The given expression is \( u^4 - u^2v^2 - 6v^4 \).

To factor this expression, we can start by making a substitution to simplify it. Let \( x = u^2 \) and \( y = v^2 \). The expression then becomes:

\[
x^2 - xy - 6y^2
\]

Next, we factor the quadratic expression \( x^2 - xy - 6y^2 \):

We need to find two numbers that multiply to \(-6y^2\) and add to \(-y\). These numbers are \(-3y\) and \(2y\).

So, we can factor the quadratic expression as:

\[
x^2 - xy - 6y^2 = (x - 3y)(x + 2y)
\]

Finally, we substitute back \( x = u^2 \) and \( y = v^2 \):

\[
u^4 - u^2v^2 - 6v^4 = (u^2 - 3v^2)(u^2 + 2v^2)
\]

So, the factored form of the expression is:

\[
(u^2 - 3v^2)(u^2 + 2v^2)
\]

Would you like further details on this process, or do you have any questions?

Here are 5 related questions to consider:
1. How can you factor the expression \( u^4 + 6u^2v^2 + 9v^4 \)?
2. What are some other methods to factor quadratic expressions?
3. How does the substitution method help in simplifying complex polynomials?
4. Can you use this factorization to solve the equation \( u^4 - u^2v^2 - 6v^4 = 0 \)?
5. How does the factorization of polynomials change when variables have different degrees?

**Tip:** When factoring polynomials, looking for substitutions that simplify the expression can make the process much easier.

Explore how to factor the polynomial u^4 - u^2v^2 - 6v^4 step by step, using substitution. This problem involves key algebraic concepts such as factoring quadratics and simplifying complex polynomials, making it suitable for high school students in Grades 10-12.

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