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!