<s>[INST] then minimum number of number of roots of $p(x) = 0$ is let $p(x)=x^6+ax^5+bx^4+x^3+bx^2+ax+1.$ given that $x=1$ is a one rot of $p(x)=0$ 
and $-1$ is not a root. then minimum number of number of roots of $p(x) = 0$ is 
Attempt: $x=0$ in not a root of $p(x)=0.$ 
So $\displaystyle \left(x^3+\frac{1}{x^3}\right)+a\left(x^2+\frac{1}{x^2}\right)+b\left(x+\frac{1}{x}\right)+1=0$
So $\displaystyle \left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)+a\left(x+\frac{1}{x}\right)^2-2+b\left(x+\frac{1}{x}\right)+1=0$
So $\displaystyle t^3+at^2+(b-3)t+1=0,$ where $\displaystyle \left(x+\frac{1}{x}\right) = t$ and $|t|\geq 2$
Could some help me to solve it, thanks
 [/INST] The minimum number of real roots is $2$ counted with multiplicity, or $1$ counted without multiplicity.
Since $p(x)$ is a reciprocal polynomial (that is, $x^6p(\frac1x)=p(x)$), its roots come in pairs $r,\frac1r$. In particular, if $1$ is a root of $p(x)$ then it is a double root. It is possible for these to be the only roots: take for example, $a=-\frac{19}{12}$ and $b=\frac1{12}$.
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