There exists a polynomial of the form
x^6 + ax + b
that has the same set of real roots as the polynomial
x^2  2x  1
Find a+b
if x^6+ax+b has the same roots of x^22x1 then x^22x1 is a factor of x^6+ax+b
thus
x^6+ax+b=(x^22x1)(x^4+px^3+qx^2+rx+s)
for some values of p,q,r,s
expanding and collecting on terms of x we get
x^6+ax+b=x^6+(p2)x^5+(q2p1)x^4+(r2qp)x^3+(s2rq)x^2+(2sr)xs
from the x^5 terms we have
p2=0 giving p=2
from the x^4 terms we have
q2p1=0
q5=0
q=5
from the x^3 terms we have
r2qp=0
r12=0
r=12
from the x^2 terms we have
s2rq=0
s29=0
s=29
this gives us
a=2sr=70
b=s=29
thus we have
x^6+ax+b=x^670x29=(x^22x1)(x^4+2x^3+5x^2+12x+29)
and the quartic factor has no real roots
thus x^670x29 has the same real roots as x^22x1
thus
a+b=7029=99=99

Posted by Daniel
on 20161210 09:04:05 