Let $$f$$ be a real polynomial of degree $$n$$ such that $$f(x) = f'(x) \cdot f''(x)$$, for all $$x \in \mathbb{R}$$. If $$f(0) = 0$$, then $$36\left(f'(2) + f''(2) + \int_0^2 f(x)\,dx\right)$$ is equal to :

The functional equation is $$f(x)=f'(x)\,f''(x)$$ and $$f$$ is a real polynomial of degree $$n$$.

Step 1 Find the degree of $$f$$.
If $$\deg f=n$$ then $$\deg f'=n-1$$ and $$\deg f''=n-2$$, so $$\deg\bigl(f'(x)\,f''(x)\bigr)=(n-1)+(n-2)=2n-3.$$ Since this equals $$\deg f=n$$, we get $$n=2n-3\;\Longrightarrow\;n=3.$$ Hence $$f$$ is a cubic polynomial.

Step 2 Write the general cubic and impose $$f(0)=0$$.
Let $$f(x)=ax^{3}+bx^{2}+cx+d.$$ Given $$f(0)=0$$, we have $$d=0$$, so $$f(x)=ax^{3}+bx^{2}+cx.$$

Step 3 Compute derivatives.
$$f'(x)=3ax^{2}+2bx+c,\qquad f''(x)=6ax+2b.$$

Step 4 Use the equation $$f=f' f''$$.
Set $$ax^{3}+bx^{2}+cx=(3ax^{2}+2bx+c)(6ax+2b).$$ Expanding the right side:

$$\begin{aligned}$$ (3ax^{2}+2bx+c)(6ax+2b) &=18a^{2}x^{3}+18abx^{2}+(4b^{2}+6ac)x+2bc. \end{aligned}

Equate coefficients with $$ax^{3}+bx^{2}+cx$$:

$$$ \begin{cases} \text{$$x^{3}$$:}& a = 18a^{2},\\[4pt] \text{$$x^{2}$$:}& b = 18ab,\\[4pt] \text{$$x$$:}& c = 4b^{2}+6ac,\\[4pt] \text{constant:}& 0 = 2bc. \end{cases} $$$

Cubic coefficient: $$a=18a^{2}\;\Longrightarrow\;a\neq0\ \Rightarrow\ 18a-1=0\;\Rightarrow\;a=\dfrac1{18}.$$

Constant term: $$0=2bc\;\Rightarrow\;b=0\ \text{or}\ c=0.$$

Quadratic coefficient: $$b=18ab=18\cdot\dfrac1{18}\,b=b,$$ always true.

Linear coefficient: $$c=4b^{2}+6ac=4b^{2}+6\!\left(\dfrac1{18}\right)c =4b^{2}+\dfrac13\,c,$$ so $$c-\dfrac13\,c=4b^{2}\;\Longrightarrow\;\dfrac23\,c=4b^{2} \;\Longrightarrow\;c=6b^{2}.$$ Combining with $$2bc=0$$ gives $$2b(6b^{2})=12b^{3}=0\;\Longrightarrow\;b=0\;\Longrightarrow\;c=0.$$

Step 5 Determine $$f(x)$$.
$$b=c=0,\qquad a=\dfrac1{18}.$$ Hence $$f(x)=\dfrac1{18}\,x^{3}.$$

Step 6 Compute the required quantities.
First derivatives: $$f'(x)=\dfrac16\,x^{2},\qquad f''(x)=\dfrac13\,x.$$

$$$ f'(2)=\dfrac16\,(2)^{2}=\dfrac{4}{6}=\dfrac23,\qquad f''(2)=\dfrac13\,(2)=\dfrac23. $$$

Integral: $$$ \int_{0}^{2}f(x)\,dx=\int_{0}^{2}\dfrac1{18}x^{3}\,dx =\dfrac1{18}\left[\dfrac{x^{4}}{4}\right]_{0}^{2} =\dfrac1{18}\cdot\dfrac{16}{4} =\dfrac{4}{18}=\dfrac{2}{9}. $$$

Step 7 Assemble the expression.
$$$ f'(2)+f''(2)+\int_{0}^{2}f(x)\,dx =\dfrac23+\dfrac23+\dfrac29 =\dfrac{4}{3}+\dfrac{2}{9} =\dfrac{12}{9}+\dfrac{2}{9} =\dfrac{14}{9}. $$$ Multiply by $$36$$: $$$ 36\left(\dfrac{14}{9}\right)=4\cdot14=56. $$$

Final answer: Option C which is: 56