Theory of Equations-1

Let the 3rd root be $\large \alpha$ .

Now by Vieta's Formula product of roots = $\dfrac{-(-12)}{3} = 4$

$\therefore 2 \times 2 \times \alpha = 4 \implies \alpha = 1$

So, required polynomial is ${(x - 2)^2} (x - 1) = 0$

$x^3 - 5x^2 + 8x - 4 = 0$

$3x^3 - 15x^2 + 24x - 12 = 0$

$p = -15$ and $q = 24$ $\implies p + q = \boxed {9}$