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June 17th, 2016, 09:19 AM  #1 
Newbie Joined: Jun 2016 From: United States Posts: 10 Thanks: 1  Binomial Factors
Hello, I have a problem (shown below), am I suppose to take out the (3x4) first? If so, will that be to the left and everything else on the righthand side? 5(3x4)²  8(3x4)(5x1) 
June 17th, 2016, 09:41 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
Yes, since there is a "3x 4" in both parts, you can use the "distributive law", (ab+ ac)= a(b+ c), to get (3x 4)[5(3x 4)+ 8(5x 1)]= (3x 4)(15x 20+ 40x 8]= (3x 4)(55x 28). (As for "left and right", if that is what you are asking about, it doesn't matter. You could also write (55x 28)(3x 4).)
Last edited by greg1313; June 17th, 2016 at 09:47 AM. 
June 17th, 2016, 09:52 AM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,934 Thanks: 1128 Math Focus: Elementary mathematics and beyond 
5(3x  4)²  8(3x  4)(5x  1) = (3x  4)[5(3x  4)  8(5x  1)] = (3x  4)(15x  20  40x + 8) = (3x  4)(25x  12) = (3x  4)(25x + 12) = (4  3x)(25x + 12) 
June 17th, 2016, 12:57 PM  #4 
Newbie Joined: Jun 2016 From: United States Posts: 10 Thanks: 1 
Wow, Okay, really good answers from both. Thanks for the help it makes a lot more sense now!


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