Values of $a$ for which $p(x)$ has a complex roots?

Question

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 $a$ )

Answer 1

HINT:

$$p(x)=(x-1)(x^3+(a+1)x^2-(a+1)x-1)$$

Rest follows very nicely.

Answer 2

Here's a different approach. (I can't say for sure whether you can or cannot do it by comparing coefficients though!)

For a degree 2 polynomial, we know that it has complex roots (assuming the coefficients are real) iff the discriminant is negative. There's a similar generalisation for high degree polynomial.

In particular, a degree 4 polynomial $$ax^4+bx^3+cx^2+dx+e$$ has discriminant $$256a^3e^3-192a^2bde^2-128a^2c^2e^2+144a^2d^2e...$$

(see https://en.wikipedia.org/wiki/Discriminant#Degree_4 for the full formula). And we know that since $x=1$ is a root there can only be a pair of complex roots so we require the discriminant to be negative.

Answer 3

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 $-4<a<0$.

Answer 4

Considering that $$p(x) = x^4 + ax^3 + bx^2 + ax + 1$$ If $x=1$ is a root, then $$p(1)=2 a+b+2=0 \implies b=-2-2a$$ and $$p(x)=(x-1)\,(x^3+(a+1) x^2-(a+1) x-1)$$ Clearly, $x=1$ is also a root for the second term making $$p(x)=(x-1)^2\,(x^2+(a+2) x+1)$$ So, there are complex roots if the discriminant $\Delta$ of the remaining quadratic equation is negative.