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

Answer 1

Avery Alexander

#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.

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Answer 2

Genesis Allen

#(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#

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Answer 3

Ethan Adkins

To divide (v^3+27)/(v+3), we can use the long division method.

First, divide v^3 by v, which gives v^2. Multiply v^2 by (v+3), which gives v^3+3v^2. Subtract this from v^3+27 to get -3v^2+27.

Next, divide -3v^2 by v, which gives -3v. Multiply -3v by (v+3), which gives -3v^2-9v. Subtract this from -3v^2+27 to get 9v+27.

Finally, divide 9v by v, which gives 9. Multiply 9 by (v+3), which gives 9v+27. Subtract this from 9v+27 to get 0.

Therefore, the division of (v^3+27)/(v+3) is v^2-3v+9.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.