Prove that $f(x)=(e^x-1)(x^2+3x-2)+x$ has exactly one positive root, exactly one negative root and one root at $x=0$. Prove that the function 
$f(x)=(e^x-1)(x^2+3x-2)+x$
has exactly one positive root, exactly one negative root and one root at $x=0$.
My work so far:
$f(0)=0$
Thus, $x=0$ is a root.
For the positive root,
$f(\frac{1}{4})=(e^{\frac{1}{4}}-1)((\frac{1}{4})^2+3(\frac{1}{4})-2)+(\frac{1}{4})
               =(e^{\frac{1}{4}}-1)(\frac{1}{16}+\frac{3}{4}-2)+\frac{1}{4}
               =-\frac{19}{16}e^{\frac{1}{4}}+\frac{23}{16}
               <0$
$f(1)=(e^1-1)((1)^2+3(1)-2)+(1)
     =(e^1-1)(1+3-2)+1=2e^1-1
     >0$
Thus, there is at least one positive root, by the intermediate value theorem. I have also done the same thing for the negative root, as $f(-4)<0$ and $f(-2)>0$.
But how do I show that there is only one positive root and only one negative root?
 A: (To state my solution more rigorously and clearly, I put my comment into an answer.)
First, it is easy to calculate:
$$
f'(x) = e^x (x^2+5x+1) - 2x -2 \\
f''(x) = (x+1)(x+6)e^x-2 \\
f'''(x)=e^x(x^2+9x+13)
$$
Second, we will check how many real roots $f''(x)=0$ has:

  
*
  
*Case $-\infty<x \leq -1$:
Calculate $f'''(x)=0$, we get the stationary points of $f''(x)$ are at
$$ x_{1} = \frac{-9 + \sqrt{29}}{2}, ~ x_{2} = \frac{-9 - \sqrt{29}}{2}$$
Then evaluate: $f''(x_1)\approx -2.55543$, $f''(x_2)\approx -1.99445$. Besides,
$$ \lim_{x \rightarrow -\infty} f''(x) = -2$$
and $f''(-1)=-2$. Thus,
$$\max_{-\infty < x \leq -1} f''(x)=f''(x_2)\approx -1.99445 < 0$$
That is, in this case, there's no real root for $f''(x)=0$.
  
*Case $x>-1$:
In this case, it is easy to check $f''(x)$ is monotone increasing and is able to go larger than 0. Thus, there's exactly one real root for $f''(x)=0, x>-1$.

Conclusion: $f''(x)=0$ has exactly one real root.


Finally, we will utilize Rolle's theorem:


(Given proper continuity and differentiability.) If $f'(x)=0$ has exactly $n$ real roots, then $f(x)=0$ at most has $n+1$ real roots. 

Why? If we can find $n+2$ real roots for $f(x)=0$, say
$$x_1<x_2<\cdots<x_{n+2}$$
according to Rolle's theorem, there will exist $c_1 \in (x_1,x_2)$, $c_2 \in (x_2,x_3)$, ..., $c_{n+1} \in (x_{n+1},x_{n+2})$, such that
$$f'(c_1) = f'(c_2) = \cdots = f'(c_{n+1}) = 0$$
It will contradict "$f'(x)=0$ has exactly $n$ real roots".

Now, we can see because $f''(x)=0$ has exactly one real root, $f'(x)=0$ will at most have two real roots, and then $f(x)=0$ will at most have three real roots. And you have found the three roots for $f(x)=0$, so there're no more roots.

Note: Be careful of the "at most" statements above. It does NOT mean the equation must have such many roots.
A: I bet, you have to find any two positive numbers a and bsuch that f(a)<0 and f(b)>0. Thus, it will prove that there is one positive root that is between a and b. 
Same logic applies for the negative root. 
A: You can also use the fact that the function is differentiable (and continuous) at all $R$, to find a contradiction: 
Suppose that $0 <a <b$ with $a$, $b$, roots of $f(x)$, with $0$, $a$, $b$, consecutive roots, then there exists $c_1\in[0,a]$ and $c_2 \in[a, b]$ such that $f(x)$ reaches maximum or minimum, as the function is twice differentiable we have:
$$f''(x)=e^x(x^2+5x+1)+e^x(2x+5)-2$$
$$f''(x)=e^x(x^2+5x+1)+2xe^x+5e^x-2$$
how $5e^x>2$, $f''(x)$  is always positive for all $x>0$ then $c_1$ and $c_2$ are both minimuns, but this implies that there is no tangent line at $a$ which contradicts the fact that $f(x)$ is differentiable at all $R$.
God bless
