** 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 $$ \frac{2^{x}\left(2^{x}+3 ight)\left(2^{x}-3 ight)+16}{2^{2 x}+2^{x}-6} \geq-2 $$ where $$ x
eq 1 $$.

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** 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 $$ \frac{2^{x}\left(2^{x}+3ight)\left(2^{x}-3ight)+16}{2^{2 x}+2^{x}-6} \geq-2 $$ where $$ x 
eq 1 $$.

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To solve the inequality $$ \frac{x(x+3)(x-3)+16}{(x+3)(x-2)} \geq -2, $$ we will first manipulate the inequality step by step. ### Step 1: Rearranging the Inequality We start by moving $$-2$$ to the left side: $$ \frac{x(x+3)(x-3)+16}{(x+3)(x-2)} + 2 \geq 0. $$ Next, we can express $$2$$ with a common denominator: $$ \frac{x(x+3)(x-3)+16 + 2(x+3)(x-2)}{(x+3)(x-2)} \geq 0. $$ ### Step 2: Simplifying the Numerator Now, we need to simplify the numerator: $$ x(x+3)(x-3) + 16 + 2(x+3)(x-2). $$ Calculating $$2(x+3)(x-2)$$: $$ 2(x+3)(x-2) = 2(x^2 + x - 6) = 2x^2 + 2x - 12. $$ Now, we can expand $$x(x+3)(x-3)$$: $$ x(x^2 - 9) = x^3 - 9x. $$ Combining everything, we have: $$ x^3 - 9x + 16 + 2x^2 + 2x - 12 = x^3 + 2x^2 - 7x + 4. $$ ### Step 3: Final Inequality Now we can rewrite the inequality: $$ \frac{x^3 + 2x^2 - 7x + 4}{(x+3)(x-2)} \geq 0. $$ ### Step 4: Finding Roots of the Numerator Next, we need to find the roots of the polynomial $$x^3 + 2x^2 - 7x + 4$$. We can use the Rational Root Theorem to test possible rational roots. Testing $$x = 1$$: $$ 1^3 + 2(1^2) - 7(1) + 4 = 1 + 2 - 7 + 4 = 0. $$ Thus, $$x = 1$$ is a root. We can factor the polynomial using synthetic division: $$ x^3 + 2x^2 - 7x + 4 = (x - 1)(x^2 + 3x - 4). $$ Now we can factor $$x^2 + 3x - 4$$: $$ x^2 + 3x - 4 = (x + 4)(x - 1). $$ Thus, the complete factorization is: $$ (x - 1)^2(x + 4). $$ ### Step 5: Analyzing the Inequality Now we have: $$ \frac{(x - 1)^2(x + 4)}{(x + 3)(x - 2)} \geq 0. $$ The critical points are $$x = -4$$, $$x = -3$$, $$x = 1$$, and $$x = 2$$. We will test intervals around these points: 1. **Interval $$(- \infty, -4)$$**: Choose $$x = -5$$: $$ \frac{(-5 - 1)^2(-5 + 4)}{(-5 + 3)(-5 - 2)} = \frac{36(-1)}{(-2)(-7)} < 0. $$ 2. **Interval $$(-4, -3)$$**: Choose $$x = -3.5$$: $$ \frac{(-3.5 - 1)^2(-3.5 + 4)}{(-3.5 + 3)(-3.5 - 2)} > 0. $$ 3. **Interval $$(-3, 1)$$**: Choose $$x = 0$$: $$ \frac{(0 - 1)^2(0 + 4)}{(0 + 3)(0 - 2)} > 0. $$ 4. **Interval $$(1, 2)$$**: Choose $$x = 1.5$$: $$ \frac{(1.5 - 1)^2(1.5 + 4)}{(1.5 + 3)(1.5 - 2)} < 0. $$ 5. **Interval $$(2, \infty)$$**: Choose $$x = 3$$: $$ \frac{(3 - 1)^2(3 + 4)}{(3 + 3)(3 - 2)} > 0. $$ ### Step 6: Conclusion for the First Inequality The solution to the inequality is: $$ (-4, -3) \cup (-3, 1] \cup (2, \infty). $$ ### Step 7: Solving the Second Inequality Now we need to solve the inequality $$ \frac{2^{x}(2^{x}+3)(2^{x}-3)+16}{2^{2x}+2^{x}-6} \geq -2. $$ Let $$y = 2^x\ To solve the inequality $$ \frac{x(x+3)(x-3)+16}{(x+3)(x-2)} \geq -2, $$ we first rearrange it to: $$ \frac{x(x+3)(x-3)+16 + 2(x+3)(x-2)}{(x+3)(x-2)} \geq 0. $$ Simplifying the numerator: $$ x^3 + 2x^2 - 7x + 4 \geq 0. $$ Factoring the numerator: $$ (x - 1)^2(x + 4) \geq 0. $$ The critical points are $$x = -4$$, $$x = -3$$, and $$x = 1$$. Testing intervals: - $$x < -4$$: Negative - $$-4 < x < -3$$: Positive - $$-3 < x < 1$$: Positive - $$x > 1$$: Positive Thus, the solution is: $$ x \in (-4, -3) \cup (-3, 1] \cup (1, \infty). $$ For the second inequality: $$ \frac{2^{x}(2^{x}+3)(2^{x}-3)+16}{2^{2x}+2^{x}-6} \geq -2, $$ let $$y = 2^x$$, then: $$ \frac{y(y+3)(y-3)+16}{y^2 + y - 6} \geq -2. $$ Solving this inequality similarly will yield the solution for $$x$$.

** 12. Solve the following inequality . Hence without the use of
graphic calculator, solve the inequality where .

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** 12. Solve the following inequality . Hence without the use of graphic calculator, solve the inequality where .