Question
$$3x^{32}+8x^2-(x^3+6x^2) = 0$$
Answer
$$ x = 0 $$
Explanation
$$ \begin{aligned} 3x^{32}+8x^2-(x^3+6x^2) &= 0&& \text{simplify left side} \\[1 em]3x^{32}+8x^2-x^3-6x^2 &= 0&& \\[1 em]3x^{32}-x^3+2x^2 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 3x^{32}-x^{3}+2x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ 3x^{32}-x^{3}+2x^{2} = x^2 \left( 3x^{30}-x+2 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The remaining roots can be found by solving equation $ 3x^{30}-x+2 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
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Polynomial Equations Solver