Find number of real solutions of $3x^5+2x^4+x^3+2x^2-x-2=0$ 
Find a number of real roots of $$f(x)=3x^5+2x^4+x^3+2x^2-x-2$$

I tried using differentiation:
$$f'(x)=15x^4+8x^3+3x^2+4x-1=0$$ and I found number of real roots of $f'(x)=0$ by drawing graphs of $g(x)=-15x^4$ and $h(x)=8x^3+3x^2+4x-1$ and obviously from graph there are two real solutions as $h(x)$ is an increasing function.
But now how to proceed?

$LHD=x^3(3x^2+2x+1)=x^3(3(x+\frac13)^2+\frac23)$
$RHD=-2x^2+x+2=-2(x-\frac14)^2+\frac{17}8$
if $x>\frac54$, obviously this equality has no solution.
if $\frac14<x<\frac54$, since LHD is monotoniously increasing, RHD is its opposite,
 so there is one solution. 
if $0<x<\frac14$ ,$LHD<1<RHD$ there is no solution.
if $-1<x<0$  ,$-2<LHD<0, 1<RHD<2$ there is no solution.
if $x<-1$  at $x=-1$, $LHD=-2$, $RHD=-1$ and since $3x^2+2x+1-(-2x^2+x+2)=5x^2+x-1$$=5(x+\frac1{10})^2-\frac1{20}>0,$ then $LHD<RHD$ 
Finally there is only one solution at $\frac14<x<\frac54$.
