Question

How do you find the quotient of `(x^3+3x^2-3x-2)div(x-1)` using long division?

Answer 1

`x^3 + 3x^2 - 3x - 2 = ( x -1)(x^2 + 4x + 1) - 1`

Explanation:

`text{ -------------------------}`
`x -1 quad text{)} quad x^3 + 3x^2 - 3x - 2`

That's a pain to format. Anyway, the first "digit", first term in the quotient, is `x^2`. We compute the digit times `x-1`, and take that away from `x^3 + 3x^2 - 3x -2`:

`text{ } x^2`
`text{ -------------------------}`
`x -1 quad text{)} quad x^3 + 3x^2 - 3x - 2`
`text{ } x^3 -x^2`
`text{ ----------------}`
`text{ } 4 x^2 - 3x - 2`

OK, back to the quotient. The next term is `4x` because that times `x` gives `4 x^2`. After that the term is `1`.

`text{ } x^2 + 4 x + 1`
`text{ -------------------------`
`x -1 quad text{)} quad x^3 + 3x^2 - 3x - 2`
`text{ } x^3 -x^2`
`text{ ----------------}`
`text{ } 4 x^2 - 3x - 2`
`text{ } 4 x^2 - 4x`
`text{ ----------------}`
`text{ } x - 2`
`text{ } x - 1`
`text{ --------}`
`text{ } -1`

We have a non-zero remainder! That says

`x^3 + 3x^2 - 3x - 2 = ( x -1)(x^2 + 4x + 1) - 1`