TITLE: How can I guarantee that my polynomial has distinct zeros?
QUESTION [0 upvotes]: I want to prove that following polynomial $f_p$ has distinct zeros
$$
f_p(x) = \begin{cases}
x^{2p+2} - abx^{2p} - (a+b)x^p-1, &\text{if $p$ is  even}\\
x^{2p+2} - abx^{2p} - 2x^{p+1} +1, & \text{if $p$ is  odd}
\end{cases}
$$
where $a$, $b$ are positive real numbers and $p>0$ is an integer. How can I prove this polynomial has distinct zeros for any arbitrary $a$,$b$ and $p$.
Thanks in advance.

REPLY [1 votes]: Standard way of doing it is creating an auxiliary polynomial in your case of $2p$ degree. Multiply this one with a hypothetical factor $(x-c)^2$. Finally compare coefficients and prove (or disprove) that such system does not have a solution. Try with smaller $p$ first until you realize what is going on then generalize.
$$(\sum\limits_{k=0}^{2p}a_{k}x^{k})(x-c)^2 = \sum\limits_{k=2}^{2p+2}a_{k-2}x^k - \sum\limits_{k=1}^{2p+1}2a_{k-1}cx^k + \sum\limits_{k=0}^{2p}a_{k}c^2x^k=$$
$$a_{2p}x^{2p+2}+(a_{2p-1}-2a_{2p}c)x^{2p+1}+\sum\limits_{k=2}^{2p}(a_{k-2}-2a_{k-1}c+a_{k}c^2)x^k+(-2a_{0}c+a_{1}c^2)x +a_{0}c^2$$
Now you have  $a_{0}c^2=1$ or $a_{0}c^2=-1$. Obviously this resolves $a_{1}$ as well. On the other side you have $a_{2p}=1$ and $a_{2p-1}-2a_{2p}c=0$ which resolves $a_{2p-1}$
As you will realize yourself soon enough $a_{k-2}-2a_{k-1}c+a_{k}c^2=0$ up to $p$ (and there is another one after it up to $2p-1$) is actually a recursion $a_{k}=-\frac{a_{k-2}}{c^2}+\frac{2a_{k-1}}{c}$ which you can solve for the given initial condition. Since you have $a_{p}$ and $a_{2p}$ you can find the condition for these two and compare if they can satisfy recursion up to $p$ and the following one from $p$ to $2p-1$ together with the initial conditions for $a_{1}$ and finishing condition for $a_{2p-1}$.
That is the plan that should either give you a possible solution or give you the reason for not having a solution.
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