Question

anyone?
\[f(x)=x^4+2x^3-4x^2-x+1\] let a,b,c,d be the roots of this ab, ac, ad, bc, bd, cd are roots of polynomial g(x) what is g(1) if g(0)=1

Answer 1

this is what i tried so far, nothing came to mind
http://openstudy.com/study?login#/updates/4f4d8bf0e4b019d0ebadd6e6

Answer 2

hello zarkon
thinking this would work out with vieta's formulas somehow but i cannot get it

Answer 3

i tried a simpler example and nothing stood out. the way i understand it is we want the sum of the coefficients, but when i tried multipling out
\[(1-ab)(1-ac)(1-ad)(1-bc)(1-bd)(1-cd)\] using worlfram, i thought i would see some nice pattern that would relate back to the elementary symmetric functions but it did not seem to work out that way

Answer 4

I don't see a nice way to do this. I'll think about it though.

Answer 5

I think ive figured out all the coefficients of g except for the x^3 and x^2. Its a real pain though, i dont know if I want to continue =/

Answer 6

-9

Answer 7

-9
c'mon now give me something to work with...

Answer 8

hello joe!

Answer 9

I'm sure there has to be a nicer way. I'm just doing a ground and pound....long tedious calculations using what we know about f and g. I get

\[g(x)=x^6+4x^5-3x^4-13x^3-3x^2+4x+1\]

I'm betting there is a slick way of doing this without capturing all the the coefficients of g. I just don't see it right now.

Answer 10

so i am assuming you found all the zeros of f somehow and then multiplied to get this?

Answer 11

Initially yes...but I have since determined the coefficients of g bases on the expansion of (x-a)(x-b)(x-c)(x-d) and abcd=1.( though I used my calculator to do all the expanding and factoring) The process is quite inelegant and I'm hoping that someone else has a more sophisticated solution.

Answer 12

where is a -9 in
\[f(x)=x^4+2x^3-4x^2-x+1\]
minus the sum of the absolute value of the coefficients?

Answer 13

I don't know.