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Mathematics 16 Online
OpenStudy (anonymous):

Factor completely 3x3 – 21x2 – 27x

OpenStudy (anonymous):

3x^3-21x^2-27x= 3x(x^2-7x-9)= =(x-(7+sqrt(85))/2) or (x-(7+sqrt(85))/2)

OpenStudy (anonymous):

sry, 3x(x-(7+sqrt(85))/2) or 3x(x-(7+sqrt(85))/2)

OpenStudy (jagatuba):

\[3x(x ^{2}-7x-9)\]

OpenStudy (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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