To find the zeros of the polynomial function f(x) = x^3 + 3x^2 + 2x + 6, we need to determine the values of x for which f(x) = 0. This involves solving the equation:
- x³ + 3x² + 2x + 6 = 0
Since this is a cubic equation, we can attempt to find its zeros using various methods, including synthetic division, factoring, or applying the Rational Root Theorem. First, we check for possible rational roots by testing integer factors of the constant term (in this case, 6).
Let’s evaluate the function at a few integer values:
- f(-3) = (-3)³ + 3(-3)² + 2(-3) + 6 = -27 + 27 – 6 + 6 = 0
Since f(-3) = 0, we have found one zero: x = -3. Next, we can perform synthetic division to divide the polynomial by (x + 3):
- Performing synthetic division using -3:
-3 | 1 3 2 6
| -3 0 -6
-----------------
1 0 2 0
The result is x² + 2, which we can set to zero to find the remaining zeros:
- x² + 2 = 0
This simplifies to:
- x² = -2
- x = ±√(-2) = ±i√2
Thus, the zeros of the function f(x) = x³ + 3x² + 2x + 6 are:
- x = -3 (a real zero)
- x = i√2 (an imaginary zero)
- x = -i√2 (another imaginary zero)
In summary, the function has one real zero at x = -3 and two complex zeros at x = i√2 and x = -i√2.