How many negative real roots does the equation $x^3-x^2-3x-9=0$ have? How many negative real roots does the equation $x^3-x^2-3x-9=0$ have ?
My approach :-
f(x)= $x^3-x^2-3x-9$
Using rules of signs, there is 1 sign change , so there can be at most 1 positive real root
f(-x)= $-x^3-x^2+3x-9$
2 sign changes here, indicating at most 2 negative real roots
I end up with following 2 possibilities:-
1)1 positive, 2 negative real roots
2)1 positive, 2 imaginary roots
how to progress further ?

Just another way - to check for negative roots, it is enough to show $x^3+x^2+9=3x$ is not possible for any positive $x$.  Using AM-GM, $x^3+(x^2+4)+5 \geqslant x^3+2\sqrt{4x^2}+5=x^3+(4x)+5>3x$, so this isn't possible.
