Answer:
The answer best describes the zeros of the polynomial function f(x)=x^3+x^2−8x−8 is third option:
The function has three real zeros. The graph of the function intersects the x-axis at exactly three locations.
Explanation:
Using Ruffini:
f(x)=x^3+x^2-8x-8
! 1 1 -8 -8
-1 !___ -1___0__8
1 0 -8 0
x^2 -8
f(x)=(x+1)(x^2-8)
Factoring the quadratic term x^2-8, using:
a^2-b^2=(a+b)(a-b)
with:
a^2=x^2→sqrt(a^2)=sqrt(x^2)→a=x
b^2=8→sqrt(b^2)=sqrt(8)→b=sqrt(4*2)→b=sqrt(4) sqrt(2)→b=2 sqrt(2)
x^2-8=(x+2 sqrt(2))(x-2 sqrt(2))
f(x)=(x+1)(x+2 sqrt(2))(x-2 sqrt(2))
Zeros of the polynomial are: x=-1, x=-2 sqrt(2), and x=2 sqrt(2)
The function has three real zeros. The graph of the function intersects the x-axis at exactly three locations.