1. **State the problem:** We have a cubic polynomial $ax^{3}+bx^{2}+cx-4$. 2. When divided by $(x+2)$, the remainder is double the remainder when divided by $(x+1)$. 3. **Recall the Remainder Theorem:** The remainder when a polynomial $f(x)$ is divided by $(x-k)$ is $f(k)$. 4. Calculate the remainder when divided by $(x+2)$, i.e., at $x=-2$: $$f(-2) = a(-2)^3 + b(-2)^2 + c(-2) - 4 = -8a + 4b - 2c - 4$$ 5. Calculate the remainder when divided by $(x+1)$, i.e., at $x=-1$: $$f(-1) = a(-1)^3 + b(-1)^2 + c(-1) - 4 = -a + b - c - 4$$ 6. According to the problem: $$f(-2) = 2 imes f(-1)$$ 7. Substitute the expressions: $$-8a + 4b - 2c - 4 = 2(-a + b - c - 4)$$ 8. Expand the right side: $$-8a + 4b - 2c - 4 = -2a + 2b - 2c - 8$$ 9. Bring all terms to one side: $$-8a + 4b - 2c - 4 + 2a - 2b + 2c + 8 = 0$$ 10. Simplify: $$(-8a + 2a) + (4b - 2b) + (-2c + 2c) + (-4 + 8) = 0$$ $$-6a + 2b + 0 + 4 = 0$$ 11. Rearrange to solve for $b$: $$2b = 6a - 4$$ $$b = \frac{6a - 4}{2} = 3a - 2$$ **Final answer:** $$b = 3a - 2$$