Question

How do you multiply `(x-2y)^2`?

Answer 1

See a solution process below:

Explanation:

This is a special form of the quadratic:

`(color(red)(a) - color(blue)(b))^2 = (color(red)(a) - color(blue)(b))(color(red)(a) - color(blue)(b)) = color(red)(a)^2 - 2color(red)(x)color(blue)(b) + color(blue)(b)^2`

Let:

• `color(red)(a) = x`

• `color(blue)(b) = 2y`

Substituting gives:

`(color(red)(x) - color(blue)(2y))^2 =>`

`(color(red)(x) - color(blue)(2y))(color(red)(x) - color(blue)(2y)) =>`

`color(red)(x)^2 - (2 xx color(red)(x) xx color(blue)(2y)) + color(blue)((2y))^2 =>`

`color(red)(x)^2 - 4color(red)(x)color(blue)(y) + color(blue)(4y)^2`

Answer 2

`x^2-4xy+4y^2`

Explanation:

We can rewrite `(x-2y)^2` as `color(blue)((x-2y)(x-2y))`. The expression I have in blue, we can use FOIL to multiply.

`(x-2y)(x-2y)`

FOIL tells us that we multiply the first terms, outside terms, inside terms and last terms respectively. We get:

• Firsts `(x*x)=x^2`
• Outsides `(x*-2y)= -2xy`
• Insides `(-2y*x)= -2xy`
• Lasts `(-2y*-2y)= 4y^2`

NOTE:Whenever we're multiplying binomials, we can use FOIL

Our new expression is as follows:

`x^2-2xy-2xy+4y^2`

We can combine like terms to get:

`x^2-4xy+4y^2`

Alternatively, we could have separated `(x-2y)(x-2y)` into two expressions if you're not a FOIL person. This is the same as doing:

`x(x-2y)-2y(x-2y)`

We can distribute the `x` and the `-2y` respectively to get:

`x^2-2xy-2xy+4y^2`

And we could combine like terms to get:

`x^2-4xy+4y^2`