How do you multiply #(0.2y+1.7)^2#?

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Answer 1 · 2016-12-24T13:08:01

This is my handwriting only
Can do this much only! I hope u find what is needed.

#=0.04y^2 +0.68y +2.89#

#(0.2y+1.7)^2#

#=(0.2y+1.7)(0.2y+1.7)#

#= 0.04y^2 +0.34y+0.34y + 2.89#

#=0.04y^2 +0.68y +2.89#

Or as shown above, using the 'short cut' identity:

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

#(0.2y+1.7)^2#

#=0.04y^2 +0.68y +2.89#

Terms are usually arranged in descending order of the powers.