<s>[INST] 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 ?
 [/INST] $x^3 - x^2 - 3x - 9=0$
$3x = x^3 - x^2 - 9$
When $x<0$ every term on the right hand side is less than 0.  If we are gong to find equality, $3x$  must be very negative.
When $-3< x < 0, |3x| > |-9|$
When $x< -3, |3x| < |x^2|$ and $|3x| < |x^3|$
There is no way that $3x$ can be sufficiently negative.
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