<s>[INST] 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?
 [/INST] 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
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