Question

I NEED IMMEDIATE HELP!!!!

Which products result in a difference of squares? Check all that apply.

[A] (x-y)(y-x)

[B] (6-y)(6-y)

[C] (3+xz)(-3+xz)

[D] (y^2-xy)(y^2+xy)

[E] (25x-7y)(-7y+25x)

[F] (64y^2+x^2)(-x^2+64y^2)

Answer 1

Hello!

The answers are:

C)
`(3+xz)(-3+xz)`

D)
`(y^2-xy)(y^2+xy)`

F)
`(64y^2+x^2)(-x^2+64y^2)`

Why?

To know which of the products results in a difference of square, we need to remember the difference of squares from:

The difference of squares form is:

`(a+b)(a-b)=a^(2)-b^(2)`

So, discarding each of the given options in order to find which products result in a difference of squares, we have:

A)

`(x-y)(y-x)=xy-x^(2)-y^(2) +yx=-x^(2) -y^(2)`

So, the obtained expression is not a difference of squares.

B)

`(6-y)(6-y)=36-6y-6y+y^(2)=y^(2)-12y+36`

So, the obtained expression is not a difference of squares.

C)

`(3+xz)(-3+xz) =(xz+3)(xz-3)=(xz)^(2)-3xz+3xz-(3)^(2)\\\\(xz)^(2)-3xz+3xz-(3)^(2)=(xz)^(2)-(3)^(2)`

So, the obtained expression is a difference of squares since it matches with the form of the difference of squares.

D)

`(y^2-xy)(y^2+xy)=(y^(2))^(2)+y^(2)*xy-y^(2)*xy-(xy)^(2) \\\\(y^(2))^(2)+y^(2)*xy-y^(2)*xy-(xy)^(2)=(y^(2))^(2)-(xy)^(2)`

So, the obtained expression is a difference of squares since it matches with the form of the difference of squares.

E)

`(25x-7y)(-7y+25x)=-175xy+(25x)^(2)+49y^(2)-175xy\\\\-175xy+(25x)^(2)+49y^(2)-175xy=(25x)^(2)+49y^(2)-350xy`

So, the obtained expression is not a difference of squares

F)

`(64y^2+x^2)(-x^2+64y^2)=(64y^2+x^2)(64y^2-x^2)\\\\(64y^2+x^2)(64y^2-x^2)=(64y^(2))^(2) -(x^(2)*64y^(2))+(x^(2)*64y^(2))-(x^(2))^(2)\\ \\(64y^(2))^(2) -(x^(2)*64y^(2))+(x^(2)*64y^(2))-(x^(2))^(2)=(64y^(2))^(2)-(x^(2))^(2)`

So, the obtained expression is a difference of squares since it matches with the form of the difference of squares.

Hence, the products that result in a difference of squares are:

C)
`(3+xz)(-3+xz)`

D)
`(y^2-xy)(y^2+xy)`

F)
`(64y^2+x^2)(-x^2+64y^2)`

Have a nice day!