Find roots of polynomial $$ p(x) = 12x^3-36x-x^5 $$

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

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 2.4495\\[1 em]x_3 &= 2.4495\\[1 em]x_4 &= -2.4495\\[1 em]x_5 &= -2.4495 \end{aligned} $$

Step 1:

Write polynomial in descending order

$$ \begin{aligned} 12x^3-36x-x^5 & = 0\\[1 em] -x^5+12x^3-36x & = 0 \end{aligned} $$

Step 2:

Factor out $ \color{blue}{ -x }$ from $ -x^5+12x^3-36x $ and solve two separate equations:

$$ \begin{aligned} -x^5+12x^3-36x & = 0\\[1 em] \color{blue}{ -x }\cdot ( x^4-12x^2+36 ) & = 0 \\[1 em] \color{blue}{ -x = 0} ~~ \text{or} ~~ x^4-12x^2+36 & = 0 \end{aligned} $$

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

Step 3:

Polynomial $ x^4-12x^2+36 $ has no rational roots that can be found using Rational Root Test, so the roots were found using quartic formulas.

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