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If I have a polynomial with real cofficients $p_x \in \mathbb{R}[x]$ , is it always possible to pick a constant $c \in \mathbb{R}$ so that $p_x + c$ only has real roots✱? Equivalently, does the following hold?

$$p_x + c \propto \prod_{j=1}^{\deg(p_x)} \left(x-a_j\right) \;\;\;\;\text{where}\;\; a \in \mathbb{R} $$

All constant polynomials are proportional to $1_x$ .

All linear polynomials are proportional to themselves.

A quadratic polynomial $ax^2+bx+c$ has only real roots if and only if $b^2 - 4ac$ is non-negative, which can be ensured by picking $c=0$ .

Geometrically, picking a constant $c$ is like picking where the x-axis is. It's definitely possible to pick a $c$ so that the line $y = c$ intersects the graph of $p_x$, but I don't have a good intuition for what the multiplicity is for the root I've created by choosing to intersect the graph of $p_x$ at some specific point.

For instance, $x^3$ has a triple root, but $x^3-1$ only has a single root. The graph of $x^3$ flattens out near the origin, but I'm not sure how to turn that into an actual argument.

Is it always possible to pick a $c$ so that the resulting polynomial has only real roots?

✱ I'm thinking of an element of $\mathbb{R}[x]$ as a formal expression with a degree and as function drawn from $\mathbb{C} \to \mathbb{C}$ that's constrained to send reals to reals, so it makes sense to talk about its complex roots. The graph of $p_x$ is just a figure in $\mathbb{R}^2$, as usual.

Greg Nisbet
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    Something like $f(x) = x^3 + x$ should be a counter-example – Winther Dec 21 '18 at 23:39
  • Alternatively, $f(x) = x^3 + c$ only holds one real solution, $-c^{\frac{1}{3}}$. – Neeyanth Kopparapu Dec 21 '18 at 23:48
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    Your polynomial would have the property that there exists $c \in \Bbb{R}$ such that all its local minima are below $c$ and all its local maxima are above $c$. It's easy to design a polynomial that does not have this property. (By the way, the wording of your question is a bit confusing: it would be clearer if you said "all the roots are real" rather than "it only has real roots".) – Rob Arthan Dec 21 '18 at 23:48
  • @NeeyanthKopparapu, $x^3 + 0$ has three real roots, counting multiplicity. – Greg Nisbet Dec 21 '18 at 23:55
  • @GregoryNisbet I'm sorry - I misunderstood the question. This is correct. Refer Winther's answer for a counterexample – Neeyanth Kopparapu Dec 21 '18 at 23:59
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    @NeeyanthKopparapu I didn’t mean to be rude. The $x^n$ cases were what I looked at an hour ago when I thought the statement was probably true. – Greg Nisbet Dec 22 '18 at 00:11

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Here is a counter-example: the function $f(x) = x^3 + x + c$ is monotonic as $f'(x) = 3x^2 + 1 > 0$ so it can only have one real root $r$. If $f$ only has real roots it must be on the form $(x-r)^3 = x^3 -3rx^2 + 3r^2x - r^3$, but this expression cannot be identical to $x^3 + x + c$ (compare the coefficients).

Winther
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    +1: another way of seeing why the real root of $f(x)$ cannot be a repeated root is to note that $f'(x) = 3x^2 + 1$ has no real roots and hence has no real roots in common with $f(x)$. – Rob Arthan Dec 22 '18 at 00:08
  • @RobArthan Ah yes that's a much more elegant way of doing it! – Winther Dec 22 '18 at 00:11