The image contains three inequalities that need to be solved. Let's go through them one by one:

1. $(x^2 - 10x + 9)(4x + 1)^2 > 0$

Here, we need to solve the inequality: $(x^2 - 10x + 9)(4x + 1)^2 > 0$

1. Factorize the quadratic expression $x^2 - 10x + 9$: $x^2 - 10x + 9 = (x - 9)(x - 1)$ So the inequality becomes: $(x - 9)(x - 1)(4x + 1)^2 > 0$

2. The term $(4x + 1)^2$ is always non-negative (since a square is always $\geq 0$). It will equal 0 only when $x = -\frac{1}{4}$. So, we just need to find where $(x - 9)(x - 1)$ is positive.

3. Solve $(x - 9)(x - 1) > 0$ using a sign chart:

   • The roots are $x = 9$ and $x = 1$, which divide the number line into intervals: $(-\infty, 1)$, $(1, 9)$, and $(9, \infty)$.
   • Test points from each interval:
     • For $x < 1$, both factors are negative, so the product is positive.
     • For $1 < x < 9$, one factor is negative and one is positive, so the product is negative.
     • For $x > 9$, both factors are positive, so the product is positive.

4. Since $(4x + 1)^2 > 0$ for all $x \neq -\frac{1}{4}$, the solution to the inequality is: $(-\infty, 1) \cup (9, \infty)$ excluding $x = -\frac{1}{4}$.

2. $(x^2 - 10x + 9)(4x + 1)^2 \leq 0$

This inequality is similar to the first, except we now have $\leq 0$.

1. Factorize the quadratic again: $(x - 9)(x - 1)(4x + 1)^2 \leq 0$.

2. As $(4x + 1)^2$ is non-negative and only zero when $x = -\frac{1}{4}$, we focus on $(x - 9)(x - 1)$:

   • $(x - 9)(x - 1) = 0$ when $x = 1$ or $x = 9$.
   • The quadratic product $(x - 9)(x - 1)$ is negative between $1$ and $9$.

3. The solution will include the points where the product equals zero ($x = 1$ or $x = 9$) and the interval where the product is negative. Therefore, the solution is: $[1, 9]$

3. $(x^2 - 6x + 8)(x^2 - 4) \leq 0$

Here, we need to solve: $(x^2 - 6x + 8)(x^2 - 4) \leq 0$

1. Factorize each quadratic:

   • $x^2 - 6x + 8 = (x - 4)(x - 2)$
   • $x^2 - 4 = (x - 2)(x + 2)$

   The inequality becomes: $(x - 4)(x - 2)(x + 2)(x - 2) \leq 0$ or, simplified: $(x - 4)(x - 2)^2(x + 2) \leq 0$

2. The term $(x - 2)^2$ is always non-negative and equals zero when $x = 2$.

3. Analyze the sign of the product $(x - 4)(x + 2)$ using a sign chart:

   • The roots are $x = -2$ and $x = 4$, dividing the number line into intervals: $(-\infty, -2)$, $(-2, 4)$, and $(4, \infty)$.
   • Test points in each interval:
     • For $x < -2$, both factors are negative, so the product is positive.
     • For $-2 < x < 4$, one factor is negative and the other positive, so the product is negative.
     • For $x > 4$, both factors are positive, so the product is positive.

4. The solution is where the product is zero or negative. The solution is: $[-2, 4]$

Summary of Solutions:

1. $(- \infty, 1) \cup (9, \infty)$, excluding $x = -\frac{1}{4}$.
2. $[1, 9]$
3. $[-2, 4]$

Would you like more details on any specific steps or clarifications?

Related Questions:

1. How do we factor quadratic expressions efficiently?
2. What is the importance of a sign chart in solving inequalities?
3. Why does $(4x + 1)^2 \geq 0$ for all real $x$?
4. How do we solve inequalities involving higher powers or more complex expressions?
5. What are the key differences between strict ($>$) and non-strict ($\geq$) inequalities?

Tip: Always test values between roots to understand how the sign of an expression changes across different intervals.