Question

$\frac{(x^2-4)^{30}(x^2-9)^9(x^2-3x+2)^{17}(3x^2+10)^{10}}{(x^2-5x+6)^{52}(x^2-25)^{60}(x^2+10)^{11}} \le 0$

Answer 1

[-3, 1] ∪ (2, 3)

Answer 2

The given inequality is $\frac{(x^2-4)^{30}(x^2-9)^9(x^2-3x+2)^{17}(3x^2+10)^{10}}{(x^2-5x+6)^{52}(x^2-25)^{60}(x^2+10)^{11}} \le 0$.

Factorize the expressions: $x^2-4 = (x-2)(x+2)$ $x^2-9 = (x-3)(x+3)$ $x^2-3x+2 = (x-1)(x-2)$ $x^2-5x+6 = (x-2)(x-3)$ $x^2-25 = (x-5)(x+5)$ $3x^2+10 > 0$ for all real $x$. $x^2+10 > 0$ for all real $x$.

Substitute the factors into the inequality: $\frac{((x-2)(x+2))^{30}((x-3)(x+3))^9((x-1)(x-2))^{17}(3x^2+10)^{10}}{((x-2)(x-3))^{52}((x-5)(x+5))^{60}(x^2+10)^{11}} \le 0$

Combine terms with the same base: $\frac{(x-2)^{30}(x+2)^{30}(x-3)^9(x+3)^9(x-1)^{17}(x-2)^{17}(3x^2+10)^{10}}{(x-2)^{52}(x-3)^{52}(x-5)^{60}(x+5)^{60}(x^2+10)^{11}} \le 0$ $\frac{(x-2)^{47}(x+2)^{30}(x-3)^9(x+3)^9(x-1)^{17}(3x^2+10)^{10}}{(x-2)^{52}(x-3)^{52}(x-5)^{60}(x+5)^{60}(x^2+10)^{11}} \le 0$

Since $(3x^2+10)^{10} > 0$ and $(x^2+10)^{11} > 0$ for all real $x$, we can divide by these terms: $\frac{(x-2)^{47}(x+2)^{30}(x-3)^9(x+3)^9(x-1)^{17}}{(x-2)^{52}(x-3)^{52}(x-5)^{60}(x+5)^{60}} \le 0$

Simplify the terms with common bases in the numerator and denominator: $\frac{1}{(x-2)^{5}} \frac{1}{(x-3)^{43}} (x+2)^{30}(x+3)^9(x-1)^{17} \frac{1}{(x-5)^{60}} \frac{1}{(x+5)^{60}} \le 0$ $\frac{(x+2)^{30}(x+3)^9(x-1)^{17}}{(x-2)^5(x-3)^{43}(x-5)^{60}(x+5)^{60}} \le 0$

Let $f(x) = \frac{(x+2)^{30}(x+3)^9(x-1)^{17}}{(x-2)^5(x-3)^{43}(x-5)^{60}(x+5)^{60}}$. The points where the denominator is zero must be excluded from the domain: $x \in \{2, 3, 5, -5\}$.

The terms with even powers are $(x+2)^{30}$, $(x-5)^{60}$, and $(x+5)^{60}$. $(x+2)^{30} \ge 0$. It is zero at $x=-2$. $(x-5)^{60} > 0$ for $x \neq 5$. $(x+5)^{60} > 0$ for $x \neq -5$.

For $x \notin \{-5, -2, 5\}$, the sign of $f(x)$ is the same as the sign of $\frac{(x+3)^9(x-1)^{17}}{(x-2)^5(x-3)^{43}}$. Let $g(x) = \frac{(x+3)^9(x-1)^{17}}{(x-2)^5(x-3)^{43}}$. The critical points for $g(x)$ are the roots of the numerator and denominator: $x=-3, 1, 2, 3$. These points divide the number line into intervals: $(-\infty, -3), (-3, 1), (1, 2), (2, 3), (3, \infty)$. The powers of the factors in $g(x)$ are all odd, so the sign of $g(x)$ changes at each critical point. Let's test a value in $(3, \infty)$, e.g., $x=4$: $g(4) = \frac{(4+3)^9(4-1)^{17}}{(4-2)^5(4-3)^{43}} = \frac{(+)^9(+)^{17}}{(+)^5(+)^{43}} = \frac{+}{+} = +$. The sign of $g(x)$ in the intervals is: $(-\infty, -3)$: $+$ $(-3, 1)$: $-$ $(1, 2)$: $+$ $(2, 3)$: $-$ $(3, \infty)$: $+$

We want $f(x) \le 0$. This occurs when $g(x) < 0$ or when $f(x)=0$.

Case 1: $g(x) < 0$. This happens for $x \in (-3, 1) \cup (2, 3)$. We must exclude the points where the original denominator is zero: $x \in \{2, 3, 5, -5\}$. The intervals $(-3, 1)$ and $(2, 3)$ do not contain $5$ or $-5$. The points $2$ and $3$ are the endpoints of the intervals, and they are excluded from the intervals. So, $x \in (-3, 1) \cup (2, 3)$ is part of the solution.

Case 2: $f(x) = 0$. $f(x) = 0$ when the numerator is zero and the denominator is non-zero. Numerator of $f(x)$ is $(x+2)^{30}(x+3)^9(x-1)^{17}$. It is zero at $x=-2, x=-3, x=1$. Denominator of $f(x)$ is $(x-2)^5(x-3)^{43}(x-5)^{60}(x+5)^{60}$. It is zero at $x=2, 3, 5, -5$. The values where the numerator is zero but the denominator is non-zero are $\{-2, -3, 1\}$. These points are solutions.

Combining the results from Case 1 and Case 2: The solution set is $((-3, 1) \cup (2, 3)) \cup \{-2, -3, 1\}$. Let's combine the intervals and points: $(-3, 1) \cup \{-3, 1\} = [-3, 1]$. So the solution is $[-3, 1] \cup (2, 3) \cup \{-2\}$. Since $-2$ is in the interval $[-3, 1]$, the union is simply $[-3, 1] \cup (2, 3)$.

We should double-check the excluded points from the original inequality denominator: $x \in \{2, 3, 5, -5\}$. Our solution set $[-3, 1] \cup (2, 3)$ does not include $2, 3, 5, -5$. The point $x=-2$ is included in $[-3, 1]$. Let's check the original inequality at $x=-2$. Numerator at $x=-2$: $( (-2)^2-4)^{30}(\dots) = (4-4)^{30}(\dots) = 0$. Denominator at $x=-2$: $((-2)^2-5(-2)+6)^{52}((-2)^2-25)^{60}((-2)^2+10)^{11} = (4+10+6)^{52}(4-25)^{60}(4+10)^{11} = (20)^{52}(-21)^{60}(14)^{11} \neq 0$. So, at $x=-2$, the inequality is $\frac{0}{\text{non-zero}} \le 0$, which is $0 \le 0$, which is true. So, $x=-2$ is indeed a solution. It is included in the interval $[-3, 1]$.

The final solution set is $[-3, 1] \cup (2, 3)$.