Can $\frac{xp(x)-ap(a)}{x-a}$ have an inside root if $p(x)$ does not?

Question

Let $p(x)$ be a polynomial that has all of its roots outside of the unit circle and let $a$ be some real number that is not a root of $p(x)$. Can the polynomial $$ \frac{xp(x)-ap(a)}{x-a}$$ have a root inside the unit circle?

Answer 1

Let :

$$Q_a(z)=\frac{zP(z)-aP(a)}{z-a}=P(z)+a\left(\frac{P(z)-P(a)}{z-a}\right)\tag{1}$$

Let $\mathbb{U}$ denote the unit disk.

Let $P(z)=z+b$, then $Q_a(z)=z+a+b$.

It suffices then to take $a$ and $b$ in such a way that :

$$|b|>1 \ \ \ \ \text{and} \ \ \ \ |a+b|<1$$

knowing that this can be done in many different ways !

(see a similar answer, though less general, in Can $\frac{zR(z)-aR(a)}{z-a}$ have an inside root if $R$ does not? ).

We could find similar counter-examples for higher degree polynomials. For example here is one for a quadratic polynomial and $a=1$ : take $P(z)=z^2-2$, whose roots are $z=\pm\sqrt{2}$, thus outside $\mathbb{U}$, whereas, according to (1) :

$$Q_a(z)=z^2-2-2\left(\frac{z^2-2+1}{z-1}\right)=z^2+z-1$$

has one of its roots $z_1:=\frac{-1+\sqrt{5}}{2}$ inside $\mathbb{U}$, the other one $z_1:=\frac{-1+\sqrt{5}}{2}$ being outside $\mathbb{U}$ (have you recognized the golden ratio ?).