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?
 A: $f'(x)=15x^4+8x^3+3x^2+4x-1>0$ 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!
A: 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.
A: $LHD=x^3(3x^2+2x+1)=x^3(3(x+\frac13)^2+\frac23)$
$RHD=-2x^2+x+2=-2(x-\frac14)^2+\frac{17}8$
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 $\frac14<x<\frac54$.
