Find the roots of the polynomial equation. x3+6x2+11x+12=0 (1 point) Responses −1±i2–√, −4 −1±i2–√, −4 1±i2–√, −4 1±i2–√, −4 −1±i3–√, −4 −1±i3–√, −4 1±i3–√, −4
To find the roots of the cubic polynomial equation x^3 + 6x^2 + 11x + 12 = 0, we can either use factorization (if possible) or apply a formula for solving cubic equations, such as Cardano's formula. However, cubic equations often have at least one real root, and that real root can sometimes be found by guessing or using the Rational Root Theorem.
In this case, we can try factorization by looking for obvious roots among the factors of the constant term, 12. Those would be ±1, ±2, ±3, ±4, ±6, and ±12.
Let's check x = -1 first because this is often a good initial guess for a polynomial with all positive coefficients and a positive constant term:
(-1)^3 + 6(-1)^2 + 11(-1) + 12 = -1 + 6 - 11 + 12 = 6 - 1 - 11 + 12 = 0
Since -1 is indeed a root, we can now factor the polynomial as follows:
(x + 1)(x^2 + bx + c) = 0
To find the quadratic factor (x^2 + bx + c), we can perform synthetic division or polynomial long division to divide x^3 + 6x^2 + 11x + 12 by x + 1:
_________|
x+1 | x^3 + 6x^2 + 11x + 12
| - (x^3 + x^2)
|______________
5x^2 + 11x
- (5x^2 + 5x)
|______________
6x + 12
- (6x + 6)
|______________
6
The resulting quadratic factor is:
x^2 + 5x + 6
Now we solve for the quadratic equation:
x^2 + 5x + 6 = 0
This factors to:
(x + 2)(x + 3) = 0
Therefore, the roots of the quadratic equation are x = -2 and x = -3. Combining this with the root we found earlier, x = -1, the complete set of roots of the cubic equation is:
x = -1, x = -2, x = -3
So, the correct answer to the question is not in the options provided, as they suggest complex roots. The roots of the given polynomial equation are solely real numbers: -1, -2, and -3.