To determine the zeros of the polynomial function [tex]\( f(x) = 4(x+5)(x+2)(x-11)^3 \)[/tex] and their multiplicities, follow these steps:
1. Identify the factors of the polynomial:
- The polynomial is given in its factored form: [tex]\( f(x) = 4(x+5)(x+2)(x-11)^3 \)[/tex].
2. Set each factor equal to zero to find the zeros:
- For the factor [tex]\((x+5)\)[/tex]:
[tex]\[
x + 5 = 0 \implies x = -5
\][/tex]
- For the factor [tex]\((x+2)\)[/tex]:
[tex]\[
x + 2 = 0 \implies x = -2
\][/tex]
- For the factor [tex]\((x-11)^3\)[/tex]:
[tex]\[
x - 11 = 0 \implies x = 11
\][/tex]
3. Determine the multiplicity of each zero:
- The factor [tex]\((x+5)\)[/tex] appears once in the polynomial, so the zero [tex]\( x = -5 \)[/tex] has a multiplicity of one.
- The factor [tex]\((x+2)\)[/tex] appears once in the polynomial, so the zero [tex]\( x = -2 \)[/tex] has a multiplicity of one.
- The factor [tex]\((x-11)\)[/tex] is raised to the third power, so the zero [tex]\( x = 11 \)[/tex] has a multiplicity of three.
4. List the zeros according to their multiplicity:
- Zero(s) of multiplicity one: [tex]\(-5, -2\)[/tex]
- Zero(s) of multiplicity two: None
- Zero(s) of multiplicity three: [tex]\(11\)[/tex]
Thus, the final answers are:
[tex]\[
\begin{align*}
\text{Zero(s) of multiplicity one:} & \quad -5, -2 \\
\text{Zero(s) of multiplicity two:} & \quad \text{None} \\
\text{Zero(s) of multiplicity three:} & \quad 11 \\
\end{align*}
\][/tex]
