 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Sum Root Value (Posted on 2016-03-05) Consider the polynomials:
R(x) = x6 – x5 – x3 – x2 - x and:
S(x) = x4 – x3 – x2 - 1
z1, z2, z3, and z4 are roots of S(x) = 0
Determine R(z1) + R(z2) + R(z3) + R(z4).

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Some thoughts Comment 2 of 2 | I was unable to finish but I thought I'd share what I started:

Dividing R(x) by S(x)

Gives R(x) = S(x)*(x^2+1) + (x^2-x+1)
Call this remainder T(x).
Then for each zn it must be that R(zn)=T(zn)

[looking at Brian's solution, which I don't fully understand, this substitution may make things simpler.]

It may not be helpful to try going further and divide S(x) by T(x) but this gives S(x) = T(x)*(x^2-2)+(-2x+1)
or
T(x) = [S(x)-(-2x+1)]/(x^2-2)
Then for each zn is must be R(zn)=(2zn-2)/(zn^2-2)
Which gives another way of rewriting R(zn) but is more complicated.

 Posted by Jer on 2016-03-05 12:02:16 Please log in:
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