what is the sum of the roots of the polynomial? x^3+2x^2-11x-12

Question

yttrium · Answer

where are you stocked?

anonymous · Answer

well I used synthetic division to solve this problem like someone else suggested, but when I did the answer was wrong so I'm confused on how to solve it, or maybe there was a error in my process... but other then that I don't really know.

hartnn · Answer

sum of roots of a cubic equation, 
$\Large ax^3+bx^2+cx+d= 0$
is just -b/a

yttrium · Answer

Oh really? @hartnn 
How about if there's no x^2 term? Is it zero?

hartnn · Answer

then b=0
sum = 0

yttrium · Answer

Ohh thanks! :))
How about the sum of roots of other degrees? Are there any shortcut?

hartnn · Answer

yes

hartnn · Answer

http://www.mathsisfun.com/algebra/polynomials-sums-products-roots.html

hartnn · Answer

$\Large ax^n +bx^{n-1}+..... = 0$
sum of roots always = -b/a

anonymous · Answer

https://www.artofproblemsolving.com/Wiki/index.php/Vieta's_formulas

yttrium · Answer

Ohhh awesome! :) Thanks

whpalmer4 · Answer

Cool!  That's today's installment of "learn something new every day on OpenStudy"...

hartnn · Answer

$Roots : p,q,r \ ax^3+bx^2+cx+d=0 \ p+q+r = -b/a \ pq+qr+pr = c/a\ pqr = - d/a$

anonymous · Answer

lol thanks to everyone :)

whpalmer4 · Answer

$$pqr = -d/a$$is obvious enough, but the others might take a bit of rumination...

hartnn · Answer

$Roots : p,q,r,s \ ax^4+bx^3+cx^2+dx+e=0 \ p+q+r+s = -b/a \ (roots ~taken ~2~at~a~time) = c/a\  (roots ~taken ~3~at~a~time) = - d/a \  (roots ~taken ~4~at~a~time) = pqrs = e/a $

hartnn · Answer

in general, such pattern is true for any value of n

yttrium · Answer

what does roots taken 2 at a time means??

hartnn · Answer

pq+pr+ps+qr+qs+rs

yttrium · Answer

Ohh okay.