget it. third, long division is a pain to type and explain here.

try this site and see if you understand

one thing this site doesn't seem to mention is that you have to leave spaces for all the powers even if they are not present

so you can't write:

Code:

```      ___________ x - y ) x^5 - y^5```
that leaves no space for the x^4, x^3, x^2 and x^1 terms, you must write it like this:

Code:

```      ___________________________________ x - y ) x^5                      - y^5```
if it helps you to visualize better, write zeroes to hold the spaces:

Code:

```      _________________________________________________ x - y ) x^5  + 0x^4  + 0x^3  + 0x^2  + 0x  - y^5```
• Jul 6th 2007, 07:12 PM
starswept
Quote:

Originally Posted by Jhevon
first, you haven't asked me too many questions. secondly, long division of polynomials is something i use to this day, and i'm past calc 3 when it comes to math, so you should know it, and never forget it. third, long division is a pain to type and explain here.

try this site and see if you understand

When I said not that I should need it, I didn't mean that I shouldn't need to be using long division in calculus, just that I shouldn't need help just to do long division (although I did just learn long division of polynomials this morning, which was my first time using long division since grade three, so I'll blame it on being rusty). I'll look over it again and see if I can get it, I think it's having the two variables in the dividend that's messing me up. And I hadn't thought of putting placeholders, I should definitely do that. Thanks :)
• Jul 6th 2007, 07:13 PM
Jhevon
Quote:

Originally Posted by starswept
When I said not that I should need it, I didn't mean that I shouldn't need to be using long division, just that I shouldn't need help just to do long division (although I did just learn long division of polynomials this morning, so who knows). I'll look over it again and see if I can get it, I think it's having the two variables in the dividend that's messing me up.

ok. don't let it mess you up. think of the y as a constant
• Jul 6th 2007, 07:24 PM
starswept
Alright, I finally got it :) I think I just wasn't used to having another variable there and got stumped. Thank you so much for your patience and all the help!