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

2 Answers
Mar 3, 2018

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

Mar 3, 2018

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