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).