OpenStudy (darkprince14):

Abstract algebra.

2 years ago
OpenStudy (darkprince14):

2 years ago
OpenStudy (amistre64):

can you define a field?

2 years ago
OpenStudy (darkprince14):

a field is commutative division ring

2 years ago
OpenStudy (amistre64):

ive got no clear thought on this. my class work never got past groups. so i lack any practical experience in the matter.

2 years ago
OpenStudy (misty1212):

HI!!

2 years ago
OpenStudy (misty1212):

i have a suggestion try it with a simple polynomial of degree say 2 and see why it works

2 years ago
OpenStudy (misty1212):

if you want a bigger hint i can provide one, but i think it will pretty much give it away

2 years ago
OpenStudy (amistre64):

already considered that, too much trouble :)

2 years ago
OpenStudy (amistre64):

half a degree is all im willing to process at the moment

2 years ago
OpenStudy (darkprince14):

Is it okay if i divide f(a)by a^n?

2 years ago
OpenStudy (misty1212):

first off the notation is a bit odd, because they use $$a_i$$ for the coefficients, and $$a$$ for the zero

2 years ago
OpenStudy (misty1212):

so lets use $$b$$ and you see that $a_0+a_1b+a_2b^2+...+a_nb^n=0$ the see what you get when you multiply this by $$b^{-n}$$ which i guess is another way of saying "divide by $$b^n$$"

2 years ago
OpenStudy (misty1212):

half a degree is better than none! it is what i have after all...

2 years ago
OpenStudy (darkprince14):

thanks:)

2 years ago
OpenStudy (misty1212):

$\color\magenta\heartsuit$

2 years ago
OpenStudy (anonymous):

consider $$g(x)=x^{-n}f(x^{-1})$$ hence $$g(a)=0\implies a^{-n}f(a^{-1})=0$$ and using the fact $$a\ne0$$ we conclude $$f(a^{-1})=0$$. so now just work backwards

2 years ago
OpenStudy (anonymous):

err $$g(x)=x^n f(x^{-1})$$ and so $$a^n f(a^{-1})=0$$

2 years ago