How do you divide # s+s^2 -8# by #x+4#?

1 Answer
Oct 17, 2015

#-2 + (s^2+s +2x)/(x+4)#

Another way of writing it: # -2 + (s^2+s)/(x+4) + (2x)/(x+4)#

Explanation:

I do not know how to set up long division on this site so I will do my best without it:

Based on the assumption you mean: #(s^2+s -8) div (x+4)#

There is nothing to associate #x# with #s# so the only thing we can reasonably divide are the constants of (-8) and 4. We would then be left over with some form of unsolvable fraction which would have to be written 'as is'.

Write #(s^2+s -8) div (x+4)" as "(s^2+s -8)/(x+4)#

considering just the constants: #(-8) div 4 = -2#

So the first part of our solution is -2

If the division gives -2 then we have to subtract
#-2 times (x+4) = -2x - 8# ................ ( 1 )
from the original expression to determine what is left. This in turn will also have to be divided by #( x + 4 )#. This process is like the old remainder system you would have been taught some years back.

so we have in effect a remainder of:

# ( (s^2 + s -8 ) - (-2)(x+4))/(x+4)#

# ( (s^2 + s -8 ) +2(x+4))/(x+4)# ....... ( 2 )

Adding the remainder to the first part we have:

#"part (1) + part(2)"-> -2 + ( (s^2 + s -8 ) +2(x+4))/(x+4)#

This would give:

#-2 + (s^2+s +2x)/(x+4)#

or # -2 + (s^2+s)/(x+4) + (2x)/(x+4)#

Addendum: Checking the solution

If my solution is correct then multiplying it by #(x+4)# should give us the original expression. This would take the form:

#(x+4)[-2+(s^2+s+2x)/(x+4)]#

#=> -2x -8 +s^2 + s +2x#

which gives us: #s^2 + s -8#
so correct!!!!