We are asked to solve the inequality:

$\left( x - 3 \right)^2 \left( x + 5 \right) \left( x + 7 \right) \leq 0$

Step 1: Identify the critical points

To find the critical points where the expression may change sign, we set the expression equal to zero:

$\left( x - 3 \right)^2 \left( x + 5 \right) \left( x + 7 \right) = 0$

The solutions to this equation are the values of $x$ that make the individual factors zero:

• $(x - 3)^2 = 0$ gives $x = 3$ (with multiplicity 2).
• $(x + 5) = 0$ gives $x = -5$.
• $(x + 7) = 0$ gives $x = -7$.

So, the critical points are $x = -7$, $x = -5$, and $x = 3$.

Step 2: Analyze the sign of the expression

We will now examine the sign of the expression in the intervals determined by the critical points: $(-\infty, -7)$, $(-7, -5)$, $(-5, 3)$, and $(3, \infty)$.

• For $x < -7$, all factors are negative, and the product of three negative terms is negative.

• For $-7 < x < -5$, $x + 7 > 0$, while $x + 5 < 0$ and $(x - 3)^2 > 0$. Hence, the product is positive.

• For $-5 < x < 3$, $x + 7 > 0$, $x + 5 > 0$, and $(x - 3)^2 > 0$. Hence, the product is positive.

• For $x > 3$, $x + 7 > 0$, $x + 5 > 0$, and $(x - 3)^2 > 0$. Hence, the product is positive.

Step 3: Include the critical points

We now check the critical points themselves:

• At $x = -7$, the expression is zero.
• At $x = -5$, the expression is zero.
• At $x = 3$, the expression is zero because $(x - 3)^2 = 0$.

Step 4: Solution

The inequality requires the product to be less than or equal to zero. This happens when the expression is either negative or zero.

From the analysis, the expression is negative in the interval $(-\infty, -7)$, and zero at $x = -7$, $x = -5$, and $x = 3$.

Thus, the solution to the inequality is:

$(-\infty, -7] \cup \{-5\} \cup \{3\}$

Would you like more details or have any questions?

Relative Questions:

1. How do multiplicities affect the sign of an expression?
2. What is the significance of the critical points in inequalities?
3. How do we analyze signs of expressions over intervals?
4. Can inequalities with even powers always include the root in the solution?
5. Why does the inequality hold at points where the expression equals zero?

Tip:

When dealing with inequalities involving factored polynomials, always test the sign of each factor in different intervals defined by the critical points.