According to remainder theorem when we divide a polynomial #P(x)# (here #P(x)=6x^3-x^2+4x+3#), by a binomial of degree one say #(x-a)#, (here #(x-3)# as we desire #P(a)# wih #a=3#)
the remainder is #P(a)#
Normally to find #P(a)#, one substitute #x# with #a#, but as we have to use Remainder theorem, we should divide #P(x)# by #(x-a)# and then the remainder would be #P(a)#. For this let us use synthetic division to divide #6x^3-x^2+4x+3# by #(x-3)#.
#3|color(white)(X)6" "color(white)(X)-1color(white)(XXX)4" "" "3#
#color(white)(1)|" "color(white)(XXxx)18color(white)(XXX)51color(white)(XX)165#
#" "stackrel("—————————————)#
#color(white)(x)|color(white)(X)color(blue)6color(white)(Xxxx)color(blue)17color(white)(XXX)color(blue)55color(white)(XX)color(red)168#
i.e. while quotient is #6x^2+17x+55#, remainder is #168#.
