Identify the multiplicities of the linear factors of h(x)=(x+4)(x−1)2(x+3)5.(1 point)

Question

Responses

The multiplicity of (x+4) is −4, the multiplicity of (x−1) is 1, and the multiplicity of (x+3) is −3.
The multiplicity of left parenthesis x plus 4 right parenthesis is negative 4 , the multiplicity of left parenthesis x minus 1 right parenthesis is 1, and the multiplicity of left parenthesis x plus 3 right parenthesis is negative 3 .

The multiplicity of (x+4) is 1, the multiplicity of (x−1) is 2, and the multiplicity of (x+3) is 5.
The multiplicity of left parenthesis x plus 4 right parenthesis is 1, the multiplicity of left parenthesis x minus 1 right parenthesis is 2, and the multiplicity of left parenthesis x plus 3 right parenthesis is 5.

The multiplicity of (x+4) is 4, the multiplicity of (x−1) is −1, and the multiplicity of (x+3) is 3.
The multiplicity of left parenthesis x plus 4 right parenthesis is 4, the multiplicity of left parenthesis x minus 1 right parenthesis is negative 1 , and the multiplicity of left parenthesis x plus 3 right parenthesis is 3.

The multiplicity of (x+4) is 0, the multiplicity of (x−1) is 2, and the multiplicity of (x+3) is 5.

Answer 1

To determine the multiplicities of the linear factors of the polynomial $h(x)=(x+4)(x-1{)}^2(x+3{)}^5$, we look at the powers of the linear factors in the expression.

1. For the factor $(x+4)$: It appears with an exponent of 1. Thus, the multiplicity is 1.

2. For the factor $(x-1{)}^2$: It appears with an exponent of 2. Hence, the multiplicity is 2.

3. For the factor $(x+3{)}^5$: It has an exponent of 5. So, the multiplicity is 5.

Putting this all together, we have:

• The multiplicity of $(x+4)$ is 1.
• The multiplicity of $(x-1)$ is 2.
• The multiplicity of $(x+3)$ is 5.

Thus, the correct response is:

The multiplicity of (x+4) is 1, the multiplicity of (x−1) is 2, and the multiplicity of (x+3) is 5.