lot the x-intercepts and make a sign chart that represents the function shown below.
f, of, x, equals, 4, left parenthesis, x, plus, 3, right parenthesis, squared, left parenthesis, x, plus, 1, right parenthesis, left parenthesis, x, minus, 1, right parenthesis
f(x)=4(x+3)
2
(x+1)(x−1)

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.

Learn how to find the x-intercepts and create a sign chart for the function f(x) = 4(x + 3)^2(x + 1)(x - 1). This problem covers key concepts in quadratic equations, polynomial functions, and sign analysis, including the zero product property. Suitable for Grades 10-12 students studying advanced algebra.

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