How do you long divide #n^3# by #n^2-4# ?

1 Answer
Dec 8, 2016

#n^3/(n^2-4) = n + (4n)/(n^2-4)#

Explanation:

The slightly tricky thing with long dividing such polynomials is that you need to include all relevant powers of #n#.

So in our example, the divisor should be treated as #n^2+0n-4#.

Then all of the terms should line up correctly and the result will be correct:

#color(white)(n^2 +0n - 4 " ") underline(color(white)(" "0) n color(white)(n^3 + 0n^2 + 0n + 0))#
#n^2 +0n -4 " " ) " " n^3 + 0n^2 + 0n + 0#
#color(white)(n^2 +0n -4 " " x " ") underline(n^3 + 0n^2 - 4n)#
#color(white)(n^2 +0n -4 " " x " " n^3 + 0n^2 +) 4n + 0#

So the skew asymptote is the function #f(n) = n#

and we can express the division as:

#n^3/(n^2-4) = n + (4n)/(n^2-4)#

graph{(y - (x^3/(x^2-4)))(y-x) = 0 [-20, 20, -10, 10]}