Given that $x=1$ is the root of $f(x)=0$ where
$$f(x)=x^6+a_1x^5+a_2x^4+x^3+a_2x^2+a_2x+1$$ and also given that $f(x+1) \ne 0$
Find Maximum number of distinct real roots $f(x)=0$ can have?
My Try: Given $x=1$ is root we have
$a_1+a_2=\frac{-3}{2}$
Any further clue?
Since the first coefficient and the constant are 1, the product of all factors must multiply out to 1 for both of them. Sign does not matter as long as they come in pairs to yield a net positive 1 on both ends of the polynomial.
Here is a made up example of 6 roots 1/5, 5, 1/4, 4, 1/2, 2
Granted that 1 is a possible real root but then you would need a -1 to match it and you would need to drop one of my example pairs and have a mismatched sign in one of the two remaining pairs.
While we can have only 6 real roots at a time, the max number or roots possible is countably infinite. For example the roots of
$x^6+121/12 x^5+289/8 x^4+667/12 x^3+289/8 x^2+121/12 x+1$
happen to be [negative]
2, 1/2, 3, 1/3, 4, 1/4 but they could be any combination of complimentary rational numbers