To solve this problem, we'll first find the x-intercepts of the function $f(x) = 4(x + 3)^2(x + 1)(x - 1)$. Then we'll create a sign chart to analyze the behavior of the function in the intervals determined by these intercepts.

Step 1: Find the x-intercepts

The x-intercepts occur where $f(x) = 0$. Therefore, we set the function equal to zero and solve for $x$:

$f(x) = 4(x + 3)^2(x + 1)(x - 1) = 0$

This equation is satisfied when any of the factors equal zero:

1. $(x + 3)^2 = 0$
   $x + 3 = 0$
   $x = -3$ (with multiplicity 2)

2. $x + 1 = 0$
   $x = -1$

3. $x - 1 = 0$
   $x = 1$

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

Step 2: Make a sign chart

We will now analyze the sign of the function in the intervals determined by these intercepts. The intervals to consider are:

• $(-\infty, -3)$
• $(-3, -1)$
• $(-1, 1)$
• $(1, \infty)$

For each interval, we choose a test point and determine the sign of $f(x)$.

1. Interval $(-\infty, -3)$:
   Choose $x = -4$.
   $f(-4) = 4(-4 + 3)^2(-4 + 1)(-4 - 1) = 4(1)^2(-3)(-5) > 0$ So, $f(x) > 0$ in $(-\infty, -3)$.

2. Interval $(-3, -1)$:
   Choose $x = -2$.
   $f(-2) = 4(-2 + 3)^2(-2 + 1)(-2 - 1) = 4(1)^2(-1)(-3) > 0$ So, $f(x) > 0$ in $(-3, -1)$.

3. Interval $(-1, 1)$:
   Choose $x = 0$.
   $f(0) = 4(0 + 3)^2(0 + 1)(0 - 1) = 4(3)^2(1)(-1) < 0$ So, $f(x) < 0$ in $(-1, 1)$.

4. Interval $(1, \infty)$:
   Choose $x = 2$.
   $f(2) = 4(2 + 3)^2(2 + 1)(2 - 1) = 4(5)^2(3)(1) > 0$ So, $f(x) > 0$ in $(1, \infty)$.

Sign Chart Summary:

• $(-\infty, -3): f(x) > 0$
• $(-3, -1): f(x) > 0$
• $(-1, 1): f(x) < 0$
• $(1, \infty): f(x) > 0$

The function is positive in the intervals $(-\infty, -3)$, $(-3, -1)$, and $(1, \infty)$, and negative in $(-1, 1)$. The x-intercepts are $x = -3$ (where the sign does not change due to the even multiplicity), $x = -1$, and $x = 1$ (where the sign does change).

Do you have any questions, or would you like a more detailed explanation on any part of this problem?

Relative questions:

1. How does the multiplicity of an x-intercept affect the behavior of a function at that point?
2. What does the sign chart tell us about the graph of the function?
3. How would the function change if the coefficient of $f(x)$ was negative?
4. Can a sign chart help in determining the relative minima and maxima of the function?
5. How does one determine the end behavior of a polynomial function?

Tip: The multiplicity of a root affects whether the graph touches or crosses the x-axis at that intercept. If the multiplicity is even, the graph touches and turns around; if odd, it crosses the x-axis.