You have the function \( g(x) = x^6 - 4x^4 + 3x^2 + 2 \), and you're given \( g = 2 \). Let's solve this step by step.

Since \( g(x) = 2 \), we substitute 2 for \( g(x) \):

\[
x^6 - 4x^4 + 3x^2 + 2 = 2
\]

Simplifying the equation:

\[
x^6 - 4x^4 + 3x^2 + 2 - 2 = 0
\]

\[
x^6 - 4x^4 + 3x^2 = 0
\]

Now, let's factor the equation. Start by factoring out \( x^2 \):

\[
x^2(x^4 - 4x^2 + 3) = 0
\]

This gives two cases to solve:

### Case 1: \( x^2 = 0 \)
In this case, \( x = 0 \).

### Case 2: \( x^4 - 4x^2 + 3 = 0 \)

Let \( u = x^2 \), so the equation becomes:

\[
u^2 - 4u + 3 = 0
\]

This is a quadratic equation. Solve it using the quadratic formula:

\[
u = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(3)}}{2(1)}
\]

\[
u = \frac{4 \pm \sqrt{16 - 12}}{2}
\]

\[
u = \frac{4 \pm \sqrt{4}}{2}
\]

\[
u = \frac{4 \pm 2}{2}
\]

So, \( u = 3 \) or \( u = 1 \).

Since \( u = x^2 \), we now have:

- \( x^2 = 3 \), which gives \( x = \pm \sqrt{3} \),
- \( x^2 = 1 \), which gives \( x = \pm 1 \).

### Final Solutions:
The solutions for \( x \) are:

\[
x = 0, \pm 1, \pm \sqrt{3}
\]

Would you like further details or have any questions?

Here are 5 related questions to explore further:
1. What happens if you change the constant term in the equation?
2. Can you solve similar equations using different factoring methods?
3. How does the quadratic formula apply to higher-degree equations?
4. What are the graphical implications of solving this function?
5. Can this equation be solved using numerical methods?

**Tip**: Always try factoring simpler terms before applying advanced techniques like the quadratic formula for efficiency!

Learn how to solve the polynomial equation g(x) = x^6 - 4x^4 + 3x^2 + 2, where g = 2. This problem covers polynomial factoring techniques, the use of the quadratic formula, and applying the zero product property. Suitable for students in Grades 10-12.

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