How do you evaluate #(3x - 5) ( 4x - 2u - 1)#?

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Answer 1 · 2017-10-30T01:46:27

FOIL method

#(3x-5)(4x-2u-1)#

distribute #3x# to everything in the second parentheses

that should give you: #12x^2-6xu-3x#

next, distribute #-5# to #(4x-2u-1)#

#-5(4x-2u-1) = 10u-20x+5#

add #12x^2-6xu-3x# and #10u-20x+5# by combining like terms

and finally you should have:

#-6ux+12x^2+10u-23x+5#