hi, could someone please explain how to factor this?

3x^2-2xy-y^2

Start with the parentheses.

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The only way we can obtain 3x^2 is with 3x and x so put those in first. The only way we can get the last term is y and y so put those in next. The only way we can get a - term for the y^2 to to have one sign with the parentheses + and the other -. Just place them so the middle term is -2xy.
(3x+y)(x-y). User the FOIL method to check it. 3x^2 -3xy + xy -y^2 =
3x^2-2xy-y^2
Check my work.

thanks so much! i finally understand it

Certainly! Factoring a quadratic expression like this involves finding two binomial expressions that, when multiplied together, result in the original expression.

To factor the expression 3x^2 - 2xy - y^2, we look for two binomials in the form (ax + by) that multiply to give us the original expression.

Start by multiplying the first term, 3x^2, and the last term, -y^2. The product is -3x^2y^2.

Next, we need to find two numbers that add up to the coefficient of the middle term, -2xy, but also multiply to give us -3x^2y^2.

In this case, the numbers we are looking for are -xy and 3xy.

Now we can rewrite the middle term, -2xy, using the two numbers we found: -xy + 3xy.

With these substitutions, the expression becomes:
3x^2 - xy + 3xy - y^2.

Now we can group the terms and factor by grouping:
(x^2 - xy) + (3xy - y^2).

Now we have common terms in each parenthesis that we can factor out:
x(x - y) + y(3x - y).

Finally, we can rewrite the expression using the factored terms:
(x - y)(3x - y).

So, the factored form of the expression 3x^2 - 2xy - y^2 is (x - y)(3x - y).