Answer:
Factored form of function: [tex]f(x)=(x+9)(x-9)(x+5)[/tex].
Zeros of function: [tex]x=-9,-5,9[/tex]
Step-by-step explanation:
We have been given a function [tex]f(x)=(x^2-81)(x+5)[/tex]. We are asked to find the factored form and zeros of our given function.
Using difference of squares [tex]a^2-b^2=(a+b)(a-b)[/tex], we can rewrite our given function as:
[tex]f(x)=(x^2-9^2)(x+5)[/tex]
[tex]f(x)=(x+9)(x-9)(x+5)[/tex]
Therefore, the factored form of our given function is [tex]f(x)=(x+9)(x-9)(x+5)[/tex].
To find the zeros of our given function, we will use zero product property. Upon equation our given function equals to zero we will get,
[tex](x+9)(x-9)(x+5)=0[/tex]
[tex](x+9)=0\text{ or }(x-9)=0\text{ or }(x+5)=0[/tex]
[tex]x+9=0\text{ or }x-9=0\text{ or }x+5=0[/tex]
[tex]x=-9\text{ or }x=9\text{ or }x=-5[/tex]
Therefore, the zeros of our given functions are [tex]x=-9,-5,9[/tex].
