Question
$$\frac{29x^8-18x^5+12x^3}{3}x^3 = 0$$
Answer
$$ x = 0 $$
Explanation
$$ \begin{aligned} \frac{29x^8-18x^5+12x^3}{3}x^3 &= 0&& \text{multiply ALL terms by } \color{blue}{ 3 }. \\[1 em]3 \cdot \frac{29x^8-18x^5+12x^3}{3}x^3 &= 3\cdot0&& \text{cancel out the denominators} \\[1 em]29x^{11}-18x^8+12x^6 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 29x^{11}-18x^{8}+12x^{6} = 0 } $, first we need to factor our $ x^6 $.
$$ 29x^{11}-18x^{8}+12x^{6} = x^6 \left( 29x^{5}-18x^{2}+12 \right) $$$ x = 0 $ is a root of multiplicity $ 6 $.
The remaining roots can be found by solving equation $ 29x^{5}-18x^{2}+12 = 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