Here are the first few "shallow diagonals" in Pascal's triangle.
We can use these shallow diagonals to make polynomials. Note that the even degree polynomials have an extra $\pm x$ at the end.
$x$
$x^2-x$
$x^3-x$
$x^4-2x^2+x$
$x^5-3x^3+x$
$x^6-4x^4+3x^2-x$
$x^7-5x^5+6x^3-x$
$x^8-6x^6+10x^4-4x^2+x$
$\dots$
Here is an explicit description (you can skip this part if you already see the pattern): In the $n$th row of Pascal's triangle (first row is $n=1$), let the leftmost number be the leftmost number of a shallow diagonal. If $n$ is odd, say $2k-1$, call the numbers in this shallow diagonal, from left to right, $a_1, a_3, \dots, a_{2k-1}$, and let $P_{2k-1}(x)=\sum\limits_{i=1}^{k} (-1)^{i-1}a_{2i-1}x^{2(k-i)+1}$. If $n$ is even, say $2k$, call the numbers in this shallow diagonal, from left to right, $a_2, a_4, \dots, a_{2k}$, and let $P_{2k}(x)=\sum\limits_{i=1}^{k} (-1)^{i-1}a_{2i}x^{2(k+1-i)}+(-1)^{k}x$.
Conjecture: For each of these polynomials, all of the roots are real, distinct, and in $(-2,2)$.
Is my conjecture true?
I've been picking random shallow diagonals, and so far I have not found a counter-example. Here are more shallow diagonals, and more even numbered shallow diagonals.
Context: I formulated this conjecture when trying to answer another question of mine: For given $n$, how to find integers $a_0, …, a_n$ so that $\sum_{k=0}^n a_k x^k$ has $n$ distinct real roots and $\sum_{k=0}^n |a_k|$ is minimized?
Best Answer
Ignoring the added $\pm x$ for the even-degree polynomials for now, your polynomials are $x\,U_{n-1}(\tfrac{x}{2})$, where $U_n$ is the degree-$n$ Chebyshev polynomial of the second kind. (See also the Wikipedia article for Fibonacci polynomials, which mentions the shallow diagonals.)
This proves your conjecture for odd $n$, as the roots of $U_{n-1}(\tfrac{x}{2})$ are $2\cos(\frac{k}{n}\pi)$ for $1\le k\le n-1$.
Next, for a monic degree-$n$ polynomial $f(x) \in \mathbb{Z}[x]$, the following are equivalent:
We compute: $$ \begin{align*} P_{2k}(x) &= x\,U_{2k-1}(\tfrac{x}{2}) \pm x\\ z^{2k}P_{2k}(z+\tfrac{1}{z}) &= z^{2k}\cdot (z + \tfrac{1}{z})\cdot \left[U_{2k-1}\left(\frac{z+\tfrac{1}{z}}{2}\right) \pm 1\right] \\ &= z^{2k}\cdot (z + \tfrac{1}{z})\cdot \left[\left(\frac{z^{2k}-\tfrac{1}{z^{2k}}}{z-\tfrac{1}{z}}\right) \pm 1\right] \\ &= (z^2+1)\cdot\left[\frac{z^{4k}-1 \pm (z^{2k+1}-z^{2k-1})}{z^2-1}\right] \\ &= (z^2+1)\cdot\frac{z^{2k-1}\pm 1}{z\pm 1} \cdot\frac{z^{2k+1}\mp 1}{z\mp 1} \tag{$*$} \end{align*} $$ which has all of its roots on the unit circle. The roots of $P_{2k}(x)$ are given by $x = 2\cos\theta$ for each root $z = e^{i\theta}$ of $(*)$. In particular, $z \ne \pm 1$ so all of the roots of $P_{2k}(x)$ are in the open interval $(-2, 2)$.
So your conjecture is true for all $n$.