To create the sign chart and find the x-intercepts of the function:

Given function:

$f(x) = (x + 5)(x + 3)(x + 1)(x - 2)^2$

Step 1: Find the x-intercepts (where $f(x) = 0$).

The x-intercepts occur when each factor is zero:

• $x + 5 = 0 \implies x = -5$
• $x + 3 = 0 \implies x = -3$
• $x + 1 = 0 \implies x = -1$
• $(x - 2)^2 = 0 \implies x = 2$ (Note: this is a double root because it's squared.)

Thus, the x-intercepts are $x = -5$, $x = -3$, $x = -1$, and $x = 2$.

Step 2: Create the sign chart.

We need to analyze the sign of $f(x)$ in different intervals based on the x-intercepts. The critical points are $-5$, $-3$, $-1$, and $2$. These points divide the real number line into the following intervals:

1. $(-\infty, -5)$
2. $(-5, -3)$
3. $(-3, -1)$
4. $(-1, 2)$
5. $(2, \infty)$

We'll check the sign of $f(x)$ in each interval by testing a value from each interval and checking the overall sign of the product of the factors. Here's the analysis:

For $(-\infty, -5)$:

Pick a test point like $x = -6$: $f(-6) = (-6+5)(-6+3)(-6+1)(-6-2)^2 = (-1)(-3)(-5)(-8)^2 = (-)(-)(-)(+) = -$ Thus, $f(x)$ is negative in this interval.

For $(-5, -3)$:

Pick a test point like $x = -4$: $f(-4) = (-4+5)(-4+3)(-4+1)(-4-2)^2 = (1)(-1)(-3)(-6)^2 = (+)(-)(-)(+) = +$ Thus, $f(x)$ is positive in this interval.

For $(-3, -1)$:

Pick a test point like $x = -2$: $f(-2) = (-2+5)(-2+3)(-2+1)(-2-2)^2 = (3)(1)(-1)(-4)^2 = (+)(+)(-)(+) = -$ Thus, $f(x)$ is negative in this interval.

For $(-1, 2)$:

Pick a test point like $x = 0$: $f(0) = (0+5)(0+3)(0+1)(0-2)^2 = (5)(3)(1)(-2)^2 = (+)(+)(+)(+) = +$ Thus, $f(x)$ is positive in this interval.

For $(2, \infty)$:

Pick a test point like $x = 3$: $f(3) = (3+5)(3+3)(3+1)(3-2)^2 = (8)(6)(4)(1)^2 = (+)(+)(+)(+) = +$ Thus, $f(x)$ is positive in this interval.

Step 3: Summarize the sign chart.

The sign of $f(x)$ in each interval is:

• $(-\infty, -5)$ → negative
• $(-5, -3)$ → positive
• $(-3, -1)$ → negative
• $(-1, 2)$ → positive
• $(2, \infty)$ → positive

Key Observations:

• At $x = -5$, $x = -3$, and $x = -1$, the function changes sign (simple roots).
• At $x = 2$, there is no sign change because it's a double root, so the function stays positive.

Graphical Summary (Sign Chart):

x:  (-∞)   -5    -3    -1    2    (+∞)
f(x):   -     0     +     0     -     0     +     0     +

This chart represents how the function behaves around its intercepts.

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Would you like further details or have any questions? Here are 5 related questions:

1. How do double roots affect the shape of the graph at the intercept?
2. Why doesn't the sign change at $x = 2$?
3. What role do the multiplicities of roots play in function behavior?
4. How would the sign chart change if one of the factors had a higher degree?
5. How can we determine the end behavior of this function?

Tip: Double roots (even powers of a factor) make the function "bounce" off the x-axis, while single roots cause the graph to cross the x-axis.