Complex Zeros (Total Zeros): 3 (since it's a cubic polynomial). Possible Real Zeros: ±1, ±3, ±9. Possible Imaginary Zeros: There is one unpaired imaginary zero. Possible Rational Zeros: ±1, ±3, ±9. Actual Zeros: -3, -√3, √3.
To find the complex zeros, real zeros, and imaginary zeros of the function f(x) = x^3 + 3x^2 + 3x + 9, we can use the Rational Root Theorem and synthetic division.
Zeros of the Function:
The given function is a cubic polynomial, so it can have up to 3 zeros.
a) Complex Zeros (Total Zeros):
The total number of complex zeros is equal to the degree of the polynomial, which is 3.
b) Possible Real Zeros:
The possible real zeros are the factors of the constant term divided by the factors of the leading coefficient. In this case, the constant term is 9, and the leading coefficient is 1.
Possible real zeros: ±1, ±3, ±9
c) Possible Imaginary Zeros:
If there are any non-real (complex or imaginary) zeros, they will come in conjugate pairs. Since the degree is 3, if there are imaginary zeros, there will be one unpaired imaginary zero.
Possible Rational Zeros:
The possible rational zeros can be found by dividing the factors of the constant term (9) by the factors of the leading coefficient (1):
Possible rational zeros: ±1, ±3, ±9
Find All Zeros:
To find the actual zeros, we can use synthetic division or other methods.
Let's use synthetic division to find the zeros:
For f(x) = x^3 + 3x^2 + 3x + 9, we can try dividing by potential zeros.
Let's start with x = -1:
-1 | 1 3 3 9
| -1 -2 -1
1 2 1
The remainder is not zero, so x = -1 is not a zero.
Now, try x = -3:
-3 | 1 3 3 9
| -3 0 -9
1 0 0
The remainder is zero, so x = -3 is a zero.
Now, we can factor the polynomial as f(x) = (x + 3)(x^2 + 0x - 3).
The remaining quadratic factor x^2 - 3 can be factored further as f(x) = (x + 3)(x - √3)(x + √3).