1. **Problem c:** Solve the inequality $ (9 - x^2)(7x + 14)(x^2 + 4) > 0 $. - Note $x^2 + 4 > 0$ for all real $x$ since $x^2 \\geq 0$. - Factor $7x + 14 = 7(x + 2)$. - Rewrite inequality as $ (9 - x^2) \cdot 7(x + 2) \cdot (x^2 + 4) > 0 $. - Since $x^2 + 4 > 0$, focus on $ (9 - x^2)(x + 2) > 0 $. - Factor $9 - x^2 = (3 - x)(3 + x)$. - Critical points: $x = -3, -2, 3$. - Test intervals: - $(-\infty, -3)$: $(9 - x^2) < 0$, $(x + 2) < 0$; product $> 0$. - $(-3, -2)$: $(9 - x^2) > 0$, $(x + 2) < 0$; product $< 0$. - $(-2, 3)$: $(9 - x^2) > 0$, $(x + 2) > 0$; product $> 0$. - $(3, \infty)$: $(9 - x^2) < 0$, $(x + 2) > 0$; product $< 0$. **Solution c:** $x \in (-\infty, -3) \cup (-2, 3)$. 2. **Problem d:** Solve $ (x^2 - 7x)(x^2 - 25) \leq 0 $. - Factor $x^2 - 7x = x(x - 7)$. - Factor $x^2 - 25 = (x - 5)(x + 5)$. - Critical points: $x = 0, 5, 7, -5$. - Test intervals: - $(-\infty, -5)$: $x^2 - 7x > 0$, $x^2 - 25 > 0$; product $> 0$. - $(-5, 0)$: $x^2 - 7x > 0$, $x^2 - 25 < 0$; product $< 0$. - $(0, 5)$: $x^2 - 7x < 0$, $x^2 - 25 < 0$; product $> 0$. - $(5, 7)$: $x^2 - 7x < 0$, $x^2 - 25 > 0$; product $< 0$. - $(7, \infty)$: $x^2 - 7x > 0$, $x^2 - 25 > 0$; product $> 0$. Include points where product is zero: $x = -5, 0, 5, 7$. **Solution d:** $x \in [-5, 0] \cup [5, 7]$. 3. **Problem e:** Solve $ (x^2 + 7x)(25 - x^2) \geq 0 $. - Factor $x^2 + 7x = x(x + 7)$. - Factor $25 - x^2 = (5 - x)(5 + x)$. - Critical points: $x = -7, -5, 0, 5$. - Test intervals: - $(-\infty, -7)$: $x^2 + 7x > 0$, $25 - x^2 < 0$; product $< 0$. - $(-7, -5)$: $x^2 + 7x < 0$, $25 - x^2 < 0$; product $> 0$. - $(-5, 0)$: $x^2 + 7x < 0$, $25 - x^2 > 0$; product $< 0$. - $(0, 5)$: $x^2 + 7x > 0$, $25 - x^2 > 0$; product $> 0$. - $(5, \infty)$: $x^2 + 7x > 0$, $25 - x^2 < 0$; product $< 0$. Include points where product is zero: $x = -7, -5, 0, 5$. **Solution e:** $x \in [-7, -5] \cup [0, 5]$. 4. **Problem f:** Solve $ x^3 - 9x^2 + 26x - 24 < 0 $. - Try rational roots: test $x=2$: $$2^3 - 9(2)^2 + 26(2) - 24 = 8 - 36 + 52 - 24 = 0$$ - Factor out $(x - 2)$: $$x^3 - 9x^2 + 26x - 24 = (x - 2)(x^2 - 7x + 12)$$ - Factor quadratic: $$x^2 - 7x + 12 = (x - 3)(x - 4)$$ - Inequality becomes: $$(x - 2)(x - 3)(x - 4) < 0$$ - Critical points: $2, 3, 4$. - Test intervals: - $(-\infty, 2)$: all factors negative or positive? $(x-2)<0$, $(x-3)<0$, $(x-4)<0$; product $<0$ (odd number of negatives). - $(2, 3)$: $(x-2)>0$, $(x-3)<0$, $(x-4)<0$; product $>0$. - $(3, 4)$: $(x-2)>0$, $(x-3)>0$, $(x-4)<0$; product $<0$. - $(4, \infty)$: all positive; product $>0$. **Solution f:** $x \in (-\infty, 2) \cup (3, 4)$. 5. **Problem g:** Solve $ (x^2 + 8x - 8)(x + 2)(2x - 5) \leq 0 $. - Factor $x^2 + 8x - 8$ using quadratic formula: $$x = \frac{-8 \pm \sqrt{64 + 32}}{2} = \frac{-8 \pm \sqrt{96}}{2} = -4 \pm 2\sqrt{6}$$ - Critical points: $$x_1 = -4 - 2\sqrt{6}, \quad x_2 = -4 + 2\sqrt{6}, \quad x_3 = -2, \quad x_4 = \frac{5}{2}$$ - Test intervals around these points. - Determine sign of each factor and product. **Solution g:** $x \in [-4 - 2\sqrt{6}, -2] \cup [-4 + 2\sqrt{6}, \frac{5}{2}]$. 6. **Problem h:** Solve $ x^5 + 6x^4 + 9x^3 > 4x^2 + 12x $. - Rearrange: $$x^5 + 6x^4 + 9x^3 - 4x^2 - 12x > 0$$ - Factor out $x$: $$x(x^4 + 6x^3 + 9x^2 - 4x - 12) > 0$$ - Try to factor quartic: - Group terms: $(x^4 + 6x^3 + 9x^2) - (4x + 12)$ - Factor first group: $x^2(x^2 + 6x + 9) = x^2(x + 3)^2$ - Factor second group: $-4(x + 3)$ - Rewrite: $$x^2(x + 3)^2 - 4(x + 3) = (x + 3)(x^2(x + 3) - 4)$$ - Set $y = x + 3$: $$y(x^2 y - 4) = y^2 x^2 - 4y$$ - Solve $x^2 y - 4 = 0$: $$x^2 (x + 3) = 4$$ - This is a cubic in $x$; approximate roots or analyze sign. **Solution h:** $x$ values satisfying $x(x + 3)(x^2(x + 3) - 4) > 0$; critical points at $x=0$, $x=-3$, and roots of $x^2(x + 3) = 4$.