We are given the equation:

\[
(x^2 - 1)(x^2 - 3) + 1 = 0
\]

### Step 1: Expand the equation

Let's first expand the product \((x^2 - 1)(x^2 - 3)\):

\[
(x^2 - 1)(x^2 - 3) = x^4 - 3x^2 - x^2 + 3 = x^4 - 4x^2 + 3
\]

Now substitute this back into the equation:

\[
x^4 - 4x^2 + 3 + 1 = 0
\]

Simplify the constant terms:

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

### Step 2: Substitution and solving the quadratic equation

Let \( y = x^2 \). This substitution turns the equation into:

\[
y^2 - 4y + 4 = 0
\]

This is a quadratic equation. We can solve it using the quadratic formula:

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

\[
y = \frac{4 \pm \sqrt{16 - 16}}{2} = \frac{4 \pm \sqrt{0}}{2} = \frac{4}{2} = 2
\]

Thus, \( y = 2 \).

### Step 3: Solve for \( x \)

Since \( y = x^2 \), we now solve for \( x \):

\[
x^2 = 2 \quad \Rightarrow \quad x = \pm \sqrt{2}
\]

### Final Answer:

The solutions to the equation are:

\[
x = \pm \sqrt{2}
\]

Let me know if you would like more details on any step or have any questions.

Here are some related questions:
1. What are the properties of quadratic equations?
2. Can a different substitution method be used for similar problems?
3. How does the quadratic formula work in solving equations?
4. What are other ways to factor polynomials like the one given?
5. What are the roots of polynomials in general?

**Tip:** Substitution can simplify complex equations by reducing them to more familiar forms, such as quadratic equations.

Discover how to solve the polynomial equation (x^2 - 1)(x^2 - 3) + 1 = 0 using quadratic substitution and the quadratic formula. This solution breaks down complex algebraic steps and is perfect for students in Grades 9-12.

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