The problem is to find the roots of the cubic equation $x^3 + 11x^2 + 28x = 0$.

1. Problem Description

The problem is to find the roots of the cubic equation $x^3 + 11x^2 + 28x = 0$.

2. Solution Steps

We are given the equation $x^3 + 11x^2 + 28x = 0$.

First, we can factor out an $x$ from each term:

$x(x^2 + 11x + 28) = 0$.

This tells us that one solution is $x=0$.

Next, we need to factor the quadratic $x^2 + 11x + 28$.

We look for two numbers that multiply to 28 and add to

1

1. These numbers are 4 and 7, since $4 \cdot 7 = 28$ and $4 + 7 = 11$.

Therefore, we can factor the quadratic as $(x+4)(x+7)$.

The equation becomes $x(x+4)(x+7) = 0$.

Setting each factor to zero gives the roots:

$x = 0$, $x+4 = 0$ which means $x = -4$, and $x+7 = 0$ which means $x = -7$.

The roots are $0, -4, -7$.

3. Final Answer

0, -4, -7