Using synthetic division, I solved the equation below, however, I'm stuck trying to write it in the following format #" dividend"= "quotient" * "divisor"+ "remainder"#? #x^3-3x^2+7x-1 -: x-3# thanks in advance

#(x^3-3x^2+7x-1)/(x-3)=(x^2(x-3)+7x-21+20)/(x-3) = #

#x^2+(7(x-3)+20)/(x-3)=x^2+7+20/(x-3)#

Now we multiply through by #(x-3)# giving

#x^3-3x^2+7x-1=(x^2+7)(x-3)+20#

Let's start with the terms: #color(blue)("dividend")# and #color(red)("divisor")#

In the expression: #color(blue)(x^3-3x^2+7x-1) div color(red)(x-3)#
#color(blue)(x^3-3x^2+7x-1)# is the #color(blue)("dividend")#
and
#color(red)(x-3)# is the #color(red)("divisor")#

Performing synthetic division:
#{: (color(white)("xxx"),,color(gray)(x^3),color(gray)(x^2),color(gray)(x^1),color(gray)(x^0)), (," | ",1,-3,7,-1), (ul(+color(white)("X")),ul(" | "),ul(color(white)("xxx")),ul(3),ul(0),ul(21)), (xx3," | ",color(lime)1,color(lime)0,color(lime)7,color(magenta)20), (,,color(gray)(x^2),color(gray)(x^1),color(gray)(x^0),color(gray)("Remainder")) :}#

#color(lime)1#, #color(lime)0#, and #color(lime)7# are the coefficients of the #color(lime)("quotient")#
and
#color(magenta)(20)# is the #color(magenta)("Remainder")#

Therefore in the form
#color(white)("XXX")color(blue)("dividend")=color(lime)("quotient") * color(red)("divisor") + color(magenta)("remainder")#
we have
#color(white)("XXX")color(blue)(""(x^3-3x^2+7x-1))=color(lime)(""(1x^2+0x+7)) * color(red)(""(x-3))+color(magenta)(20)#