#Someone explain vietas formula
40 messages · Page 1 of 1 (latest)
Hey @prisma forge, Thanks for sharing your math problem with us. While you wait for a Helper to help you, we want to share some vital information with you.
● Please take a moment to read the helpee guidelines. This will make sure that your post follows the helpee rules of our community.
● Please don't ping <@&775784618955505685>, <@&1283689826742440016>, <@&819616364188139550>, or <@&624327278137966593> for help because their job is to take care of the server's administrative tasks, not to answer queries directly. However, if you have a problem with how a Helper is acting, you can ping a Helper Moderator.
● It's always very useful if you can show us the work you've done so far. This makes it easier for our Helpers to find mistakes and help you get to the right answer.
someone ping me
For c, you either multiply it out or use vietas
You use vietas to find the coefficients of polynomials
Given the roots
All coefficients
-b/a = sum of all roots
Where b is the coefficient of the 2nd term
so if you had a polynomial of x^4 and had (1 + i) and (1 - i) i could find the other 2?
Why would you need vietas to find that out?
because the other 2 we dont know? therefore youd have to multiply out and divide then use quadratic formula?
oh, so if you had all 4 how would you find the coefficent of x^3,2,1 etc?
Yeah
how if you did have those 4.
Using the formulas
and for others like x^5 onwards would be different equations?
The first 3 stay the same
But it adds another one
And the last one is negated
Product of all roots
Which makes sense
Since the roots are factored as (x-a)
(-1)^n is odd if the power is odd
Please thank the helpers who assisted you by clicking the buttons below. You can thank each helper only once. Once you're done, click "Close Post" to close this thread.
Thank you for your feedback! 1significantfiguresSIGMASIGMABOY has been awarded 1
. They now have 21
. They have 3
daily left for today.