Real Numbers are Roots $r, s$. 
Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$, and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$. Find the sum of all possible values of $|b|$.

Using Vieta's Formulas,
$r+s+x_1$ $=0$ $\Rightarrow x_1$ $=-r-s$, where $x_1$ is the third root. Similarly, $x_2=-r-s-1$ $=x_1-1$, where $x_2$ is the third root of $q(x)$.
I have the list here:

$a_n = a_n$

$a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)$
$a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)$
$\vdots$
$a_0 = (-1)^n a_n r_1r_2\cdots r_n$
Obviously, $b = a_0$ so for $p(x)$:
$$b = (-1)^{3}(1)(r)(s)(-1)(r + s) = (r)(s)(r+s) = r^2s + s^2r$$
For $q(x)$ then, 
$$a_0 = (b + 240) = (-1)^{3}(1)(r+4)(s-3)(-1)(r + s + 1) = (r+4)(s-3)(r + s + 1) $$
$$=  r^2s - 3r^2 + rs^2 + 2rs - 15r + 4s^2 -8s - 12$$
$$b = r^2s - 3r^2 + rs^2 + 2rs - 15r + 4s^2 -8s - 252$$
But that leaves an awfully weird system of equations.
PLEASE DO NOT GIVE ME A FULL ANSWER, Just help.

Since $r,s$ are two roots of $x^3+ax+b=0$, we have
$$ r^3+ar+b=0,s^3+as+b=0 \tag{1}$$
and hence
$$ (r^3-s^3)+a(r-s)=0. $$
Assuming $r-s\neq0$, we have
$$ r^2+rs+s^2+a=0.\tag{2}$$
Similarly since $r+4,s-3$ are two roots of $x^3+ax+b+240=0$, we have
$$ (r+4)^3+a(r+4)+b+240=0,(s-3)^3+a(s-3)+b+240=0\tag{3}$$
and hence
$$ [(r+4)^3-(s-3)^3]+a(r-s+7)=0.$$
Assuming $r-s+7\neq 0$, we have
$$ (r+4)^2+(r+4)(s-3)+(s-3)^2+a=0.\tag{4}$$
From (4)-(3) gives
$$ 13+5r-2s=0 $$
from which we find that 
$$s=\frac{1}{2}(5r+13).\tag{5}$$
Putting this in (2), we obtain $a$ in terms of $r$. 
Putting them in the first equation of (1), we have $b$ in terms of $r$.
Putting $s,a,b$ in the first equation of (3), we will have an equation of $r$ which you can get the values $r$. I think you can do the rest.
