Polynomial mania

Let $\quad f(x) = (x-\alpha)(x^3+3x^2+9x+r)$

$\Rightarrow (x-\alpha)(x^3+3x^2+9x+r) = x^4 +\left \lfloor \dfrac{2p}{5} \right \rfloor x^3 + 12x^2+4qx + r$

$x^4 + (3-\alpha) x^3 + (9-3\alpha)x^2+(r-9\alpha)x- \alpha r \\ = x^4 +\left \lfloor \dfrac{2p}{5} \right \rfloor x^3 + 12x^2+4qx + r$

Equating the coefficients:

$\begin{cases} \left \lfloor \dfrac{2p}{5} \right \rfloor = 3 - \alpha &...(1) \\ 9-3\alpha = 12 &...(2) \\ 4q = r-9\alpha &...(3) \\ r = - \alpha r &...(4) \end{cases}$

From $\text{ Eq.2 }$ and $\text{ Eq.4 }$ : $\quad \Rightarrow \alpha = -1$

From $\text{ Eq.2 }$ : $\quad \left \lfloor \dfrac{2p}{5} \right \rfloor = 4 \quad \Rightarrow 10 \le p \le 12$

From $\text{ Eq.3 }$ :

$q = \dfrac{r+9}{4} \quad \Rightarrow \dfrac{0+9}{4} < q < \dfrac{7+9}{4} \quad \Rightarrow 3 \le q < 4 \quad \Rightarrow q = 3$

From $\text{ Eq.3 }$ : $\quad r = 4q-9 = 3$

Therefore,

$\left \lfloor \dfrac {p+q+r-1}{3} \right \rfloor = \left \lfloor \dfrac {p +5}{3} \right \rfloor \quad \Rightarrow \left \lfloor \dfrac {15}{3} \right \rfloor \le \left \lfloor \dfrac {p+q+r-1}{3} \right \rfloor \le \left \lfloor \dfrac {17}{3} \right \rfloor \\ \Rightarrow \left \lfloor \dfrac {p+q+r-1}{3} \right \rfloor = \boxed{5}$