How do I find the quotient and remainder using synthetic division?

Let's take this as an example:

#f(x) = x^3 - x^2 + x - 6#

To find factors, notice that #6# has factors of #pm1,pm2,pm3,pm6#. Pick one and try synthetic division on it, and if you pick the right one (meaning that it divides), it'll give a remainder of #0#.

I pick #2#, so I am assuming that #x-2# divides #x^3 - x^2 + x - 6#. This means the result should be of the form:

#color(blue)((x^3 - x^2 + x - 6)/(x-2) = q(x) + r(x)#

where #q(x)# is the quotient and #r(x)# is the remainder.

For synthetic division, we only use coefficients. So, we start with:

#color(white)([(color(black)(2),color(black)(|),color(black)(1),color(black)(-1),color(black)(1),color(black)(-6)),(color(black)(-),color(black)(""),color(black)("_"),color(black)("_"),color(black)("_"),color(black)("_")),(color(black)(""),color(black)(""),color(black)(""),color(black)(""),color(black)(""),color(black)(""))])#

If you don't have a term, use #0#. So if you had #x^3 - 6#, use #"1 0 0 "-"6"#.

The general process is:

• Bring down the first coefficient.
• Multiply it by the test factor (in this case, #2#), and store the result under the next coefficient.
• Subtract the result from this coefficient.
• Repeat steps 2 and 3 until you've subtracted the result from the last coefficient.

The result is one degree lower.

#=> color(white)([(color(black)(2),color(black)(|),color(black)(1),color(black)(-1),color(black)(1),color(black)(-6)),(color(black)(-),color(black)(""),color(black)("_"),color(black)(ul(2)),color(black)("_"),color(black)("_")),(color(black)(""),color(black)(""),color(black)(1),color(black)(-3),color(black)(""),color(black)(""))])#

#" "#

#=> color(white)([(color(black)(2),color(black)(|),color(black)(1),color(black)(-1),color(black)(1),color(black)(-6)),(color(black)(-),color(black)(""),color(black)("_"),color(black)(ul(2)),color(black)(ul(-6)),color(black)("_")),(color(black)(""),color(black)(""),color(black)(1),color(black)(-3),color(black)(7),color(black)(""))])#

#" "#

#=> color(white)([(color(black)(2),color(black)(|),color(black)(1),color(black)(-1),color(black)(1),color(black)(-6)),(color(black)(-),color(black)(""),color(black)("_"),color(black)(ul(2)),color(black)(ul(-6)),color(black)(ul(14))),(color(black)(""),color(black)(""),color(black)(\mathbf(1)),color(black)(\mathbf(-3)),color(black)(\mathbf(7)),color(black)(\mathbf(-20)))])#

So, our result is #color(blue)(x^2 - 3x + 7 - 20/(x-2))# for #x ne 2#. What we have is:

#\mathbf(q(x) = x^2 - 3x + 7)#
#\mathbf(r(x) = -20/(x-2))#

If #r(x) = 0#, then #q(x)# is a quotient that divides #f(x) = x^3 - x^2 + x - 6#.