The given function is:

$f(x) = 4x^4 + 49x^2 + 12$

We need to find the zeros of this polynomial. This is a quartic equation, but it can be simplified by using a substitution method. Let’s solve this step-by-step.

Step 1: Substitution

Let $u = x^2$, which simplifies the equation to a quadratic in $u$:

$4u^2 + 49u + 12 = 0$

Step 2: Solve the quadratic equation

We can solve this quadratic using the quadratic formula:

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

where $a = 4$, $b = 49$, and $c = 12$.

Substitute these values:

$u = \frac{-49 \pm \sqrt{49^2 - 4(4)(12)}}{2(4)}$

$u = \frac{-49 \pm \sqrt{2401 - 192}}{8}$

$u = \frac{-49 \pm \sqrt{2209}}{8}$

$u = \frac{-49 \pm 47}{8}$

Step 3: Solve for $u$

We now have two solutions for $u$:

1. $u_1 = \frac{-49 + 47}{8} = \frac{-2}{8} = -\frac{1}{4}$
2. $u_2 = \frac{-49 - 47}{8} = \frac{-96}{8} = -12$

Step 4: Back-substitute $u = x^2$

Now that we have the values for $u$, we back-substitute $u = x^2$ to solve for $x$.

For $u_1 = -\frac{1}{4}$:

$x^2 = -\frac{1}{4} \quad \Rightarrow \quad x = \pm \frac{i}{2}$

For $u_2 = -12$:

$x^2 = -12 \quad \Rightarrow \quad x = \pm 2i\sqrt{3}$

Step 5: Final answer

The zeros of the function are:

$x = \left\{ \frac{i}{2}, -\frac{i}{2}, 2i\sqrt{3}, -2i\sqrt{3} \right\}$

Do you need any further explanation or have any questions about the process?

Here are 5 related questions to explore:

1. How do you solve quartic equations in general cases?
2. What is the significance of using substitution in solving quartic equations?
3. Can this method be applied to equations that don't involve squares or higher powers?
4. What does it mean when the solutions to a polynomial are imaginary?
5. How would the graph of the function $f(x) = 4x^4 + 49x^2 + 12$ look based on its zeros?

Tip: When working with higher-degree polynomials, substitution can help reduce the equation to a simpler quadratic form.