How do you evaluate #(9b ^ { 3} + 52b ^ { 2} + 107b + 60) \div ( 9b + 7)#?

1 Answer
George C.
Feb 1, 2017

#(9b^3+52b^2+107b+60)/(9b+7)=b^2+5b+8+4/(9b+7)#

Explanation:

Note that #60# is not evenly divisible by #7#, so there will be a remainder.

One way of performing the division is to split the numerator into multiples of the divisor like this:

#(9b^3+52b^2+107b+60)/(9b+7)#

#=((9b^3+7b^2)+(45b^2+35b)+(72b+56)+4)/(9b+7)#

#=(b^2(9b+7)+5b(9b+7)+8(9b+7)+4)/(9b+7)#

#=((b^2+5b+8)(9b+7)+4)/(9b+7)#

#=b^2+5b+8+4/(9b+7)#

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