The graph of f(x) = x3 + 6x2 + 12x + 8 is shown. based on the graph, how many real number solutions does the equation x3 + 6x2 + 12x + 8 = 0 have?

SEE ATTACHED IMAGE TO OBSERVE THE GRAPH OF THE FUNCTION.
 For this case, the first thing we should see are the cut points with the x axis.
 We note that the graph cuts to the x-axis at x = -2
 Therefore, x = -2 is the real solution to the polynomial.
 Also this function:
 x3 + 6x2 + 12x + 8
 It can be rewritten as:
 (x + 2) ^ 3
 From where we conclude that its roots are:
 x = -2 (with multiplicity 3)
 
 the equation x3 + 6x2 + 12x + 8 = 0 have:
 x = -2
 As a real solution with multiplicity 3.

You didn't show the graph.
Anyway, that equation is a perfect cube and equals
(x +2)^3
or (x +) * (x +) * (x +)
Therefore, it has 3 real number solutions and ALL of those solutions equal -2
 

Lol i had a teacher named miss walker back then