Find roots of polynomial $$ p(x) = -r^4-r^3+6r^2 $$

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

$$ \begin{aligned}r_1 &= 0\\[1 em]r_2 &= 2\\[1 em]r_3 &= -3 \end{aligned} $$

Step 1:

Factor out $ \color{blue}{ -r^2 }$ from $ -r^4-r^3+6r^2 $ and solve two separate equations:

$$ \begin{aligned} -r^4-r^3+6r^2 & = 0\\[1 em] \color{blue}{ -r^2 }\cdot ( r^2+r-6 ) & = 0 \\[1 em] \color{blue}{ -r^2 = 0} ~~ \text{or} ~~ r^2+r-6 & = 0 \end{aligned} $$

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

Step 2:

The solutions of $ r^2+r-6 = 0 $ are: $ r = -3 ~ \text{and} ~ r = 2$.

You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.

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