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By real polynomial, I mean a univariate polynomial with real coefficients. We know that polynomials with real coefficients either have only real numbers as roots or non-real complex roots that come as conjugate pairs. This means any real polynomial can be factored into 1st degree real polynomials with real roots and (in case they have complex roots) 2nd degree real polynomials with two non-real complex roots which are conjugates of each other.

I can easily show that a 2nd degree real polynomial with two non-real conjugate complex roots can be vertically shifted such that it would have two real roots. (You can shift the parabola represented by the function so that it crosses the x-axis.)

I want to reach a general conclusion that this is true for any real polynomial because I feel like it should be true, but it seems to be a lot harder than to show it for individual parabolas in the coordinate plane.

Is my assumption true and if so, how could we prove it?

Rócherz
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jacob78
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    It is easy to visually construct valleys and hills of a possible polynomial which do not share real solution space; consider a valley whose lowest point is higher than a hill not adjacent to the valley. – abiessu Dec 04 '23 at 06:49
  • To clarify, @abiessu's comment is a suggested strategy for proving that the conclusion does not actually hold when $n\ge4$ is even. – Greg Martin Dec 04 '23 at 07:58
  • @Greg, I think you need degree at least five for that. – Gerry Myerson Dec 05 '23 at 07:48

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$p(x)= x^3+x = x(x^2+1)$ would be a concrete counter-example. It is strictly monotonically increasing on all of $\mathbb R$. Therefore, we have the following properties of the shifted polynomial $p_C(x) := p(x) + C$, for any $C \in \mathbb R$:

  • $p'_C=p'$, hence $p_C$ is stricty monotonically increasing as well.
  • Because of the monotonicity, the graph of $p_C(x)$ will cross the $x$-axis exactly once, giving you a single real root $x_0$ of $p_C(x)$.
  • That root has multiplicity of $1$ - if it didn't, we would have $p'_C(x_0) = 0$, which contradicts the strict monotonicity. Therefore, there must be two other roots, distinct from $x_0$. Since there are no other real roots, those other roots have to be non-real.
DominikS
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  • I think $p(x)=3x^5-65x^3+540x$ is a counterexample. Degree five, $p'$ vanishes at $\pm2$ and at $\pm3$, but no horizontal line crosses it more than three times. – Gerry Myerson Dec 04 '23 at 08:28
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    Ah yes, thanks! Have deleted the conjecture. :) – DominikS Dec 04 '23 at 08:30
  • Another example are polynomial functions of the form $y=x^4+r$, for some positive real $r$. We can easily see that there can not be any real zeroes. Applying derivative calculus, we get a derivative of $y'=4x^3$ The derivative has exactly one real zero, $0$. This means there is only one vertex, and shifting the curve down gets at most two intersections with the x-axis, and thus, at most two real solutions for the shifted function, even though there are always four complex zeroes if the vertex is not at ($0,0$) – Michael Ejercito Feb 16 '24 at 03:02