anonymous
  • anonymous
Factor completely: 3a^2-20ab+12b^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[3a ^{2}-20ab+12b ^{2}\]
jim_thompson5910
  • jim_thompson5910
multiply the first and last coefficients 3*12 = 36
jim_thompson5910
  • jim_thompson5910
find two numbers that multiply to 36 and that add to -20

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More answers

anonymous
  • anonymous
18 and 2
jim_thompson5910
  • jim_thompson5910
actually -18 and -2 -18 times -2 = +36 -18 plus -2 = -20
anonymous
  • anonymous
What is the next step? I'm confused on the different variables.
jim_thompson5910
  • jim_thompson5910
\[\Large 3a ^{2}\color{red}{-20ab}+12b ^{2}\] \[\Large 3a ^{2}\color{red}{-18ab}\color{red}{-2ab}+12b ^{2}\] now factor by grouping
anonymous
  • anonymous
(3a^2-18ab) (-2ab+12b^2)
jim_thompson5910
  • jim_thompson5910
don't forget the plus in between (3a^2-18ab) + (-2ab+12b^2)
jim_thompson5910
  • jim_thompson5910
now factor each group
anonymous
  • anonymous
3a(a-6b) 2b(-1a+6b)
anonymous
  • anonymous
did i do it wrong? i feel like the 2nd group is incorrect.
jim_thompson5910
  • jim_thompson5910
(3a^2-18ab) + (-2ab+12b^2) 3a(a-6b) + (-2ab+12b^2) 3a(a-6b) -2b(a - 6b) there's one more step
anonymous
  • anonymous
What would the next step be?
anonymous
  • anonymous
oh i know it :D
anonymous
  • anonymous
(3a-2b) (a-6b)
jim_thompson5910
  • jim_thompson5910
good
anonymous
  • anonymous
Thanks for the help @jim_thompson5910 :)
jim_thompson5910
  • jim_thompson5910
no problem

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