(Formatting copied from another answer by Truong-Son R.)
How do you use synthetic division to divide (2x^3 + x^2 - 2x + 2) by x+2?
First, you let the coefficients of each degree be used in the division (2, 1, -2, 2).
Then, dividing by x +2 implies that you use -2 in your upper left (if it was x-2, put 2). So, draw the bottom and right sides of a square, put -2 inside it, and then write "2" " 1" " -2" " 2" to the right.
"-2" || "2 " "1 " "-2 " "2 "
+
"" " "-----
First, bring the 2 down to the bottom, and multiply it by the -2. Put that -4 below 1.
"-2" || "2 " "1 " "-2 " "2 "
+ "" " " "" "-4"
"" " "-----
"" " " "2"
Then add it up:
"-2" || "2 " "1 " "-2 " "2 "
+ "" " " "" "-4"
"" " "-----
"" " " "2" "" " -3"
Repeat a few times once you've figured out the simple pattern ("divisor"*"new sum", put result under next-lowest degree term, add to get another "new sum", repeat).
"-2" || "2 " "1 " "-2 " "2 "
+ "" " " "" "-4" "" " 6"
"" " "-----
"" " " "2" "" " -3" "" " 4"
"-2" || "2 " "1 " "-2 " "2 "
+ "" " " "" "-4" "" " 6" " -8"
"" " "-----
"" " " "2" "" " -3" "" " 4"" -6"
You know you can stop when you reach the far right and you have no spot left below the original dividend (2, 1, -2, 2). to insert a product.
Since you started with a cubic, the answer is a quadratic. Thus, you have:
2x^2 -3x + 4 (r=-6)
Indeed, if you multiply them together and add the remainder, you get the original back:
(2x^2 - 3x + 4)(x+2) = 2x^3 - 3x^2 + 4x^2 + 4x - 6x + 8 +(-6) = 2x^3 + x^2 - 2x + 2
So you would have:
(2x^3 + x^2 - 2x + 2)/(x+2) = 2x^2-3x+4 + (-6)/(x+2)
or
(2x^3 + x^2 - 2x + 2)/(x+2) = 2x^2-3x+4 - 6/(x+2)
