I'll assume you want the divisibility to be in $\mathbb{Q}[x]$. Write $\displaystyle f(x) = \prod_{i=1}^{k} (x - a_i)^{m_i}$ where the $a_i$ are distinct (in $\mathbb{C}$).
Lemma: $\displaystyle \frac{f'(x)}{f(x)} = \sum_{i=1}^{k} \frac{m_i}{x - a_i}$.
This is a direct consequence of the product rule and you do not need to use any calculus to prove it. (That is, you don't need to take logarithms. The derivative can be defined as a totally formal operation on polynomials and this identity holds in $\mathbb{Q}(x)$.)
If $f'(x)$ divides $f(x)$, then $\frac{f'(x)}{f(x)} = \frac{n}{x - a}$ for some $a$. Now, the functions $\frac{1}{x - a}$ are linearly independent (you also do not need to use any calculus to prove this) in $\mathbb{C}(x)$. It follows that this is possible if and only if $f$ has $a$ as a repeated root of multiplicity $n$, so $a$ is the only root.
It follows that $f$ has the form $c(px - q)^n$ for some integers $c, p, q$ with $a, p \neq 0$. (I'm ignoring the degenerate cases.)
I think I got the proof that no such real polynomial with degree $ \geq 6$ exists.
Let $n \geq 6$
Suppose for contradiction that $z_1,\ldots,z_n \in \mathbb R-\{0\}^n$ are such that $(X-z_1)...(X-z_n)=X^n+\sum_{k=1}^{n-1}z_iX^{n-i}$
Then three useful identities appear $$\sum_{k=1}^{n}z_k=-z_1 \; \; \; \;(1)$$
$$\sum_{\large1\leq i<j \leq n}z_iz_j=z_2 \; \; \; \;(2)$$
$$\prod_{k=1}^n z_k=(-1)^n z_n \; \; \; \;(3)$$
Since $$(\sum_{k=1}^{n}z_k)^2=\sum_{k=1}^{n}z_k^2+2\sum_{\large1\leq i<j \leq n}z_iz_j$$it follows that$$z_1^2=2z_2+\sum_{k=1}^{n}z_k^2$$
Hence $$0< \sum_{k=2}^{n}z_k^2=-2z_2 \; \; \; \;(4)$$ and $$0<\sum_{k=3}^{n}z_k^2=1-(z_2+1)^2 \; \; \; \;(5) $$
$(4)$ and $(5)$ imply $$\; \; \; \;-2<z_2<0 \; \; \; \;(6)$$
thus $(6)$ and $(4)$ imply $$0<\sum_{k=2}^{n}z_k^2 < 4 \; \; \; \; (7)$$
Also $(6)$ and $(5)$ imply $$0<\sum_{k=3}^{n}z_k^2 \leq 1 \; \; \; \; (8)$$
By AM-GM, $$\left(|z_3|^2\ldots|z_{n-1}|^2 \right)^{1/(n-3)} \leq \frac{1}{n-3}\sum_{k=3}^{n-1}z_k^2 \leq \frac{1}{n-3}\sum_{k=3}^{n}z_k^2$$
Hence
$$|z_3|^2\ldots|z_{n-1}|^2 \leq \left(\frac{1}{n-3}\sum_{k=3}^{n}z_k^2\right)^{n-3} $$
Squaring, $$|z_3|\ldots|z_{n-1}| \leq \left(\frac{1}{n-3}\sum_{k=3}^{n}z_k^2\right)^{\large \frac{n-3}{2}} \leq_{ \large (8)} \dfrac{1}{{(n-3)}^{(n-3)/2}} \; \; \; \; (9)$$
By triangle inequality $(1)$, and Cauchy-Schwarz
$$2|z_1| \leq \sum_{k=2}^{n}|z_k| \leq \sqrt{n-1} \sqrt{\sum_{k=2}^{n}z_k^2} $$
Hence by $(7)$,
$$|z_1| \leq \sqrt{n-1} \; \; \; \; (10)$$
Rewriting $(6)$ as $$|z_2|\lt2 \; \; \; \; (11) $$
Recalling $(3)$ (with $z_n$ cancelled from both sides) and putting together $(9)$, $(10)$ and $(11)$, we have
$$1=|z_1||z_2||z_3|\cdots|z_{n-1}| < \dfrac{ 2\sqrt{n-1}}{{(n-3)}^{(n-3)/2}}$$
This inequality fails for $n\geq 6$.
Contradiction.
I can't prove anything for $n=5$ so maybe the conjecture doesn't hold.
Best Answer
No polynomial $p(x)$ with integer coefficients such that $p(0)=3$ and $p(1)=11$ has integer roots. Let $n\in\Bbb Z$. If $n$ is even, then $n\equiv0\pmod2$ and therefore $p(n)\equiv p(0)(=3)\pmod 2$. So, $p(n)$ is odd. And if $n$ is even, then $n\equiv1\pmod2$ and therefore $p(n)\equiv p(1)(=11)\pmod2$. So, again, $p(n)$ is odd. Since $p(n)$ is always odd, it cannot be equal to $0$.