Question

How do you divide `(x^3+9x^2+23x+15)div(x+5)` using synthetic division?

Answer 1

`(x^3+9x^2+23x+15)div(x+5)=color(green)(x^2+4x+3)color(white)("XXX")"Remainder: "color(green)(0)`
(see synthetic division method below)

Explanation:

Set up a table with the coefficients of the dividend on the first row (`color(red)("row A")` below) leaving a little extra space at the beginning of the line.

Leave a blank line (to be filled in as we go along).

Write the negative of the divisor at the far left of the third line (`color(red)("row C")` below).

You should have something that looks like:
`color(white)("XXXX")color(cyan)(x^3)color(white)("XX")color(cyan)(x^2)color(white)("XX")color(cyan)(x^1)color(white)("XX")color(cyan)(x^0)`
`color(white)("XXx")|color(white)("XX")1color(white)("XX")9color(white)("XX")23color(white)("XX")15color(white)("X")color(red)(" row A")`
`underline(color(white)("xxx")|color(white)("XXXXXXXXXXXX"))color(red)(" row B")`
`-5color(white)("x")|color(white)("XXXXXXXXXXXXXX")color(red)("row C")`

For each coefficient column in turn from left to right:
* Add the value in `color(red)("row B")` to the (coefficient) value in `color(red)("row A")` and write the result in `color(red)("row C")`
* Multiply the negative divisor (on the far left of `color(red)("row C")`) by the value just written in `color(red)("row C")` and writer the product in ** the next column** of `color(red)("row B")`

After processing the first column, your result should look like:
`color(white)("XXx")|color(white)("XX")1color(white)("XX")9color(white)("XX")23color(white)("XX")15color(white)("X")color(red)(" row A")`
`underline(color(white)("XXx")|color(white)("XXxx")-5color(white)("XXXXXXX"))color(red)(" row B")`
`-5color(white)("x")|color(white)("X")1color(white)("XXXXXXXXXXXx")color(red)("row C")`

...and after processing all the columns, like:
`color(white)("XXx")|color(white)("XX")1color(white)("XX")9color(white)("XX")23color(white)("XX")15color(white)("X")color(red)(" row A")`
`underline(color(white)("XXx")|color(white)("XXxx")-5color(white)("x")-20color(white)("x")-15)color(red)(" row B")`
`-5color(white)("x")|color(white)("X")1color(white)("XX")4color(white)("XXX")3color(white)("XX")0color(white)("X")color(red)("row C")`

The final result in `color(red)("row C")` gives the coefficients of the resulting quotient plus (as the last value in `color(red)("row C")` the remainder:
`-5color(white)("x")|color(white)("X")color(cyan)(underline(1color(white)("XX")4color(white)("XXX")3))color(white)("XX")color(brown)(0)color(white)("XXXXXXX")color(red)("row C")`
`color(white)("XXXX")color(cyan)(x^2color(white)("XX")x^1color(white)("XX")x^0)color(white)("XX")color(brown)("Remainder")`