answer : option C
[tex]f(x) = x^3 + x^2 - 8x - 8[/tex]
Lets find the number of zeros by factoring
[tex]0 = x^3 + x^2 - 8x - 8[/tex]
Group first two terms and last two terms
[tex]0 = (x^3 + x^2) (- 8x - 8)[/tex]
Factor out GCF from each group
[tex]0 = x^2(x + 1)-8(x + 1)[/tex]
[tex]0 = (x^2-8)(x + 1)[/tex]
Now we set each factor =0 and solve for x
0 = (x^2-8) and (x + 1)=0
[tex]x^2 = 8[/tex] and x = -1
[tex]x= 2+-\sqrt{2}[/tex] and x= -1
So we have three real zeros
The function has three real zeros. The graph of the function intersects the x-axis at exactly three locations.
