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 are two zeros. the multiplicity of the smallest zero is. the multiplicity of the largest zero is. (simplify your answers.)
b. there is one zero. the multiplicity of the zero is. (simplify your answer.)
c. there are three zeros. the multiplicity of the smallest zero is. the multiplicity of the other zero is. the multiplicity of the largest zero is. (simplify your answers.)

Explanation:

Step1: Factor the polynomial

Factor \(f(x)=x^{3}+8x^{2}-4x - 32\) by grouping.
Group the terms: \((x^{3}+8x^{2})-(4x + 32)\).
Factor out the greatest - common factor from each group: \(x^{2}(x + 8)-4(x + 8)=(x + 8)(x^{2}-4)\).
Then factor \(x^{2}-4\) as a difference of squares: \((x + 8)(x - 2)(x+2)\).

Step2: Find the zeros

Set \(f(x)=0\).
\((x + 8)(x - 2)(x + 2)=0\).
Using the zero - product property \(a\cdot b\cdot c = 0\) implies \(a = 0\) or \(b = 0\) or \(c = 0\).
\(x+8=0\) gives \(x=-8\); \(x - 2=0\) gives \(x = 2\); \(x+2=0\) gives \(x=-2\).

Step3: Determine the multiplicities

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

Answer:

C. There are three zeros. The multiplicity of the smallest zero (\(x=-8\)) is 1. The multiplicity of the other zero (\(x=-2\)) is 1. The multiplicity of the largest zero (\(x = 2\)) is 1.