anonymous
  • anonymous
Factor completely 3x3 – 21x2 – 27x
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
3x^3-21x^2-27x= 3x(x^2-7x-9)= =(x-(7+sqrt(85))/2) or (x-(7+sqrt(85))/2)
anonymous
  • anonymous
sry, 3x(x-(7+sqrt(85))/2) or 3x(x-(7+sqrt(85))/2)
jagatuba
  • jagatuba
\[3x(x ^{2}-7x-9)\]

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jagatuba
  • jagatuba
It is all about what you can evenly pull out of the expression. Start with x. How many x's can you pull from each term? Since 27x only has one x you can only pull one from the other two terms as well so that leaves you with: \[x(3x ^{2}-21x-27)\] Now what is the biggest factor that you can pull out of the numbers in each term. Since 3 is the smallest number it cannot be bigger than that. So check. Are the other numbers 21 and 27 divisibile by 3? Yep so lets pull 3 from each term: \[3/3=1, -21/3=-7, -27/3=-9\] SO now we have the factor that we pulled out (3x) times what we have left or: \[3x(x ^{2}-7x-9)\]

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