roots of f (x)= $ x^3 +ax^2 + 6x - 1$ If function $$f (x)=  x^3 +ax^2 + 6x - 1$$ has a critical point at $x=-2$ then find nature of roots  $f (x)=0$ has ?
Using the fact that $ f '(-2) =0$ we get a=9/2,
So now I need to decide nature of roots of 
   $$f (x)=  x^3 +9/2x^2 + 6x - 1$$
I used descarte rule of change of signs
 So using descarte of signs we see that equation has at most 2 negative rea roots and at most $1$ positive real root.
But we know other possibility that it can also have $2$ complex roots and $1$ real root. 
I am having problem to decide which case it is. 
Is there some other tool to check nature of roots?
 A: You used Descartes Rule of Signs to tell you that your equation has:
a) At most $1$ positive real root
b) At most $2$ negative real roots
However, the full rule tells us that it has:
a) At most $1$ positive real root OR $1-2n$ positive roots ($n=1,2,...$). The only valid results from this give us at most $1$ positive real root.
b) At most $2$ negative real roots OR $2-2n$ negative real roots ($n=1,2,...$). The only valid results from this give us either $2$ negative real roots OR $0$ negative real roots.
You are then supposed to take this information and find all combination of roots it can have. In your case you have a third degree polynomial so the total number of roots must be $3$ and we can determine that:
a) It has $1$ positive real root and $2$ negative real roots, OR
b) It has $1$ positive real root and $2$ complex roots

To determine which case is true, you can calculate the discriminant ($\Delta$) of this polynomial (in your case it comes out as $-\frac{567}{2}$. For a cubic you can use the value of the discriminant as follows:
1) $\Delta>0\implies$the equation has $3$ distinct real roots
2) $\Delta<0\implies$the equation has $1$ real root and $2$ complex conjugate roots
3) $\Delta=0\implies$at least $2$ roots coincide, and they are all real. It may be that the equation has a double real root and another distinct single real root; alternatively, all three roots coincide yielding a triple real root.
NOTE: For a cubic of the form $ax^3+bx^2+cx+d$ the discriminant is given by:$$\Delta=b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$$
A: You're right that $a = 9/2$.But that actually doesn't matter when determining the number of real zeros, just the fact that it is positive. Now, $f(x) = x^3 + \frac{9}{2}x^2 + 6x -1$. We can see that the sign of the successive terms only changes once. From that we can determine there is only a single positive real root. We can similarly determine the negative real roots from the fact $f(-x) = -x^3 + \frac{9}{2}x^2 - 6x - 1$, with two sign changes indicating two or zero negative real roots.
We can further use the rational root test with synthetic division (exercise to the reader) to determine there are no rational real roots. In fact, the only real root is $r = \frac{1}{2} \left(-3+\sqrt[3]{13-2 \sqrt{42}}+\sqrt[3]{13+2 \sqrt{42}}\right)$, according to Mathematica. 
The other two roots are complex, but this cannot be determined from Descartes's Rule alone. We can only determine there is either $1$ positive and $2$ negative real roots or $1$ positive and $2$ complex roots. 
