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How do you long divide #(2n^3 + 0n^2 - 14n + 12) /(n + 3)#?

1 Answer

Aug 7, 2015

#2(n-2)(n-1)#

Explanation:

Assume #n+3# is a factor for the numerator and infer the other factor:
#2n^3-14n+12=(n+3)(an^2+bn+c)=#
#an^3+(b+3a)n^2+(c+3b)n+3c#

This gives the result:
#a=2#
#b+3a=b+6=0=>b=-6#
#c+3b=c-18=-14=>c=4#
#3c=12#

Therefore #n+3# is a factor and we have:
#(2n^3-14n+12)/(n+3)=(cancel((n+3))(2n^2-6n+4))/cancel(n+3)=#
#2(n^2-3n+2)=2(n-2)(n-1)#

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