Answer:
[tex]-9, -9, 9, 9, 9[/tex]
Step-by-step explanation:
The given expression is
[tex]f(x)=3(x+9)^{2} (x-9)^{3}[/tex]
From the expression, we observe that [tex](x+9)[/tex] has a multiplicity of 2 and [tex](x-9)[/tex] has a multiplicity of 3.
So, for the first factor, there's one number that makes such factor equal to zero, and that is -9, because
[tex]x+9=0\\x=-9[/tex]
We know that this factor has a multiplicity of 2, so its zeros are -9 and -9.
We do the same process for [tex](x-9)[/tex]
[tex]x-9=0\\x=9[/tex]
Its multiplicity is 3, so the zeros are 9, 9 and 9.
Therefore, all the zeros of the polynomial function are
[tex]-9, -9, 9, 9, 9[/tex]
