Prove that the equation 2x^3+12x^2+18x+24= 0 has one real solution.

First, we'll simplify the given equation, by dividing by
2:

x^3 + 6x^2 + 9x + 12 = 0

To
determine the number of real roots of the equation, we'll have to create the Rolle's
string.  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.

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

We'll evaluate the limits of the function, if x
approaches to + and - infinite.

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 varies 1 time, between -infinite and
-3:

Therefore, the equation will have one
real root that belongs to the range: (-infinite ;
-3).