Method of FTA:
$$P(\overline z)=\sum_{k=0}^{2n+1}a_k\overline z^k=\sum_{k=0}^{2n+1}\overline a_k\overline{z^k}=\sum_{k=0}^{2n+1}\overline{a_kz^k}=\overline{\sum_{k=0}^{2n+1}a_kz^k}=\overline{P(z)}$$
which states $z$ is a root for $P(z)=0$ iff its complex conjugate $\bar z$ is. According to FTA, there are odd number of roots for a polynomial of odd degree. That implies there must be one single root $z$ satisfying $z=\bar z$, hence the real root.
Method of IVT:
$$\frac{P(x)}{x^{2n+1}}=1+\sum_{k=0}^{2n}a_k\frac{x^k}{x^{2n+1}}=1+\sum_{k=0}^{2n}a_kx^{k-(2n+1)}$$
For any $\varepsilon>0$, there exists $N>0$ such that for all $|x|>N$, $\left|\sum_{k=0}^{2n}a_kx^{k-(2n+1)}\right|<\varepsilon$. Hence for $x>N$, we have $P(x)>x^{2n+1}-\varepsilon x^{2n+1}>0$ and similarly for $x<-N$, we have $P(x)<0$. Then IVT implies there exists some $y$ such that $P(y)=0$.
(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.
Best Answer
If you know the derivative is positive everywhere, then you know the function has at most one root -- for example, you can reason that if it had two different roots, then the mean value theorem says that $f'(x)=0$ for some $x$ between those roots, which you know isn't the case.
Unfortunately, in this case you need a little more finesse than that, because $f'(1)=0$, as André observed. However, the argument still shows that there can't be two roots on the same side of $1$ -- and you can easily calculate that $f(1)>0$, so another application of the MVT shows that there can't be a root greater than $1$.
But you need to argue separately that the function has at least one root. The intermediate value theorem will do that for you if you show that the function has both a positive and a negative value.