What are the complex zeros of the polynomial function f(x) = x^3 + x^2 – 6x – 6?

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To find the complex zeros of the polynomial function f(x) = x3 + x2 – 6x – 6, we can start by applying the Rational Root Theorem and synthetic division to identify any potential rational roots.

After evaluating several possible rational roots, we find that x = 2 is a root. We can perform synthetic division of the polynomial by (x – 2): 

    2 |  1   1  -6  -6  
      |      2   6   0  
    -------------------  
      |  1   3   0  -6

This means we can rewrite the polynomial as:

  • f(x) = (x – 2)(x2 + 3x + 3)

Next, we need to find the zeros of the quadratic x2 + 3x + 3 using the quadratic formula:

x = (-b ± √(b2 – 4ac)) / 2a

Where a = 1, b = 3, and c = 3. Plugging in these values:

x = (-3 ± √(32 – 4 * 1 * 3)) / (2 * 1)

This simplifies to:

x = (-3 ± √(9 – 12)) / 2

Now, simplifying further:

x = (-3 ± √(-3)) / 2

Since the square root of a negative number gives us imaginary numbers, we can express the complex roots as:

  • x = (-3 + i√3) / 2
  • x = (-3 – i√3) / 2

In conclusion, the polynomial function f(x) = x3 + x2 – 6x – 6 has one real zero, x = 2, and two complex zeros: x = (-3 + i√3) / 2 and x = (-3 – i√3) / 2. This provides a full view of the zeros of the polynomial, combining both real and complex roots.

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