Let's analyze the polynomial function \( f(x) \) given by:
[tex]\[ f(x) = 2(x+4)^2(x-5)^2(x-7)^3 \][/tex]
To find the zeros and their multiplicities, we will examine each factor in the polynomial.
1. Identifying Zeros:
- From the factor \((x+4)^2\), set \( x+4 = 0 \):
[tex]\[ x = -4 \][/tex]
- From the factor \((x-5)^2\), set \( x-5 = 0 \):
[tex]\[ x = 5 \][/tex]
- From the factor \((x-7)^3\), set \( x-7 = 0 \):
[tex]\[ x = 7 \][/tex]
2. Determining Multiplicities:
- The factor \((x+4)^2\) has a zero at \( x = -4 \) with multiplicity 2.
- The factor \((x-5)^2\) has a zero at \( x = 5 \) with multiplicity 2.
- The factor \((x-7)^3\) has a zero at \( x = 7 \) with multiplicity 3.
3. Categorize Each Zero According to Its Multiplicity:
- Zero(s) of multiplicity one:
There are no zeros in the polynomial \( f(x) \) with a multiplicity of one.
- Zero(s) of multiplicity two:
The zeros \( x = -4 \) and \( x = 5 \) each have a multiplicity of two.
- Zero(s) of multiplicity three:
The zero \( x = 7 \) has a multiplicity of three.
In summary:
- Zero(s) of multiplicity one: \(\) (no zeros)
- Zero(s) of multiplicity two: \(-4, 5\)
- Zero(s) of multiplicity three: \(7\)
So the answer is:
- Zero(s) of multiplicity one:
- Zero(s) of multiplicity two: \(-4, 5\)
- Zero(s) of multiplicity three: [tex]\(7\)[/tex]
