Question
$$4^2x\cdot25^2+3^2x\cdot\frac{125}{2^3}x\cdot5^2 = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -\dfrac{ 128 }{ 45 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 4^2x\cdot25^2+3^2x\cdot\frac{125}{2^3}x\cdot5^2 &= 0&& \text{multiply ALL terms by } \color{blue}{ 8 }. \\[1 em]84^2x\cdot25^2+83^2x\cdot\frac{125}{2^3}x\cdot5^2 &= 8\cdot0&& \text{cancel out the denominators} \\[1 em]80000x+28125x^2 &= 0&& \text{simplify left side} \\[1 em]28125x^2+80000x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 28125x^{2}+80000x = 0 } $, first we need to factor our $ x $.
$$ 28125x^{2}+80000x = x \left( 28125x+80000 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ 28125x+80000 = 0$.
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