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?

First, let us calculate few of $f$'s derivatives:
\begin{align}
f'(x) &= 15x^4+8x^3+3x^2+4x-1,\\
f''(x) &= 2(30x^3+12x^2+3x+2),\\
f'''(x) &= 6(30x^2+8x+1) > 0,\ \forall x\in\mathbb R.
\end{align}
Thus, $f'$ is convex, and $f''$ strictly increasing. Since $\lim_{x\pm\infty} f''(x) = \pm\infty$, $f''$ has exactly one root $c$, and this is global minimum of $f'$. Depending on $f'(c)$, $f'$ has $0$, $1$ or $2$ roots. Since, $f(-1) > f(0)$, there is some point at which $f'$ is negative and we conclude that $f'$ has exactly $2$ roots, $x_1$ and $x_2$, corresponding to local maximum and local minimum of $f$. 
Let's take a short break to sum up what we have: we know what $f$ is increasing on intervals $(-\infty,x_1)$ and $(x_2,+\infty)$, while decreasing on $(x_1,x_2)$.
Note that we now have enough data to conclude that $f$ has either $1$ or $3$ roots and it depends on whether local extremes have the same or opposite sign. We will now prove that $f(x_1), f(x_2) < 0$ and conclude that $f$ has exactly one root, somewhere on interval $(x_2,+\infty)$.
So, we know that $f'(x_i) = 0$ and by polynomial division we have $$f(x_i) = \frac 1{75}(f'(x_i)(15x+2)+2(7x_i^3+42x_i^2-34x_i-74)) = \frac 2{75}(7x_i^3+42x_i^2-34x_i-74)$$
so let us define $g(x) = 7x^3+42x^2-34x-74$.
There are two things we want to prove:


*

*$\ g(x)<0$ for $|x|< 1,$

*$f'(x)>0$ for $|x|\geq 1.$
This will give us that $g(x_i)<0$ and consequently that $f(x_i) < 0$, which would finish our proof.
To prove 1., calculate $g''(x) = 42(x+2)$, meaning that $g$ is convex on $[-1,1]$, so the maximal value on that segment will be achieved at $\pm 1$. Since $g(-1) = -5$ and $g(1) = -59$, we conclude that $g(x) < 0$ on $(-1,1)$.
To prove 2., if $x\geq 1$, $f'(x) \geq 15+8+3+4-1 > 0$. If $x\leq 1$, consider $$h(x) = f'(-x) = 15x^4-8x^3+3x^2-4x-1 = x^3(15x-8)+(3x^2-4x-1).$$
Both $x^3(15x-8)$ and $3x^2-4x-1$ are strictly increasing on $[1,+\infty)$, while $h(1) = 5$. This means that $h(x) > 0$ on $[1,+\infty)$, i.e. $f'(x)>0$ on $(-\infty,-1]$, which proves 2.
