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 ?
 A: $$x^3-x^2-3x-9=(x-3)(x^2+2x+3)$$
Now you just have to compute the discriminant of the quadratic term to conclude that the roots are imaginary.
A: 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 $-\infty < x < 0.$
A: 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.
A: 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 $0$.
A: $x^3 - x^2 - 3x - 9=0$
$3x = x^3 - x^2 - 9$
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.
