How do you multiply #(6g + 4) ( g - 20)#?

2 Answers
alyssialui
Sep 30, 2016

#= 6g^2 -116g -80#

Explanation:

  1. Expand the brackets by multiplying each term in the first bracket with each term in the second bracket. (eg. #(a+b)(c+d) = ac + ad + bc + bd#) The signs for each term should be maintained.
    Therefore,
    #(6g + 4)(g - 20)#
    #= (6g)(g) + (6g)(-20) + (4)(g) + (4)(-20)#
    #= 6g^2 -120g + 4g -80#
  2. Group like terms
    #= 6g^2 -116g -80#
Shell
Sep 30, 2016

#6g^2-116g-80#

Explanation:

#(6g+4)(g-20)#

Multiply each term in the first binomial by each term in the second.
Some teachers refer to this as FOIL: multiply the First 2 terms, then the Outer 2 terms, then the Inner 2 terms, then the Last 2 terms.

#(6g*g) +(6g*-20) +(4*g)+(4*-20)#

#6g^2-120g+4g-80#

#6g^2-116g-80#

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