<s>[INST] Values of $a$ for which $p(x)$ has a complex roots? Let $p(x) = x^4 + ax^3 + bx^2 + ax + 1$. Suppose $x = 1$ is a root of $p(x)$, then find the range of values of $a$ such that $p(x)$ has complex (non-real) roots?
My approach: Using the fact that $1$ is a root, I am able to deduce $b = -2(a+1)$. Then I wrote $$p(x) = (x-1)(x-z)(x-\bar{z})(x-k)$$ where $z = c + id, d\neq 0$ and $k$ is some other real roots. Now I plan to expand and compare the coefficients but have been unable to do so in a fruitful manner (there just too many variables to get to a proper estimate of range of $a$ )
 [/INST] When a polynomial is "palindromic", reading the same way forwards and backwards, it must also have palindromic factors.  Thus
$x^4+ax^3+bx^2+ax+1=(x^2+cx+1)(x^2+dx+1)$
If $x=1$ is to be a root then it must give $c=-2$ or $d=-2$ in one of the factors, as no other choice has the root $x=1$.  Taking $c$ as $-2$ we then get the complex roots if $-2<d<2$, corresponding to a negative discriminate for the $x^2+dx+1$ factor.
Expanding the product $(x^2+cx+1)(x^2+dx+1)$ and equating this to $x^4+ax^3+bx^2+ax+1$ then implies $a=c+d=d-2$ so the solution to the problem is $-4<a<0$.
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