Question

20 POINTS

Answer 1

answer : option C

$f(x) = x^3 + x^2 - 8x - 8$

Lets find the number of zeros by factoring

$0 = x^3 + x^2 - 8x - 8$

Group first two terms and last two terms

$0 = (x^3 + x^2) (- 8x - 8)$

Factor out GCF from each group

$0 = x^2(x + 1)-8(x + 1)$

$0 = (x^2-8)(x + 1)$

Now we set each factor =0 and solve for x

0 = (x^2-8)             and   (x + 1)=0

$x^2 = 8$            and x = -1

$x= 2+-\sqrt{2}$     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.