To find the complex zeros of the polynomial function $f(x) = x^3 + 8$, we can follow these steps:

Step 1: Rewrite the polynomial using the sum of cubes formula

The expression $x^3 + 8$ can be factored using the sum of cubes formula: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ Here, $x^3 + 8 = x^3 + 2^3$, so we set $a = x$ and $b = 2$.

Step 2: Apply the formula

Using the sum of cubes formula: $x^3 + 8 = (x + 2)(x^2 - 2x + 4)$ Now we have factored $f(x) = x^3 + 8$ as $f(x) = (x + 2)(x^2 - 2x + 4)$.

Step 3: Find the roots of each factor

1. For the first factor, $x + 2 = 0$: $x = -2$ This gives us one real root, $x = -2$.

2. For the second factor, $x^2 - 2x + 4 = 0$, we solve for $x$ using the quadratic formula: $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$ Simplifying this: $x = \frac{2 \pm \sqrt{4 - 16}}{2} = \frac{2 \pm \sqrt{-12}}{2}$ Since $\sqrt{-12} = 2i\sqrt{3}$, we have: $x = \frac{2 \pm 2i\sqrt{3}}{2} = 1 \pm i\sqrt{3}$

Step 4: Write the final answer

The complex zeros of $f(x) = x^3 + 8$ are: $x = -2, \quad x = 1 + i\sqrt{3}, \quad x = 1 - i\sqrt{3}$

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Related Questions

1. How can we determine if a polynomial has real or complex roots?
2. What are other techniques to factor a polynomial of higher degree?
3. How does the sum of cubes formula differ from the difference of cubes formula?
4. Why do complex roots of polynomials with real coefficients come in conjugate pairs?
5. How does the quadratic formula help in finding complex roots?

Tip

Always check if a polynomial can be factored using special formulas like sum of cubes, difference of cubes, or difference of squares before using other methods like the quadratic formula.