How do you multiply #(x ^ { 2} + 8) ^ { 4} 2x#?

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Answer 1 · 2018-01-04T20:12:20

#2x^9+64x^7+786x^5+41689x^3+8192x#

First, we need to expand the #(x^2+8)^4#, which is

#(x^2+8) xx (x^2+8) xx (x^2+8) xx (x^2+8)#

#(x^2+8)^2 xx (x^2+8)^2#

#(x^4+8x^2+8x^2+64) xx (x^4+8x^2+8x^2+64)#

#(x^4+16x^2+64) xx (x^4+16x^2+64)#

#x^8+16x^6+64x^4+16x^6+256x^4+1024x^2+64x^4+1024x^2+4096#

#x^8+32x^6+384x^4+2048x^2+4096#

Our new expression is #(x^8+32x^6+384x^4+2048x^2+4096)2x# which can be simplified to

#2x^9+64x^7+786x^5+41689x^3+8192x#