we are given
![f(x)=x^3+x^2-8x-8](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E3%2Bx%5E2-8x-8)
We can use Descarte's sign rule to find number of real roots
Positive real roots:
![f(x)=x^3+x^2-8x-8](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E3%2Bx%5E2-8x-8)
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)
![f(-x)=(-x)^3+(-x)^2-8(-x)-8](https://tex.z-dn.net/?f=f%28-x%29%3D%28-x%29%5E3%2B%28-x%29%5E2-8%28-x%29-8)
![f(-x)=-x^3+x^2+8x-8](https://tex.z-dn.net/?f=f%28-x%29%3D-x%5E3%2Bx%5E2%2B8x-8)
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