We use long division method to divide #3c^5+5c^2+c+5# by #(c+2)#
#color(white)(xxxxxxx)3c^4-6c^3+12c^2-19c+39#
#color(white)(xx) c+2| bar(3c^5+0c^4+0c^3+5c^2+c+5)#
#color(white)(xxxxxxx)ul(3c^5+6c^4)color(white)(xx)color(red)(darr)# - subtracting
#color(white)(xxxxxxxxx)-6c^4+0c^3#
#color(white)(xxxxxxx**x)ul(-6c^4-12c^3)#
#color(white)(xxxxxxxxxxxxxxx)12c^3+5c^2#
#color(white)(xxxxxxxxxxxxxxx)ul(12c^3+24c^2#
#color(white)(xxxxxxxxxxxxxxxxxx)-19c^2color(white)(xx)+c#
#color(white)(xxxxxxxxxxxxxxxxxxx)ul(-19c^2-38c)#
#color(white)(xxxxxxxxxxxxxxxxxxxxxxxxx)39c+5#
#color(white)(xxxxxxxxxxxxxxxxxxxxxXXX) ul(39c+78)#
#color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxx)-73#
WE can check it using Remainder theorem, as #3c^5+5c^2+c+5# is divided by #c+2#,
we should get a remainder #3(-2)^5+5(-2)^2-2+5=3xx(-32)+5xx4-2+5=-96+20-2+5=-73#
and hence #(3c^5+5c^2+c+5)/(c+2)=3c^4-6c^3+12c^2-19c+39-73/(c+2)#
