Question

How do you use synthetic division to divide `3x^3 - x^2 + 8x-10` by `x-1`?

Answer 1

This will be a bit hard to write out here, but let's see. I'm assuming that this is a typo. You may have meant `3x^3`. If it was `3x^2`, you wouldn't need synthetic division and you could just factor this to get `x = 1, x = -5`.

Assuming `3x^3`... First, you let the coefficients of each degree be used in the division (`3, -1, 8, -10`).

Then, dividing by `x - 1` implies that you use `1` in your upper left (if it was `x+1`, put `-1`). So, draw the bottom and right sides of a square, put `1` inside it, and then write `"3" " -1" " 8" " -10"` to the right.

`"1" ||` `"3 " "-1 " "8 " "-10 "`
`+`
`"" " "` `-----`

First, bring the `3` down to the bottom, and multiply it by the `1`. Put that `3` below `-1`.

`"1" ||` `"3 " "-1 " "8 " "-10 "`
`+` `"" " " "" "3"`
`"" " "` `-----`
`"" " " "3"`

Then add it up:

`"1" ||` `"3 " "-1 " "8 " "-10 "`
`+` `"" " " "" "3"`
`"" " "` `-----`
`"" " " "3" "" " 2"`

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).

`"1" ||` `"3 " "-1 " "8 " "-10 "`
`+` `"" " " "" "3" "" " 2"`
`"" " "` `-----`
`"" " " "3" "" " 2" "" " 10"`

(All that really happened here was `1*2 = 2` and `8 + 2 = 10`.)

`"1" ||` `"3 " "-1 " "8 " "-10 "`
`+` `"" " " "" "3" "" " 2" " 10"`
`"" " "` `-----`
`"" " " "3" "" " 2" "" " 10" " 0"`

(Then to finish it up, `1*10 = 10`, and `-10 + 10 = 0`.)

You know you can stop when you reach the far right and you have no spot left below the original dividend (`"3" " -1" " 8" " -10"`) to insert a product.

Since presumably you started with a cubic, the answer is a quadratic. Thus, you have:

`3x^2 + 2x + 10x^0 + (r=0)`

Indeed, if you multiply them together, you get the original back:

`(3x^2 + 2x + 10)(x-1) = 3x^3 - 3x^2 + 2x^2 - 2x + 10x - 10 = 3x^3 - x^2 + 8x - 10`

So the answer would be `3x^2 + 2x + 10`. In a general case, you could write it like this:

Let `f(x) = ax^3 + bx^2 + cx + d` and `g(x) = x - h`. If the solution is `h(x)`, and a remainder is `r(x)`, then:

`h(x) = f(x)/g(x) + (r(x))/(g(x))`

So you would have:
`h(x) = "answer" = (3x^3 - x^2 + 8x - 10)/(x-1) color(blue)(= 3x^2 + 2x + 10 + (r(x))/(x-1))`

where `r(x) = 0`.

---

Had you meant `3x^2 - x^2 + 8x - 10`, you could have just done the quadratic equation. Or, factor.

`3x^2 - x^2 + 8x - 10 = 2x^2 + 8x - 10 => (2x - 2)(x + 5) = 0`
`=> x = 1, x = -5`