<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] Given the posted question's precalculus tag, I normally would not have posted the following answer.  However, since Siong Thye Goh has given a precalculus answer, I will give an answer that involves derivatives, which is a concept from Calculus (AKA Real Analysis).  Note that I am generally ignorant of the precalculus tools available to attack such a problem.  Therefore, there may be other precalculus methods of attack besides that given in the answer by Siong Thye Goh.


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

The posted question is equivalent to asking how many times $f(x)$ crosses the $x$-axis, for $-\infty < x < 0.$  The behavior of $f(x)$ may be analyzed by analyzing its first and second derivatives, and by noticing that $f(x)$ is a continuous function.
Without belaboring the underlying theorems, what this implies is that if there is an interval $[a,b]$ such that $a < b$, and if there exists $x_1, x_2$ both in $[a,b]$ such that $f(x_1) < 0 < f(x_2)$, then there must exist at least one value $x_0$ in $[a,b]$ such that $f(x_0) = 0$.  Note that it doesn't matter whether $x_1 < x_2$ or $x_1 > x_2$.
$f'(x) = 3x^2 - 2x - 3$ and $f''(x) = 6x - 2$.
$\displaystyle f'(x) = 0 \implies ~x ~= 
~\left(\frac{1}{6}\right) 
~\left[ ~2 ~\pm ~\sqrt{4 + 36} ~\right] ~= 
~\left(\frac{1}{3}\right) ~\left[ ~1 ~\pm ~\sqrt{10} ~\right].$
Since the posted question is concerned only about negative roots, attention can be confined to 
$\displaystyle (x_3) ~= ~\left(\frac{1}{3}\right) ~\left[ ~1 ~- ~\sqrt{10} ~\right].$
Since $f''(x) < 0$, for all $x < 0$, $f(x)$ is maximized at $x = x_3$, for all $x < 0$.
$\displaystyle (x_3)^3 ~= ~\left(\frac{1}{27}\right)
~\left[ ~1 - 3\sqrt{10} + 30 - 10\sqrt{10} ~\right]
~= ~\left(\frac{1}{27}\right)
~\left[ 31 - 13\sqrt{10} ~\right].$
$\displaystyle (x_3)^2 ~= ~\left(\frac{1}{9}\right)
~\left[ ~1 - 2\sqrt{10} + 10 ~\right]
~= ~\left(\frac{1}{27}\right)
~\left[ 33 - 6\sqrt{10} ~\right].$
$\displaystyle (x_3) ~= ~\left(\frac{1}{3}\right)
~\left[ ~1 - \sqrt{10} ~\right]
~= ~\left(\frac{1}{27}\right)
~\left[ 9 - 9\sqrt{10} ~\right].$
Therefore,
$$f(x_3) = \left(\frac{1}{27}\right) \times 
~\left\{ ~\left[ 31 - 13\sqrt{10} ~\right] 
~- ~\left[ 33 - 6\sqrt{10} ~\right]
~- ~\left[ 27 - 27\sqrt{10} ~\right]
~- ~\left[ 243 ~\right]
~\right\}.
$$
Simplifying,
$$f(x_3) = \left(\frac{1}{27}\right) \times 
 ~\left[ -272 + 20\sqrt{10} ~\right] ~< ~0.$$
In summary, for $x$ restricted to negative values, $f(x)$ is maximized at $x = x_3$, with $f(x_3) < 0$.  Therefore, there is no value of $x < 0$ such that $f(x) \geq 0.$  Therefore, the function $f(x)$ does not cross the $x$-axis in the interval $-\infty < x < 0.$
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