This is inspired by [this old question](http://mathoverflow.net/questions/219315/aligned-roots-of-irreducible-polynomials), which may provide a bit more background. But the two present questions seem somewhat more fundamental to me.


Let $p$ be an irreducible polynomial with integer coefficients.



> 
> 1. Is it possible that three of the roots of $p$ are collinear on a line in "general position", i.e. which is neither horizontal nor vertical nor through the origin?
> 
> 
> 


Further, for $\alpha\in\mathbb R$, denote by $N\_p(\alpha)$ the number of zeros of $p$ with real part $\alpha$. 



> 
> 2. If $N\_p(\alpha)>1$, is it true that $N\_p(\alpha)$ is always a power of $2$?
> 
> 
> 



Collinear Galois conjugates
The answer to Q1 is yes. For example, $p(x) = x^6 + 45x^4 + 122x^3 + 504x^2 + 1740x + 2213$ is a polynomial with three roots on the line $y = 2x+3$ (and the other three on the line $y = -2x-3$). Take your favorite irreducible cubic with real roots $f(x)$ (mine is $x^{3} - 3x + 1$) and let $\alpha\_{1}$, $\alpha\_{2}$, $\alpha\_{3}$ be the roots. Now, choose rational numbers $m$ and $b$ - we'll make a degree 6 polynomial whose roots (thought of in the form $x + iy$) lie on the line $y = mx + b$. (I chose $m = 2$ and $b = 3$.) Let $p(x) = f\left(\frac{x-bi}{1+mi}\right) f\left(\frac{x+bi}{1-mi}\right)$. If $z$ is a root of $p(x)$ coming from the first factor, then $\frac{z-bi}{1+mi} = \alpha$ for one of the real roots $\alpha$ of the cubic $f(x)$ and then $z = \alpha + (m \alpha + b)i$ lies on the line $y = mx+b$.


The answer to Q2 is no. You can take an irreducible cubic $f(x)$ with all roots real and negative. Then, $f(x^{2})$ can be an irreducible polynomial with all imaginary roots and so $N\_{p}(0) = 6$. (For example, you can take $f(x) = x^{3} + 6x^{2} + 9x + 3$. Then $f(x^{2})$ is irreducible by Eisenstein's criterion.)


