$$(x+5)(x-4)\cdot(4+x)\cdot0 = 0$$
Answer
The equation has an infinite number of solutions.
Explanation
$$ \begin{aligned} (x+5)(x-4)\cdot(4+x)\cdot0 &= 0&& \text{simplify left side} \\[1 em](x^2-4x+5x-20)\cdot(4+x)\cdot0 &= 0&& \\[1 em](x^2+x-20)\cdot(4+x)\cdot0 &= 0&& \\[1 em](4x^2+x^3+4x+x^2-80-20x)\cdot0 &= 0&& \\[1 em](x^3+5x^2-16x-80)\cdot0 &= 0&& \\[1 em]0x^3+0x^2+0x+0 &= 0&& \\[1 em]0 &= 0&& \\[1 em] \end{aligned} $$
Since the statement $ \color{blue}{ 0 = 0 } $ is TRUE for any value of $ x $, we conclude that the equation has infinitely many solutions.
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