How do you expand #(6y^4-1)^2#?

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Answer 1 · 2018-06-13T12:19:02

#color(blue)[(6y^4-1)^2=(36*y^8-12y^4+1)]#

Note that:

#color(red)[(a+b)^2=a^2+2ab+b^2]#

#color(red)[(a-b)^2=a^2-2ab+b^2]#

now lets expand #(6y^4-1)^2#

#color(blue)[(6y^4-1)^2=(36*y^8-12y^4+1)]#

Answer 2 · 2018-06-13T12:22:11

#(6y^4-1)^2 = 36y^8 - 12y^4 + 1#

The square of a sum can be expanded as

#(a+b)^2 = a^2+2ab+b^2#

Which means that you have to square both terms, and add twice their product.

The two terms are #6y^4# and #-1#. Their squares are #36y^8# and #1#.

Their product is #-6y^4#, so twice their product is #-12y^4#

Now we only need to sum everything together to get

#(6y^4-1)^2 = 36y^8 - 12y^4 + 1#