Math Help Forum

Math Help Forum Feed Site Feed

Go Back   Math Help Forum > University Math Help > Linear and Abstract Algebra
Reply
 
Thread Tools Display Modes
  #1  
Old November 7th, 2009, 07:10 AM
Newbie
 
Join Date: Nov 2009
Posts: 4
Country:
Thanks: 3
Thanked 0 Times in 0 Posts
ohadch is on a distinguished road
Default problem

consider the equation 2z^7-5z^5+z-9=0 .
The equation has 7 zeros. Find the product of these Zeros and their sum .
Reply With Quote
Advertisement
 
  #2  
Old November 7th, 2009, 10:25 AM
MHF Contributor
 
Join Date: Oct 2009
Posts: 1,164
Thanks: 52
Thanked 399 Times in 377 Posts
tonio is just really nicetonio is just really nicetonio is just really nicetonio is just really nicetonio is just really nice
Default

Quote:
Originally Posted by ohadch View Post
consider the equation 2z^7-5z^5+z-9=0 .
The equation has 7 zeros. Find the product of these Zeros and their sum .

For any polynomial a_0 + a_1x+...+a_nx^n, Vieta's formulas (which are usually studied in high school but mainly for the quadratic equation. Their generalization for any degree is a nice exercise in symmetric polynomials) tell us that the sum of its roots equals -\frac{a_{n-1}}{a_n}, and their product equals (-1)^n\frac{a_0}{a_n}

Tonio
Reply With Quote
The following users thank tonio for this useful post:
Donate to MHF
Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off
Forum Jump


All times are GMT -7. The time now is 04:51 PM.


Powered by vBulletin® Version 3.7.3
Copyright ©2000 - 2009, Jelsoft Enterprises Ltd.
SEO by vBSEO 3.2.0 ©2008, Crawlability, Inc.
©2005 - 2009 Math Help Forum


Math Help Forum is a community of maths forums with an emphasis on maths help in all levels of mathematics.
Register to post your math questions or just hang out and try some of our math games or visit the arcade.