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
 A: We assume you want real roots (complex roots are 6 exactly).
$x=1$ is one of the root of the main equation so $t=2$ is the root of the second, write
$$\displaystyle t^3+at^2+(b-3)t+1-2a=0~~;~~~t\neq0~~(x\neq-1)$$
$$(t-2)(t^2+mt+n)=0~~;~~~t\neq0$$
this concludes that $n=a-\dfrac12$ and $m=a+2$.
$t\neq0$ says $n=a-\dfrac12\neq0$ and for $t^2+mt+n=t^2+(a+2)t+(a-\dfrac12)=0$ we see
$$t_1=\dfrac{-(a+2)+\sqrt{a^2+6}}{2}~~~,~~~t_2=\dfrac{-(a+2)-\sqrt{a^2+6}}{2}$$
if $a>-\dfrac14$ then $t_2<-2$ thus $t+\dfrac1t=t_2$ has two roots.
if $a<-\dfrac52$ then $t_1>2$ thus $t+\dfrac1t=t_1$ has two roots.
for $-\dfrac52<a<-\dfrac14$ is not determined if the eqution has other root(s) or not.
A: 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 $b=\frac1{12}$.
A: This answer assumes that you want to find the minimum number of the real roots of $p(x)$.
Since $x=1$ is a root of $p(x)$, we have
$$p(1)=0\iff 2a+2b+3=0\iff b=\frac{-2a-3}{2}$$
from which we can write
$$p(x)=(x-1)^2\left(x^4+(a+2)x^3+\left(a+\frac 32\right)x^2+(a+2)x+1\right)$$
So, we see that the number of the real roots of $p(x)$ is equal to or more than $2$.
By the way, for $a=-2$, we have
$$p(x)=(x-1)^2\left(x^4-\frac 12x^2+1\right)=(x-1)^2\left(\left(x^2-\frac 14\right)^2+\frac{15}{16}\right)$$
of which $x=-1$ is not a root.
The number of the real roots of $p(x)$ for $a=-2$ is $2$.
Therefore, the minimum number of the real roots of $p(x)$ is $\color{red}{2}$.
