Answer:
x=-4 and x=2
Step-by-step explanation:
You are given the function
[tex]f(x)=\dfrac{x^2+2x-8}{x^2-2x-8}[/tex]
The zeros of the function are those value of x, for which [tex]f(x)=0[/tex]
First, find x for which function is undefined
[tex]x^2-2x-8\neq0\\ \\x^2-4x+2x-8\neq 0\\ \\x(x-4)+2(x-4)\neq 0\\ \\(x-4)(x+2)\neq 0\\ \\x\neq 4\text{ and }x\neq -2[/tex]
Now find points at which f(x)=0:
[tex]f(x)=0\Rightarrow x^2+2x-8=0\\ \\x^2+4x-2x-8=0\\ \\x(x+4)-2(x+4)=0\\ \\(x-2)(x+4)=0\\ \\x=2\text{ or }x=-4[/tex]
For both these values (x=-4 and x=2) function f(x) is defined, then x=-4 and x=2 are zeros of the function.
