$$\frac{3x^5+4x^4-2x^3}{x^3+2x+1} = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -\dfrac{ 2 }{ 3 }-\dfrac{\sqrt{ 10 }}{ 3 } & x_3 = -\dfrac{ 2 }{ 3 }+\dfrac{\sqrt{ 10 }}{ 3 } \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{3x^5+4x^4-2x^3}{x^3+2x+1} &= 0&& \text{multiply ALL terms by } \color{blue}{ x^3+2x+1 }. \\[1 em](x^3+2x+1)\frac{3x^5+4x^4-2x^3}{x^3+2x+1} &= (x^3+2x+1)\cdot0&& \text{cancel out the denominators} \\[1 em]3x^5+4x^4-2x^3 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 3x^{5}+4x^{4}-2x^{3} = 0 } $, first we need to factor our $ x^3 $.
$$ 3x^{5}+4x^{4}-2x^{3} = x^3 \left( 3x^{2}+4x-2 \right) $$$ x = 0 $ is a root of multiplicity $ 3 $.
The remaining roots can be found by solving equation $ 3x^{2}+4x-2 = 0$.
$ 3x^{2}+4x-2 = 0 $ is a quadratic equation.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this equation.
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