Question

I NEED IMMEDIATE HELP!!!!

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!