Location of a root of a cubic polynomial For $\alpha\in(0,\frac12)$, $\beta\in(0,\infty)$, $N\in\mathbb N\backslash\{0\}$ and $n\in\{0,\ldots,N\}$, how can I prove that exactly one zero of the cubic polynomial
$$
(N+2\beta)x^3-(N+n+3\beta)x^2+(n+\beta+N\alpha-N\alpha^2)x+n\alpha^2-n\alpha
$$
lies in $[\alpha,1-\alpha]$?
 A: Evaluating at $\alpha$, we have:
$$\begin{align*}
f(\alpha) &= (N+2\beta)\alpha^3 - (N+n+3\beta)\alpha^2 + (n+\beta+N\alpha-N\alpha^2)\alpha + n\alpha^2 - n\alpha\\
&= (N+2\beta - N)\alpha^3 + (-N-n-3\beta+N+n)\alpha^2 + (n+\beta-n)\alpha\\
&= 2\beta\alpha^3 - 3\beta\alpha^2 + \beta\alpha\\
&= \alpha\beta(2\alpha^2 -3\alpha + 1).
\end{align*}$$
Evaluating at $1-\alpha$ gives
$$\begin{align*}
f(1-\alpha) &= (N+2\beta)(1-3\alpha + 3\alpha^2-\alpha^3) - (N+n+3\beta)(1-2\alpha+\alpha^2)\\
&\qquad \mathop{+} (n+\beta+N\alpha-N\alpha^2)(1-\alpha) + n\alpha^2 - n\alpha\\
&= (-N-2\beta +N)\alpha^3 + (3N+6\beta - N-n-3\beta -N-N+n)\alpha^2\\
&\qquad \mathop{+}(-3N-6\beta+2N+2n+6\beta-n-\beta+N-n)\alpha\\
&\qquad \mathop{+} (N+2\beta-N-n-3\beta+n+\beta)\\
&= -2\beta\alpha^3 +3\beta\alpha^2-\beta\alpha\\
&= -\alpha\beta(2\alpha^2 - 3\alpha + 1).
\end{align*}$$
So, unless $2\alpha^2-3\alpha+1$ is $0$, the two values have opposite signs. But the roots of $2x^2-3x+1$ are $1$ and $\frac{1}{2}$, so $\alpha$ cannot be a root.
Thus, there is at least one root for the polynomial in $[\alpha,1-\alpha]$ (in fact, in $(\alpha,1-\alpha)$. 
Since $f(x)$ has opposite signs on $\alpha$ and on $1-\alpha$, if $f(x)$ has more than one (distinct) root on $[\alpha,1-\alpha]$, then it must have three distinct roots in the interval (why?). Can all three roots be in that interval? 
