View Full Version : Roots of x^3 + 216
Alkatran
Sep18-04, 01:59 PM
I need to factorise x^3 + 216 to include (x+6):
(x + 6) \ (x^3 + 216) (lim where x approaches -6)
I broke it down to (x + 6i)(x^2 - 36i) ... but that's no good.
arildno
Sep18-04, 02:05 PM
Well, shouldn't polynomial division work fine?
Pyrrhus
Sep18-04, 02:07 PM
Remember
X^3 + Y^3 = (X+Y)(X^2-XY+Y^2)
Pyrrhus
Sep18-04, 02:08 PM
He has
\frac{x+6}{x^3 + 216}
\frac{x+6}{x^3 + 6^3}
Alkatran
Sep18-04, 06:31 PM
I got it fine before I even came back. I was trying to remember how to do long division and the cube rule that you guys posted. The start of the year, still getting restarted...
vBulletin® v3.8.7, Copyright ©2000-2012, vBulletin Solutions, Inc.