How do you multiply #(9-2x^4)^2#?

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Answer 1 · 2018-04-05T02:15:30

#81-36x^4+4x^8#

#(9-2x^4)^2#

=#(9-2x^4)(9-2x^4)#

=#81-18x^4-18x^4+4x^8#

=#81-36x^4+4x^8#

Answer 2 · 2018-04-05T02:40:40

#(9-2x^4)^2=color(blue)(4x^8-36x^4+81#

Multiply/Simplify/Expand:

#(9-2x^4)^2#

Use the square of a difference:

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

where:

#a=9#, and #b=2x^4#

Plug in the known values.

#(9-2x^4)^2=9^2-(2*9*2x^4)+(2x^4)^2#

Simplify #9^2# to #81#.

#(9-2x^4)^2=81-(2*9*2x^4)+(2x^4)^2#

Apply the multiplicative distributive property: #(ab)^m=a^mb^m"#

#(9-2x^4)^2=81-(2*9*2x^4)+2^2*(x^4)^2#

Apply power rule: #(a^m)^n=a^(m*n)#

#(9-2x^4)^2=81-(2*9*2x^4)+2^2*x^8#

Simplify.

#(9-2x^4)^2=81-36x^4+4x^8#

Rearrange the equation in descending order.

#(9-2x^4)^2=4x^8-36x^4+81#