Question
Find roots of polynomial $$ p(x) = 32y^3-6y-3y^2 $$
Answer
The roots of polynomial $ p(y) $ are:
$$ \begin{aligned}y_1 &= 0\\[1 em]y_2 &= 0.4824\\[1 em]y_3 &= -0.3887 \end{aligned} $$Explanation
Step 1:
Write polynomial in descending order
$$ \begin{aligned} 32y^3-6y-3y^2 & = 0\\[1 em] 32y^3-3y^2-6y & = 0 \end{aligned} $$Step 2:
Factor out $ \color{blue}{ y }$ from $ 32y^3-3y^2-6y $ and solve two separate equations:
$$ \begin{aligned} 32y^3-3y^2-6y & = 0\\[1 em] \color{blue}{ y }\cdot ( 32y^2-3y-6 ) & = 0 \\[1 em] \color{blue}{ y = 0} ~~ \text{or} ~~ 32y^2-3y-6 & = 0 \end{aligned} $$One solution is $ \color{blue}{ y = 0 } $. Use second equation to find the remaining roots.
Step 3:
The solutions of $ 32y^2-3y-6 = 0 $ are: $ y = \dfrac{ 3 }{ 64 }-\dfrac{\sqrt{ 777 }}{ 64 } ~ \text{and} ~ y = \dfrac{ 3 }{ 64 }+\dfrac{\sqrt{ 777 }}{ 64 }$.
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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