Roots of cubic equation If$\frac{1+\alpha}{1-\alpha},\frac{1+\beta}{1-\beta},\frac{1+\gamma}{1-\gamma}$ are the roots of the cubic equation $f(x)=0$ where $\alpha,\beta,\gamma$ are the real roots of the cubic equation $3x^3-2x+5=0$,then find the number of negative real roots of the equation $f(x)=0$.
My attempt:I tried finding out solution of $3x^3-2x+5=0$ to get $\alpha,\beta,\gamma$ by rational root method and hit and trial method but could not get them.Is my approach correct? Or Descretes rule is to be applied?
Can someone help me solve this question? 

Note that $\frac{1+\alpha}{1-\alpha}<0$ is equivalent to $1+\alpha$ and $1-\alpha$ having opposite signs, which happens when $\alpha>1$ or $\alpha <-1$. So all we have to do is count the number of roots of $3x^2-2x+5$ in $[-1,1]$.
Let $p(x) = 3x^3-2x+5$. Now $p(-1)=4$ and $p(1)=6$. Moreover, $p'(x)=9x^2-2$, which has roots at $x=\pm\sqrt{2}/3$. Also, $p''(x)=18x$, which has the sign the same as $x$, so $x=-\sqrt{2}/3$ is a local maximum of $p$ while $x=\sqrt{2}/3$ is a local minimum of $p$. In particular, the minimum of $p$ on $[-1,1]$ is the minimum of $p(-1)=4$, $p(1)=6$, and $p(\sqrt{2}/3)$. Now $p(\sqrt{2}/3)=3(\sqrt{2}/3)^3-2\sqrt{2}/3+5=-(4/9)\sqrt{2}+5>0$. Therefore, $p$ has no roots on $[-1,1]$, so the answer to your question is simply the number of real roots of $p$, which is $1$.
