Question

How do you use synthetic division to divide `x^3- 6x^2+ 11x - 6` by `x-1`?

Answer 1

Synthetic division is generally used, in finding zeroes (or roots) of polynomials.

Explanation:

Well since the polynomial `x^3-6x^2+11x-6=(x-1)(x-2)(x-3)`

Thus the quotient of division with `x-1` is `(x-2)(x-3)` and the remainder is `0`

Answer 2

See the explanation section.

Explanation:

Divide `x^3 -6x^2 +11 x - 6` by `x-1`.

First, you let the coefficients of each degree to be used in the division (`1, -6, 11, -6`).

Then, dividing by `x-1` implies that you use `1` in your upper left. So, put `1`, then a line (I've used a double line), and then write `" 1" " -6" " 11" " -6 "` to the right.

`"1 " ||` `" 1" " -6" " 11" " -6 "`
`+`
`" " " "` `-----`

First, bring the first `1` down to the bottom, and multiply it by the `1`. Put that `1` below the `-6`.

`"1 " ||` `" 1" " -6" " 11" " -6 "`
`+` `" "" " " 1"`
`" " " "` `-----`
`" "" " " 1"`

Then add `-6+1 = -5` Put that under the `1`

`"1 " ||` `" 1" " -6 " " 11" " -6 "`
`+` `" "" " " 1"`
`" " " "` `-----`
`" " " " " 1" " -5"`

Multiply `1 xx -5 = -5` and put the `-5` under the `11`. Then add:

`"1 " ||` `" 1" " -6 " " 11 " " -6 "`
`+` `" "" " " 1 " " -5"`
`" " " "` `--------`
`" " " " " 1 " " -5 " " 6"`

Now `1 xx 6 = 6`, so we put `6` under `-6` and add::

`"1 " ||` `" 1" " -6 " " 11 " " -6 "`
`+` `" "" "" 1 " " -5 " " 6"`
`" " " "` `--------`
`" "" " " 1 " " -5 " " 6 " |"0 "`

The bottom row ignoring the last number gives us the coefficients of the quotient.
The last number on the bottom row is the remainder (and it is also `P(1)`).

So the division gives us:

`(x^3 -6x^2 +11 x - 6) /(x-1) = x^2-5x+6`

You can check the answer by multiplyng:

`(x-1)( x^2-5x+6)` to make sure we get `x^3 -6x^2 +11 x - 6`.

(I've used Synthetic Division Formatting by Truong-Son R.)