How do you divide #(3x^4 + 2x^3 - 11x^2 - 2x + 5)/ (x^2 - 2) # using polynomial long division?

#3x^4+2x^3-11x^2-2x+5|x^2-2#
you ask what is it #(3x^4)/x^2# and you get #3x^2#

(REMEMBER #3x^2#)

now yow duplicate #3x^2# to #x^2-2 # and get #3x^4-6x^2#
and do:
#3x^4+2x^3-11x^2-2x+5#
#-#
#3x^4+0x^3-6x^2+0x+0#
#=#
#0+2x^3-5x^2-2x+5#
so we have now #2x^3-5x^2-2x+5# which is very exiting because we don't have the power of 4 anymore!

now we do the same again, try to follow:

First step (write it down):
#2x^3-5x^2-2x+5|x^2-2#

Second step (divided strongest power):
#(2x^3)/(x^2)=2x#

(REMEMBER #2x#)

Third step (Second step times the divider):
#2x*(x^2-2)=2x^3-4x#

Forth step (which is first step minus third step):
#2x^3-5x^2-2x+5#
#-#
#2x^3+0x^2-4x+0#
#=#
#0-5x^2+2x+5#

--and agian--

First step (write it down):
#-5x^2+2x+5|x^2-2#

Second step (divided strongest power):
#(-5x^2)/(x^2)=-5#

(REMEMBER #-5#)

Third step (Second step times the divider):
#-5*(x^2-2)=-5x^2+10#

Forth step (which is first step minus third step):
#-5x^2+2x+5#
#-#
#-5x^2+10#
#=#
#0+2x-5#

All REMEMBERs are the result:
#3x^2+2x-5#

BUT in that case, #2x-5# could not be divided so it remains as #(2x-5)/(x^2-2)#

SOOOOOOOOOOOOOOOOOOOOOOO :)

#(3x^4+2x^3-11x^2-2x+5)/(x^2-2)=3x^2+2x-5+(2x-5)/(x^2-2)#

Lets check:

#(3x^2+2x-5+(2x-5)/(x^2-2))(x^2-2)=#
#=3x^4-6x^2+2x^3-4x-5x^2+10+2x-5=#
#=3x^4+2x^3-11x^2-2x+5#

So it's fine :)