Vtyjkyuk

Let $p(x)$ be a real coefficient polynomial. Assume that there exits $ain mathbbR$ s.t. $p(a)neq 0$ but $p'(a) = p''(a) = 0$. Show that $p(x)$ has at least one non-real root.

This is a problem from today's exam, and actually, I already have a solution. However, I can't understand the solution, so I tried to do it my self in the following direction:

Assume that $p(x)$ is a monic polynomial. We can just set $p(x) = (x-a)^3q(x) + B$ for some $Bneq 0$. If $p$ has only real roots, we have $p(x) = (x-alpha_1)cdots (x-alpha_n)$ for some $alpha_1leq cdots leq alpha_n$. Then we have $(x-a)^3q(x)+B = (x-alpha_1)cdots(x-alpha_n)$, and I can' proceed from here.

Here's the original solution:

Solution:
Observe that if $q(z)$ is a real-rooted polynomial with distinct roots, then by Rolle’s theorem $q'(z)$ is also real-rooted (since it has degree one less than the degree of $q$) and has the property that between every two roots of $q'$ there is a root of $q$.
Since polynomials with distinct roots are dense in the set of real-rooted polynomials, this implies that if $q$ is any real-rooted polynomial and $q'(z)$ has a double root at $z$ then $q(z) = 0$.

For the given polynomial $p'(z)$ has a double root at $a$, but $p(a) neq 0$, so $p$ cannot be real-rooted.

(Original image here.)

Honestly, I find the "original solution" to be quite hand-waving the density argument.

For an alternative direct proof, consider that translation along the $,x,$ axis preserves the nature of the roots (real vs. complex) of a real polynomial, so it can be assumed WLOG that $,a=0,$, then:

$$p(x)=a_nx^n+a_n-1x^n-1+ldots+a_3x^3+a_0 quadquad stylefont-family:inherittextwith ;;a_n ne 0;;stylefont-family:inherittextand;; p(0)=a_0 ne0$$

If the roots of $,p(x),$ are $,x_i ne 0,$ then the polynomial having as roots $,y_i=1 / x_i,$ is:

$$q(y)=y^npleft(frac1yright)=a_0y^n+a_3y^n-3+ldots+a_n-1y+a_n$$

By Vieta's relations $,sum_i y_i = 0,$ and $,sum_i lt j y_iy_j = 0,$, so $,sum_i y_i^2 = left(sum_i y_iright)^2-2sum_i lt j y_iy_j=0,$. If all $,y_i,$ were real, that would imply $,y_i=0,$, but $,a_n ne 0,$ so none of the roots can be $,0,$. It follows that not all $,y_i,$ can be real, and therefore not all of $,x_i,$ are real.