$$x \cdot \frac{x+2}{x-5}(x+3) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -2 & x_3 = -3 \end{matrix} $$
Explanation
$$ \begin{aligned} x \cdot \frac{x+2}{x-5}(x+3) &= 0&& \text{simplify left side} \\[1 em]\frac{x^2+2x}{x-5}(x+3) &= 0&& \\[1 em]\frac{x^3+5x^2+6x}{x-5} &= 0&& \text{multiply ALL terms by } \color{blue}{ x-5 }. \\[1 em](x-5)\frac{x^3+5x^2+6x}{x-5} &= (x-5)\cdot0&& \text{cancel out the denominators} \\[1 em]x^3+5x^2+6x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{3}+5x^{2}+6x = 0 } $, first we need to factor our $ x $.
$$ x^{3}+5x^{2}+6x = x \left( x^{2}+5x+6 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ x^{2}+5x+6 = 0$.
$ x^{2}+5x+6 = 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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