How do you divide #(v^3+27)/(v+3)#?

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Answer 1 · 2015-08-05T21:53:55

#(v^3+27)/(v+3)=v^2-3v+9#

Assume #v+3# is a factor for #v^3+27# and from this infer the remaining factor. This gives:
#v^3+27=(v+3)(v^2-3v+9)#

Therefore:
#(v^3+27)/(v+3)=v^2-3v+9#

Answer 2 · 2015-08-06T01:11:54

#v^3 + 27# is of the form:

#a^3 pm b^3#

Thus, factoring it asks for:

#(a + b)(a^2 - ab + b^2)#
#= a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3 = a^3 + b^3#

or:

#(a - b)(a^2 + ab + b^2)#
#= a^3 + a^2b + ab^2 - a^2b - ab^2 - b^3 = a^3 - b^3#

#a = v#
#b = 3#

So you get, with the first one:

#= (v + 3)(v^2 - 3v + 9)#

#-> ((v + 3)(v^2 - 3v + 9))/(v+3) = color(blue)(v^2 - 3v + 9)#

No guessing necessary.