I have a function\begin{align*}Z(t) &= 4(1-\alpha)(3-5\alpha)t^6+12(1-\alpha)^2t^5+8(1-\alpha)(2\alpha-1)t^4\\&\quad -2(1-\alpha)t^3+3(3-8\alpha+6\alpha^2)t^2-2(1-\alpha)(2\alpha-1)t\\&\quad +(1-\alpha)(2\alpha-1)\end{align*}defined on $t\geq 1$. I have proved that for $\alpha\in(1/2,3/5]$, $Z(t)\geq 0$. And I numerically observed that for $\alpha\in(3/5,1)$, $Z(t)$ has exactly one solution on $t\geq 1$. It is easy to check that $Z(1)=3(8-19\alpha+12\alpha^2)>0$, since $8-19\alpha+12\alpha^2$ is convex in $\alpha$ and achieves its global minimum at $\alpha=19/24$ with the corresponding value $23/16$. Also, it is easy to verify that $Z(t)\rightarrow-\infty$ as $t\rightarrow\infty$. So it is clear that there exists at least one solution for $Z(t)=0$ on $t\geq 1$. However, how to prove that such a solution is unique???
This is a high-order polynomial, and I used Mathematica, not that useful...Is there some "Buffalo Ways" to prove this property?