Answer:
The zeros of the function:
[tex]x=2,\:x=-4[/tex]
Step-by-step explanation:
Given the function
[tex]f\left(x\right)=x^2+2x-8[/tex]
In order to determine the zeros of the function, substitute f(x) = 0
[tex]x^2+2x-8\:=0[/tex]
Breaking the expression x²+2x-8=0 into groups
[tex]\:\left(x^2-2x\right)+\left(4x-8\right)=0[/tex]
Factor out x from x²-2x: x(x-2)
Factor out 4 from 4x-8: 4(x-2)
so
[tex]x\left(x-2\right)+4\left(x-2\right)=0[/tex]
Factor out common term: x-2
[tex]\left(x-2\right)\left(x+4\right)=0[/tex]
Using the zero factor principle
if ab=0, then a=0 or b=0 (or both a=0 and b=0)
[tex]x-2=0\quad \mathrm{or}\quad \:x+4=0[/tex]
solving x+2 = 0
x+2 = 0
x = -2
solving x+4=0
x+4 = 0
x = -4
Therefore, the zeros of the function:
[tex]x=2,\:x=-4[/tex]
