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.
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).