To factor the expression $x^2 + 2xy - 3y^2$, we'll use the method of factoring a trinomial.

The general form of a trinomial is $ax^2 + bx + c$, and we want to express it as a product of two binomials.

1. Identify coefficients:

   • $a = 1$ (the coefficient of $x^2$),
   • $b = 2y$ (the coefficient of $xy$),
   • $c = -3y^2$ (the constant term).

2. Find two numbers that multiply to $a \times c = 1 \times (-3y^2) = -3y^2$ and add to $b = 2y$:

   • We need two numbers that multiply to $-3y^2$ and add up to $2y$. The pair that works here is $3y$ and $-y$, because: $3y \times (-y) = -3y^2 \quad \text{and} \quad 3y + (-y) = 2y.$

3. Rewrite the middle term $2xy$ using $3y$ and $-y$: $x^2 + 3xy - xy - 3y^2.$

4. Group the terms in pairs: $(x^2 + 3xy) - (xy + 3y^2).$

5. Factor out the common factors from each group: $x(x + 3y) - y(x + 3y).$

6. Factor out the common binomial $(x + 3y)$: $(x - y)(x + 3y).$

Thus, the factored form of $x^2 + 2xy - 3y^2$ is: $(x - y)(x + 3y).$

Let me know if you'd like a more detailed breakdown, or if you have any other questions!

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Here are five related questions to explore further:

1. What is the method for factoring quadratic trinomials when the leading coefficient is not 1?
2. How would you factor a difference of squares expression?
3. What is the process for factoring by grouping when there are four terms?
4. How do you solve a quadratic equation after factoring it?
5. How do you recognize when factoring is not possible over the integers?

Tip: Always check for a greatest common factor (GCF) before factoring, as it can simplify the problem.