Respuesta :
we are given
[tex]f(x)=x^3+x^2-8x-8[/tex]
We can use Descarte's sign rule to find number of real roots
Positive real roots:
[tex]f(x)=x^3+x^2-8x-8[/tex]
we can see that number of sign changes in this function is 1
so, number of positive real root =1
Negative real roots:
Firstly, we will find f(-x)
[tex]f(-x)=(-x)^3+(-x)^2-8(-x)-8[/tex]
[tex]f(-x)=-x^3+x^2+8x-8[/tex]
we can see that number of sign changes in this function is 2
so, number of negative real root =2
so, total number of real roots = number of positive real roots + number of negative real roots
total number of real roots =1+2
total number of real roots =3
Since, the degree of this polynomial is 3
so, maximum number of roots must be 3
We know that all roots are also called x-intercept because it crosses x-axis at that value
so, function will cross x-axis thrice
so, option-A.......Answer