jmc

Let the two common roots be $a$ and $b.$ Let the roots of the first cubic be $a,$ $b,$ and $c,$ and let the roots of the second cubic be $a,$ $b,$ and $d.$ Subtracting the cubics, we get $$4x^2+(q-r)=0.$$The roots of this quadratic are $a$ and $b,$ so $a+b=0.$

Then by Vieta's formulas, $a+b+c=-5$ and $a+b+d=-1.$ Then $c=-5$ and $d=-1,$ so $c+d=\boxed{-6}.$