Решите неравенство методом интервалов. (9x^2-4)(16-x^2)(2x^2+3) больше 0

[latex](9x^2-4)(16-x^2)(2x^2+3)\ \textgreater \ 0\\\\(3x-2)(3x+2)(4-x)(4+x)(2x^2+3)\ \textgreater \ 0\\\\(3x-2)(3x+2)(x-4)(x+4)(2x^2+3)\ \textless \ 0\\\\2x^2+3\ \textgreater \ 0\; \; pri\; \; x\in R\\\\+++(-4)---(-\frac{2}{3})+++(\frac{2}{3})---(4)+++\\\\x\in (-4,-\frac{2}{3})\cup (\frac{2}{3},4)[/latex]