How do you divide #(x^2 - 2x - 15)/(x + 3)# using polynomial long division?

Given: #(x^2-2x-15)/(x+3)........................(1)#

Not all questions permit this approach of solution!

Consider #color(white)(..)x^2-2x-15#

This can be factored into:

#(x-5)(x+3).............................(2)#

Substitute expression (2) into expression (1)

#((x-5)(x+3))/(x+3)#

Write as: #(x+3)/(x+3) xx (x-5)#

But #(x+3)/(x+3)# has the value of 1 giving

#1xx (x-5)#

#x-5#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Foot note")#

Consider: #(x+3)/(x+3)#

If you were investigating values then this produces a problem.
You are not mathematically allowed to divide by 0.

So #(x+3)/(x+3)# is 'Undefined' at #x=-3#

For this very reason #0/0# does #underline(color(red)("not equal 1"))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Given: #(x^2-2x-15) -:(x+3)#

#color(white)("ddddddd.dd") x^2-2x-15#
#color(magenta)(x)(x+3)-> ul(x^2+3x larr" Subtract")#
#color(white)("dddddddddd") 0 color(white)(",") -5x-15#
#color(magenta)(-5)(x+3)-> color(white)("d") ul(-5x-15 larr" Subtract")#
#color(white)("dddddddddddddd") 0+0#

#(x^2-2x-15) -:(x+3) = color(magenta)(x-5)#