Find roots of polynomial $$ p(x) = -4r^3-3r^2+12r $$

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

$$ \begin{aligned}r_1 &= 0\\[1 em]r_2 &= 1.3972\\[1 em]r_3 &= -2.1472 \end{aligned} $$

Step 1:

Factor out $ \color{blue}{ -r }$ from $ -4r^3-3r^2+12r $ and solve two separate equations:

$$ \begin{aligned} -4r^3-3r^2+12r & = 0\\[1 em] \color{blue}{ -r }\cdot ( 4r^2+3r-12 ) & = 0 \\[1 em] \color{blue}{ -r = 0} ~~ \text{or} ~~ 4r^2+3r-12 & = 0 \end{aligned} $$

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

Step 2:

The solutions of $ 4r^2+3r-12 = 0 $ are: $ r = -\dfrac{ 3 }{ 8 }-\dfrac{\sqrt{ 201 }}{ 8 } ~ \text{and} ~ r = -\dfrac{ 3 }{ 8 }+\dfrac{\sqrt{ 201 }}{ 8 }$.

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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