How do you multiply #(3t-2)(4t+2)#?

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Answer 1 · 2018-04-03T02:48:30

The answer is #12t^2-2t-4#.

First,Multiply the equation #( 4t +2) # by #(3t)# and #( 4t +2) # by #(-2)#.

Converting it into an equation,

#rArr# #3t(4t+2) -2(4t+2)#

#rArr# #12t^2+6t -8t-4#

#rArr# #12t^2+(6-8)t-4#

#rArr# #12t^2-2t-4#

#:.# the answer is #12t^2-2t-4#.

Let's hope this helps you:)