Question

find the zeros for the polynomial function and give the multiplicity for each zero. state whether the graph crosses the x - axis or touches the x - axis and turns around at each zero.
f(x)=x^{3}+8x^{2}-4x - 32
determine the zero(s), if they exist.
the zero(s) is/are - 8, - 2,2
(type integers or decimals. use a comma to separate answers as needed.)
determine the multiplicities of the zero(s), if they exist. select the correct choice below and, if necessary, fill in the answer box(es) within your choice.
a. there is one zero. the multiplicity of the zero is
(simplify your answer.)
b. there are two zeros. the multiplicity of the smallest zero is
the multiplicity of the largest zero is
(simplify your answers.)
c. there are three zeros. the multiplicity of the smallest zero is
the multiplicity of the other zero is 1. the multiplicity of the largest zero is 1. the multiplicity of the smallest zero is 1
(simplify your answers.)
determine the behavior of the function at each zero. select the correct choice below and, if necessary, fill in the answer boxes within your choice.
a. the graph crosses the x - axis at x =
the graph turns around at x =
(type integers or decimals. simplify your answers. use a comma to separate answers as needed.)
b. the graph touches the x - axis and turns around at all zeros.
c. the graph crosses the x - axis at all zeros.

Explanation:

Step1: Factor the polynomial

We have \(f(x)=x^{3}+8x^{2}-4x - 32\). Group the terms: \((x^{3}+8x^{2})-(4x + 32)=x^{2}(x + 8)-4(x + 8)=(x + 8)(x^{2}-4)\). Then, factor \(x^{2}-4=(x + 2)(x - 2)\). So \(f(x)=(x + 8)(x + 2)(x - 2)\).

Step2: Find the zeros

Set \(f(x)=0\). Then \((x + 8)(x + 2)(x - 2)=0\). By the zero - product property, \(x+8=0\) gives \(x=-8\), \(x + 2=0\) gives \(x=-2\), and \(x - 2=0\) gives \(x = 2\).

Step3: Determine multiplicities

Since each factor \((x + 8)\), \((x + 2)\) and \((x - 2)\) has an exponent of 1, the multiplicity of \(x=-8\), \(x=-2\) and \(x = 2\) is 1.

Step4: Determine the behavior at the zeros

If the multiplicity of a zero is odd, the graph of the function crosses the \(x\) - axis at that zero. Since the multiplicities of \(x=-8\), \(x=-2\) and \(x = 2\) are all 1 (odd), the graph crosses the \(x\) - axis at \(x=-8\), \(x=-2\) and \(x = 2\).

Answer:

The zeros are \(x=-8,-2,2\). The multiplicity of each zero is 1. The graph crosses the \(x\) - axis at \(x=-8\), \(x=-2\) and \(x = 2\).