Find all the zeros of the equation.

Question

\[-x^{3}-3 x^{2}=6 x+4\]

Answer 1

To find the zeros of the equation, we need to solve the equation $-x^{3}-3 x^{2}-6 x-4=0$.

One way to find the zeros is by using factoring. Let's rearrange the equation by moving all the terms to one side:
\[x^{3}+3x^{2}+6x+4=0.\]
Since all the coefficients of the terms are positive, there are no positive zeros. The possible negative zeros are $-1, -2, -4$. We can check if any of these are zeros by substituting them into the equation. If the substitution yields 0, then that number is a zero of the equation.

For $x=-1$:
\[(-1)^{3}+3(-1)^{2}+6(-1)+4 = -1+3-6+4 = 0,\]
so $x=-1$ is a zero of the equation.

For $x=-2$:
\[(-2)^{3}+3(-2)^{2}+6(-2)+4 = -8+12-12+4 = -4,\]
so $x=-2$ is not a zero of the equation.

For $x=-4$:
\[(-4)^{3}+3(-4)^{2}+6(-4)+4 = -64+48-24+4 = -36,\]
so $x=-4$ is not a zero of the equation.

Therefore, the only zero of the equation is $x=-1$.