How do you factor $x^{2} - 2 x y - 3 y^{2}$ $x^2-2xy-3y^2$ ?

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How do you factor $x^{2} - 2 x y - 3 y^{2}$ $x^2-2xy-3y^2$ ?

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Answer 1 · 2018-03-26T22:03:56

The factored form is $(x - 3 y) (x + y)$ $(x-3y)(x+y)$ .

Explanation:

You can treat it like a quadratic, where instead of constant numbers, there are $y$ $y$ terms.

First, find two numbers that multiply to $- 3$ $-3$ (the $c$ $c$ value of the "quadratic") and add up to $- 2$ $-2$ (the $b$ $b$ value of the "quadratic").

These two numbers are $- 3$ $-3$ and $1$ $1$ . Split the middle term into these two numbers. Then, factor the first two and last two terms separately, then combine them:

$= x^{2} - 2 x y - 3 y^{2}$ $color(white)=x^2-2xy-3y^2$

$= x^{2} + x y - 3 x y - 3 y^{2}$ $=x^2+xy-3xy-3y^2$

$= x (x + y) - 3 x y - 3 y^{2}$ $=color(red)x(x+y)-3xy-3y^2$

$= x (x + y) - 3 y (x + y)$ $=color(red)x(x+y)color(blue)-color(blue)(3y)(x+y)$

$= (x - 3 y) (x + y)$ $=(color(red)xcolor(blue)-color(blue)(3y))(x+y)$

This is the factored form. Hope this helped!

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How do you factor $x^{2} - 2 x y - 3 y^{2}$ $x^2-2xy-3y^2$ ?

1 Answer

Explanation:

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How do you factor $x^{2} - 2 x y - 3 y^{2}$ $x^2-2xy-3y^2$ ?