To find the complex zeros of the polynomial function $f(x) = x^4 + 6x^2 + 8$, we can start by making a substitution to simplify the equation.

Step 1: Substitution

Let $y = x^2$. This turns the original equation into:

$f(y) = y^2 + 6y + 8$

This is now a quadratic equation in terms of $y$.

Step 2: Solve the quadratic equation

To solve the quadratic equation $y^2 + 6y + 8 = 0$, we can use the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where $a = 1$, $b = 6$, and $c = 8$.

$y = \frac{-6 \pm \sqrt{6^2 - 4(1)(8)}}{2(1)}$

$y = \frac{-6 \pm \sqrt{36 - 32}}{2}$

$y = \frac{-6 \pm \sqrt{4}}{2}$

$y = \frac{-6 \pm 2}{2}$

Thus, the two solutions for $y$ are:

$y_1 = \frac{-6 + 2}{2} = -2, \quad y_2 = \frac{-6 - 2}{2} = -4$

Step 3: Return to $x^2$

Recall that $y = x^2$, so now we solve for $x$ in each case:

1. $x^2 = -2$
   Taking the square root of both sides: $x = \pm \sqrt{-2} = \pm i\sqrt{2}$

2. $x^2 = -4$
   Taking the square root of both sides: $x = \pm \sqrt{-4} = \pm 2i$

Step 4: Complex zeros

The complex zeros are: $x = \pm i\sqrt{2}, \quad x = \pm 2i$

Thus, the complex zeros are: $x = i\sqrt{2}, \quad x = -i\sqrt{2}, \quad x = 2i, \quad x = -2i$

Step 5: Write the polynomial as a product of linear factors

The polynomial can be written as a product of linear factors based on its zeros:

$f(x) = (x - i\sqrt{2})(x + i\sqrt{2})(x - 2i)(x + 2i)$

This is the factorized form of the polynomial $f(x) = x^4 + 6x^2 + 8$.

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Do you want more details or have any questions?

Here are some related questions for further practice:

1. Find the complex zeros of $f(x) = x^4 + 5x^2 + 6$.
2. Factor $f(x) = x^4 + 2x^2 + 1$ completely over the complex numbers.
3. Solve $f(x) = x^4 + 4x^2 + 4$ for its complex zeros.
4. Write $f(x) = x^4 + 3x^2 + 2$ as a product of linear factors.
5. Determine the complex roots of $f(x) = x^4 + 7x^2 + 12$.

Tip: When factoring higher-degree polynomials, look for ways to reduce them into simpler quadratic equations via substitution, especially when there is a clear pattern like $x^4$ and $x^2$.