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?

$f'(x)=15x^4+8x^3+3x^2+4x-1>0$ for $x>\frac{1}{2}$ and $f\left(\frac{1}{2}\right)<0$.
Hence, since $\lim\limits_{x\rightarrow+\infty}f(x)=+\infty$, we see that there is unique root for $x>\frac{1}{2}$.
Now, prove that $f(x)<0$ for all $x\leq\frac{1}{2}$.
For example, for $0\leq x\leq\frac{1}{2}$ we have
$$3x^5+2x^4+x^3+2x^2-x-2=$$
$$=\left(3x^5-\frac{3}{4}x^3\right)+\left(2x^4-x^3\right)+(2x^2-x)+\left(\frac{11}{4}x^3-2\right)<0.$$
For $x<0$ we can replace $x$ at $-x$ and it's enough to prove that
$$-3x^5+2x^4-x^3+2x^2+x-2<0$$ for all $x>0$ or
$$3x^5-2x^4+x^3-2x^2-x+2>0$$ or
$$3x^5-3x^4-3x^3+3x^2+x^4+4x^3-5x^2-x+2>0$$ or
$$3x^2(x+1)(x-1)^2+x^4-2x^2+1+4x^3-3x^2-x+1>0$$ or
$$3x^2(x+1)(x-1)^2+(x^2-1)^2+4x^3-3x^2-x+1>0$$ or
$$3x^2(x+1)(x-1)^2+(x^2-1)^2+(x-0.5)^2+4x^3-4x^2+0.75>0$$ and it remains to prove that
$$4x^3-4x^2+0.75>0,$$
which is AM-GM:
$$4x^3-4x^2+0.75=2x^3+2x^3+0.75-4x^2\geq3\sqrt[3]{(2x^3)^2\cdot0.75}-4x^2=\left(3\sqrt[3]3-4\right)x^2>0,$$
which gives that our equation has one real root.
Done!
