dentify the multiplicities of the linear factors of f(x)=(x+4)(x−1)2(x+3)5

Question

. Then find the zeros of f(x). State whether the function crosses the x-axis or bounces off the x-axis at each of the zeros.

Answer 1

To identify the multiplicities of the linear factors of the function \( f(x) = (x + 4)(x - 1)^2(x + 3)^5 \), let's break down each factor:

1. Factor \( (x + 4) \):

   • This factor has a multiplicity of 1.

2. Factor \( (x - 1)^2 \):

   • This factor has a multiplicity of 2.

3. Factor \( (x + 3)^5 \):

   • This factor has a multiplicity of 5.

Now, summarizing the multiplicities of the linear factors:

• \( x + 4 \) has a multiplicity of 1.
• \( x - 1 \) has a multiplicity of 2.
• \( x + 3 \) has a multiplicity of 5.

Next, we find the zeros of \( f(x) \):

• To find the zeros, we set each linear factor equal to zero:

1. For \( x + 4 = 0 \):

   • \( x = -4 \)

2. For \( x - 1 = 0 \):

   • \( x = 1 \)

3. For \( x + 3 = 0 \):

   • \( x = -3 \)

Thus, the zeros of \( f(x) \) are:

1. \( x = -4 \)
2. \( x = 1 \)
3. \( x = -3 \)

Next, we analyze whether the function crosses the x-axis or bounces off it at each of the zeros:

• A function crosses the x-axis at a zero if the multiplicity of that zero is odd.
• A function bounces off the x-axis at a zero if the multiplicity of that zero is even.

Now, looking at the multiplicities:

1. For \( x = -4 \) (multiplicity 1):

   • Odd multiplicity → crosses the x-axis.

2. For \( x = 1 \) (multiplicity 2):

   • Even multiplicity → bounces off the x-axis.

3. For \( x = -3 \) (multiplicity 5):

   • Odd multiplicity → crosses the x-axis.

In summary:

• The zeros of \( f(x) \) are \( x = -4 \) (cross), \( x = 1 \) (bounce), \( x = -3 \) (cross).