Question

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

Answer 1

[-3, 1] ∪ (2, 3) ∪ {4}

Answer 2

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

Factorize the polynomials:

$x^2-9 = (x-3)(x+3)$

$x^2-3x+2 = (x-1)(x-2)$

$x^2-5x+6 = (x-2)(x-3)$

$x^2-25 = (x-5)(x+5)$

$x^2+10$ is always positive for real $x$.

Substitute the factors into the inequality:

$\frac{(x-4)^{30}((x-3)(x+3))((x-1)(x-2))^{17}}{((x-2)(x-3))^{52}((x-5)(x+5))^{60}(x^2+10)^{11}} \le 0$

Combine terms with the same base:

$\frac{(x-4)^{30}(x-3)^1(x+3)^1(x-1)^{17}(x-2)^{17}}{(x-2)^{52}(x-3)^{52}(x-5)^{60}(x+5)^{60}(x^2+10)^{11}} \le 0$

Since $(x^2+10)^{11} > 0$, we can ignore it for the sign analysis.

$\frac{(x-4)^{30}(x-3)(x+3)(x-1)^{17}(x-2)^{17}}{(x-2)^{52}(x-3)^{52}(x-5)^{60}(x+5)^{60}} \le 0$

Simplify by canceling terms with common bases (subtracting exponents):

$\frac{(x-4)^{30}(x+3)(x-1)^{17}}{(x-2)^{52-17}(x-3)^{52-1}(x-5)^{60}(x+5)^{60}} \le 0$

$\frac{(x-4)^{30}(x+3)^1(x-1)^{17}}{(x-2)^{35}(x-3)^{51}(x-5)^{60}(x+5)^{60}} \le 0$

Let $f(x) = \frac{(x-4)^{30}(x+3)^1(x-1)^{17}}{(x-2)^{35}(x-3)^{51}(x-5)^{60}(x+5)^{60}}$.

The critical points are the values of $x$ where the numerator or denominator is zero: $-5, -3, 1, 2, 3, 4, 5$.

The points where the denominator is zero must be excluded from the solution: $x \neq 2, 3, 5, -5$.

We analyze the sign of $f(x)$ using the wavy curve method based on the critical points and the powers of the factors:

Critical points in increasing order: $-5, -3, 1, 2, 3, 4, 5$.

Powers of the corresponding factors:

$(x+5)^{60}$ (even power)

$(x+3)^1$ (odd power)

$(x-1)^{17}$ (odd power)

$(x-2)^{35}$ (odd power)

$(x-3)^{51}$ (odd power)

$(x-4)^{30}$ (even power)

$(x-5)^{60}$ (even power)

Let's determine the sign of $f(x)$ in the interval $(5, \infty)$. For $x > 5$, all factors are positive, so $f(x) > 0$.

Now, we move from right to left across the critical points, changing the sign if the power of the corresponding factor is odd, and keeping the sign if the power is even.

• At $x=5$ (power 60, even): sign does not change. Interval $(4, 5)$: $+$.

• At $x=4$ (power 30, even): sign does not change. Interval $(3, 4)$: $+$.

• At $x=3$ (power 51, odd): sign changes. Interval $(2, 3)$: $-$.

• At $x=2$ (power 35, odd): sign changes. Interval $(1, 2)$: $+$.

• At $x=1$ (power 17, odd): sign changes. Interval $(-3, 1)$: $-$.

• At $x=-3$ (power 1, odd): sign changes. Interval $(-5, -3)$: $+$.

• At $x=-5$ (power 60, even): sign does not change. Interval $(-\infty, -5)$: $+$.

Summary of the sign of $f(x)$:

$f(x) > 0$ for $x \in (-\infty, -5) \cup (-5, -3) \cup (1, 2) \cup (3, 4) \cup (4, 5) \cup (5, \infty)$.

$f(x) < 0$ for $x \in (-3, 1) \cup (2, 3)$.

We want $f(x) \le 0$. This includes intervals where $f(x) < 0$ and points where $f(x) = 0$.

From the sign analysis, $f(x) < 0$ for $x \in (-3, 1) \cup (2, 3)$.

Now consider where $f(x) = 0$. This happens when the numerator is zero and the denominator is non-zero.

Numerator is $(x-4)^{30}(x+3)(x-1)^{17}$. It is zero when $x=4$, $x=-3$, or $x=1$.

Denominator is $(x-2)^{35}(x-3)^{51}(x-5)^{60}(x+5)^{60}$. It is zero when $x=2, 3, 5, -5$.

The values where the numerator is zero are $\{-3, 1, 4\}$.

The values where the denominator is zero are $\{2, 3, 5, -5\}$.

The values where $f(x)=0$ are the values from $\{-3, 1, 4\}$ that are not in $\{2, 3, 5, -5\}$. These are $\{-3, 1, 4\}$.

So, $x=-3, 1, 4$ are solutions.

Combining the intervals where $f(x) < 0$ and the points where $f(x) = 0$:

Solution set = $(-3, 1) \cup (2, 3) \cup \{-3, 1, 4\}$.

This can be written as $[-3, 1] \cup (2, 3) \cup \{4\}$.

The points where the original denominator is zero ($x \in \{2, 3, 5, -5\}$) are excluded from the solution, which is consistent with our solution set.

The final answer is $[-3, 1] \cup (2, 3) \cup \{4\}$.