Find roots of polynomial $$ p(x) = -x^3-3x^2 $$

The roots of polynomial $ p(x) $ are:

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= -3 \end{aligned} $$

Step 1:

Factor out $ \color{blue}{ -x^2 }$ from $ -x^3-3x^2 $ and solve two separate equations:

$$ \begin{aligned} -x^3-3x^2 & = 0\\[1 em] \color{blue}{ -x^2 }\cdot ( x+3 ) & = 0 \\[1 em] \color{blue}{ -x^2 = 0} ~~ \text{or} ~~ x+3 & = 0 \end{aligned} $$

One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.

Step 2:

To find the last zero, solve equation $ x+3 = 0 $

$$ \begin{aligned} x+3 & = 0 \\[1 em] x & = -3 \end{aligned} $$

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