How do you long divide #n^3# by #n^2-4# ?
1 Answer
Dec 8, 2016
Explanation:
The slightly tricky thing with long dividing such polynomials is that you need to include all relevant powers of
So in our example, the divisor should be treated as
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
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]}