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 ?

Actually, in working through my Calculus oriented answer, I accidentally thought of a somewhat convoluted precalculus approach to solving the problem.

$f(x)= x^3-x^2-3x-9.$ 
How many negative real roots does the equation $f(x) = 0$ have ?

$$\text{Let} ~~g(x) = x^3 - x^2 - 3x + 3 = (x - 1) \times (x^2 - 3).\tag1$$

Part of the inspiration for this method, was to construct $g(x)$ such that $f(x) - g(x)$ equals a constant and where $g(x)$ is easy to factor.  I noted that the two leftmost coefficients of $f(x)$ followed the pattern $[(+1), (-1)],$ so I constructed $g(x)$ so that the next two coefficients of $g(x)$ would follow the pattern $(-3) \times [(+1), (-1)].$
The other part of the inspiration for this method, was my observation in my other posted answer that the absolute value of the rightmost coefficient of $f(x)$, namely $|-9| = 9$ seemed inordinately large.

Then, with $g(x)$ specified by (1) above, you have that $f(x) = g(x) - 12.$ 
Therefore, it is sufficient to show that on the interval $-\infty < x < 0$, 
the maximum value achieved by $g(x)$ is less than $12$.
Given how $g(x)$ is factored in (1) above, the only values of $x < 0$ for which $g(x)$ will be $ > 0$ will be $-\sqrt{3} < x < 0.$
However, for $-\sqrt{3} < x < 0$, a routine examination of 
$|g(x)| = |x - 1| \times |x^2 - 3|$ shows that

*

*$|x - 1| = 1 + (-x) < 1 + (\sqrt{3}) ~< ~3 ~< ~4.$

*$|x^2 - 3| = 3 - x^2 \leq 3.$
Therefore, the product of the two factors above must be strictly less than $(4 \times 3)$.  Therefore, the maximum value for $g(x)$ on the interval $-\sqrt{3} < x < 0$ must be less than $(12)$.  Since this is the only interval for which $x < 0$ and $g(x) > 0$, the maximum value of $g(x)$ on the interval $-\infty < x < 0$ must be less than $(12)$.
Therefore, since $f(x) = g(x) - 12$, you must have that the maximum value of $f(x)$ on the interval $-\infty < x < 0$ must be less than $0$.
