What is the number of real solutions of the equation x^3+6x^2+9x+12=0 ?

We'll create the Rolle's string to determine the number of
real roots of the equation. According to Rolle's string, between 2 consecutive roots of
derivative, we'll find a real root of the equation, if and only if the product of the
values of derivatives, is negative.

 First, we'll check the
continuity of the function. The Rolle's string could be applied if and only if the
polynomial function is continuous.

lim f(x) = lim
(x^3+6x^2+9x+12) = + infinite, for x approaches to
+infinite.

To determine the Rolle's string we need to
determine the roots of the 1st derivative of the
function.

f'(x) =
(x^3+6x^2+9x+12)'

f'(x) = 3x^2 + 12x +
9

We'll put f'(x) = 0

3x^2 +
12x + 9 = 0

We'll divide by
3:

x^2 + 4x + 3 = 0

x1 = -1
and x = -3

Now, we'll calculate the values of the function
for each value of the roots of the derivative.

f(-inf.) =
lim f(x) =  -inf

f(+inf.) = lim f(x) =
+inf.

f(-1) =
-1+6-9+12=8

f(-3) = -27 + 54 - 27 + 12 =
12

The values of the function represents the Rolle's
string.

-inf.  12  8  +inf.

We
notice that the sign is changing 1 time:

- from -inf. to
12

Therefore, the equation will have 1 real
roots, located  in the interval: (-inf. ; -3).