To find the zeros of the function \( f(x) = x^2 + 8x + 4 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 8 \), and \( c = 4 \).
First, we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 8^2 - 4 \cdot 1 \cdot 4 = 64 - 16 = 48 \]
Now, substituting into the quadratic formula:
\[ x = \frac{-8 \pm \sqrt{48}}{2 \cdot 1} \]
Next, simplify \( \sqrt{48} \):
\[ \sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3} \]
Now substitute this back into the equation:
\[ x = \frac{-8 \pm 4\sqrt{3}}{2} \]
Now simplify further:
\[ x = \frac{-8}{2} \pm \frac{4\sqrt{3}}{2} \] \[ x = -4 \pm 2\sqrt{3} \]
Thus, the zeros of the function \( f(x) = x^2 + 8x + 4 \) expressed in simplest radical form are:
\[ x = -4 \pm 2\sqrt{3} \]
Therefore, the corresponding answer from the options is:
\[ x = -4 \pm 2\sqrt{3} \]