Finding the constant term of a degree $3$ polynomial Let $p(x)=x^3+ax^2+bx+c$, where $a,b,c$ are real constants. If $p(-3)=p(2)=0$ and $p'(-3)<0$, which of the following is a possible value of $c$? 
A) $-27$ 
B) $-18$ 
C) $-6$ 
D) $-3$ 
E) $-\dfrac{1}{2}$

$\textbf{My attempt at this problem}$: 
  I drew a rough sketch of the curve on the $xy$-plane. A portion of the degree $3$ curve has a local max to the left of $-3$ and a local min between $-3$ and $2$. When $x> 2$, $p$ is a strictly increasing function, no longer intersecting the $x$-axis. 
Since $p(-3)=p(2)=0$, we plug in these integers to get 
  $$
p(-3)=(-3)^3 + a(-3)^2+b(-3)+c=0, \\ 
p(2) =(2)^3 + a(2)^2+b(2)+c=0.  
$$

There is another root when $x<-3$. 
Is there a trick to solve this problem without so much calculation? 

$p$ has three roots, $-3$, $2$ and $x_0$, say. We have 
$$ p(x) = x^3 + ax^2 + bx + c = (x+3)(x-2)(x-x_0). $$
So $c = 6x_0$. As, as you said, $x_0 < -3$, A) is the only possibility.
