** 12. Solve the following inequality \( \frac{x(x+3)(x-3)+16}{(x+3)(x-2)} \geq-2 \). Hence without the use of graphic calculator, solve the inequality \( \frac{2^{x}\left(2^{x}+3\right)\left(2^{x}-3\right)+16}{2^{2 x}+2^{x}-6} \geq-2 \) where \( x \neq 1 \).

To tackle the first inequality \( \frac{x(x+3)(x-3)+16}{(x+3)(x-2)} \geq-2 \), we first simplify the left-hand side. Rewrite it as \( \frac{x(x+3)(x-3)+16 + 2(x+3)(x-2)}{(x+3)(x-2)} \geq 0 \). This will combine terms to allow us to investigate when the numerator is non-negative. By expanding and simplifying the numerator, we find \( x^3 - 9x + 12 \). You can then find the roots of this polynomial using synthetic division or the Rational Root Theorem, which means testing factors of the constant term over the leading coefficient. After solving for \( x \), you would test intervals based on these roots against the inequality \( x^3 - 9x + 12 \geq 0 \). For the second inequality \( \frac{2^{x}\left(2^{x}+3\right)\left(2^{x}-3\right)+16}{2^{2 x}+2^{x}-6} \geq-2 \), we recognize the terms \( 2^x = y \). Hence, rewrite it as \( \frac{y(y+3)(y-3)+16}{y^2+y-6} \geq-2 \). The same method applies: simplify the numerator, find roots, and check the sign of the expression across intervals identified by those roots. Remember, when dealing with rational inequalities, always consider where the denominator may be zero and exclude those values from your solution set. Lastly, double-check your intervals; it's easy to flip signs if you miss a negative factor! Good luck!