How do you divide #(3b ^ { 5} - 9b ^ { 4} + 3b ^ { 2} ) -: 6b ^ { 2}#?

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Answer 1 · 2017-05-04T21:17:20

#=(b^3-3b^2+1)/2#

all terms can be factorised as multiples of #3b^2#:

#3b^5 = b^3*3b^2#

#9b^4 = 3*b^2*3b^2#

#3b^2 = 1*3b^2#

#6b^2 = 2*3b^2#

#(3b^5 - 9b^4 + 3b^2) = 3b^2(b^3-3b^2+1)#

#(3b^2(b^3-3b^2+1))/(3b^2(2))#

#=(cancel(3b^2)(b^3-3b^2+1))/(cancel((3b^2))(2))#

#=(b^3-3b^2+1)/2#