How do you multiply #(3n-2)(2n+5)#?

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Answer 1 · 2018-06-11T06:41:02

#6n^2+11n-10#

As per the question, we have

#(3n-2)(2n+5)#

#=(3nxx2n)+(3nxx5)+(-2xx2n)+(-2xx5)#

#=6n^2+15n+(-4n)+(-10)#

#=6n^2+15n-4n-10#

#=6n^2+11n-10#

Hence, the answer.