SOLUTION: Indentify the number of positive, negative, and imaginary roots, all rational roots , and the total amount of roots. x^3+x-6=0

The polynomial on the left side is of degree 3 - so the total amount of the roots is 3.


It includes real and complex number roots.


The derivative of the left side polynomial is  3x^2 + 1; from the first glance,

it is clear that this derivative function is always positive - so the original 

polynomial function on the left side is monotonically increasing.


It means, that there is ONLY ONE real root.  Hence, other two roots are complex numbers.


At x= 0, the left side function has the value of -6, and it rises monotonically to +infinity as  x --> oo   - 

    hence, the real root is a positive real number. 


Then, according to the Rational Root theorem, rational roots can be only among these numbers (divisors of 6): 1, 2, 3 or 6.


But easy mental check shows that the number 2 is not the root, and, moreover, at x= 2 the left side function is just positive, 
so the only real root is IRRATIONAL number between 1 and 2.


So, the required analysis is COMPLETED, and its results are as follows:


    - total number of roots is 3;

    - there is only one real root;

    - there are two complex roots;

    - the real root is positive;

    - the number of positive real roots is 1;

    - the number of negative real roots is zero;

    - the number of rational real roots is zero.