Question

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

Answer 1

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`.