# How to prove that my polynomial has distinct roots?

I want to prove that the polynomial

$$f_p(x) = x^{2p+2} – abx^{2p} – 2x^{p+1} +1$$

has distinct roots. Here $a$, $b$ are positive real numbers and $p>0$ is an odd integer. How can I prove that this polynomial has distinct roots for any arbitrary $a$,$b$ and $p$.

#### Solutions Collecting From Web of "How to prove that my polynomial has distinct roots?"

Let $c$ denote $ab$. Note that
\begin{equation*}
f(x) = (x^{p+1}-1)^2 – cx^{2p}
\end{equation*}
and
\begin{equation*}
f'(x) = 2(p+1)x^p(x^{p+1}-1) – 2pcx^{2p-1}
\end{equation*}

Then, $f(x) = 0 \iff$
\begin{equation*}
c = \dfrac{(x^{p+1}-1)^2}{x^{2p}} = \varphi(x)\ (\text{say})
\end{equation*}
and $f'(x) = 0 \iff$
\begin{align*}
c & = \dfrac{(p+1)x^p(x^{p+1}-1)}{px^{2p-1}} = \dfrac{(p+1)x}{p}\dfrac{x^{p+1}-1}{x^p} \iff\\
c & = \left(\dfrac{p+1}{p}\right)x \sqrt{\varphi(x)}.
\end{align*}

Thus, $f(x)$ and $f'(x)$ vanish for the same $x$ if and only if for some root $x$ of $f(x)$,
\begin{align*}
c & = \left(\dfrac{p+1}{p}\right)x \sqrt c \iff\\
x & = \dfrac{p\sqrt c}{p+1}.
\end{align*}

Thus, for every $p$ and $c = ab$, if such an $x$ is a root, it is a multiple root.

Now, when does such a root $x$ exist? Let $x = t$ be one such. Then $c = \left(\dfrac{p+1}{p} \right)^2 t^2$. Then, since $f(t) = 0$, we have (from the original form of the equation):
\begin{align*}
t^{2p+2} – \left(\dfrac{p+1}{p}\right)^2 t^{2p + 2} – 2t^{p+1} + 1 = 0\\
-\dfrac{(2p + 1)}{p^2}t^{2p + 2} – 2t^{p+1} + 1 = 0\\
(2p + 1)t^{2(p + 1)} + 2p^2 t^{p + 1} – p^2 = 0.
\end{align*}

When treated as a quadratic equation in $t^{p+1}$, the discriminant is
\begin{equation*}
4p^4 + 4p^2(2p + 1) = 4p^2(p + 1)^2,
\end{equation*}
and therefore, the solutions are
\begin{equation*}
t^{p+1} = -p, \dfrac{p}{2p + 1}.
\end{equation*}
That is,
\begin{equation*}
t = (-p)^{\frac 1 {p + 1}}, \left(\dfrac p {2p + 1} \right)^{\frac 1 {p + 1}}.
\end{equation*}

But substituting the same $c$ in $f'(t) = 0$, we get
\begin{align*}
& 2(p+1)t^p(t^{p+1}-1)-2p\left(\dfrac{p+1}{p}\right)^2t^{2p+1}=0\\
& p(t^{p+1}-1)-(p+1)t^{p+1}=0 \implies\\
& t = (-p)^{\frac{1}{p+1}}
\end{align*}

Thus, only the first of the previous two solutions satisfies both equations.

Then, $c = \left( \dfrac{p+1}{p} \right)^2 t^2$ gives us
\begin{equation*}
\boxed{c= \dfrac{(p+1)^2}{(-p)^{\frac{2p}{p+1}}}}.
\end{equation*}

Thus, the equation has multiple roots exactly when $c$ and $p$ are related as above.

Note that for odd values of $p$, $c$ will be a real number if and only if $p$ is of the form $4k + 1$, and then, $c < 0$. If, as stated in the question, $c = ab$ is a positive real number, the equation will have distinct roots.

Example

For $p = 1$, $f(x) = x^4 – cx^2 – 2x^2 + 1$ and $f'(x) = 4x^3 – 2cx – 4x$.

Then, $f(x) = 0$ and $f'(x) = 0$ imply that $c = \left(\dfrac{x^2 – 1}{x}\right)^2$ and $c = 2x\left(\dfrac{x^2-1}{x}\right)$ respectively. Thus, if $x$ is a multiple root, then $x = \dfrac{\sqrt c}{2}$.

Taking $t$ to be such a root, so that $c = 4t^2$, and substituting in $f(t) = 0$, we get
\begin{align*}
t^4 – 4t^4 – 2t^2 + 1 = 0\\
3t^4 + 2t^2 – 1 = 0.
\end{align*}
Thus, $t^2 = -1, \dfrac 1 3$, of which only the first one satisfies $f'(t)=0$. Thus, $c = -4$.

For $c = 4$, $x^4 + 2x^2 + 1 = 0$ has roots $\pm i, \pm i$.

The claim is false. Set $ab:=(\frac{729}{16})^\frac{1}{3}$ and $p:=2$

Then $f(x)=x^6-abx^4-2x^3+1$ has a double root at $x=-2^{\frac{1}{3}}$

With the help of Mathematica the discriminant is given by:

$$\Delta = \left \{ \begin{array}{cc} (4c)^{p+1} \left[ p^{p} c^{(p+1)/2}+(p+1)^{p+1} \right]^{2} & \text{odd } p \\[5pt] (4c)^{p+1} [(p+1)^{2(p+1)}-p^{2p}c^{p+1}] & \text{even } p \end{array} \right. \\$$

$\Delta \neq 0 \,$ if $\, \left \{ \begin{array}{ll} \text{odd } p=4n-1 , & c\neq 0 \\[5pt] \text{odd } p=4n-3, & c^{(p+1)/2} \neq 0, \displaystyle -\frac{(p+1)^{p+1}}{p^{p}} \\[5pt] \text{even } p, & c^{p+1} \neq 0, \displaystyle \frac{(p+1)^{2(p+1)}}{p^{2p}} \end{array} \right.$