Question

How do you use synthetic division to divide `(2x^3 + x^2 - 2x + 2)` by `x+2`?

Answer 1

(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)`