Question

How do you divide `(x^5 + x^3 − 9)/(x-1)`?

Answer 1

`(x^5+x^3-9)/(x-1) = (x^4+x^3+2x^2+2x+2)-7/(x-1)`

Explanation:

You can long divide the coefficients, not forgetting to include `0`'s for any missing powers of `x` ...

In our example we find:

`x^5+x^3-9 = (x-1)(x^4+x^3+2x^2+2x+2)-7`

or if you prefer:

`(x^5+x^3-9)/(x-1) = (x^4+x^3+2x^2+2x+2)-7/(x-1)`

That is: `(x^5+x^3-9)` divided by `(x-1)` is `(x^4+x^3+2x^2+2x+2)` with remainder `-7`.

If all we wanted to know was the remainder, then we could have substituted `x=1` into `x^5+x^3-9` to get `1+1-9 = -7`

Alternative Method

Alternatively, start writing out the answer linearly, evaluating the partial sums as you go along and choosing the next term accordingly:

`x^5+x^3-9`

`(x-1)(x^4 color(white)(xxxxxxxxxxxxxxx)` gives `x^5-x^4`

`(x-1)(x^4+x^3 color(white)(xxxxxxxxxxx)` gives `x^5-x^3`

`(x-1)(x^4+x^3+2x^2 color(white)(xxxxxxx)` gives `x^5+x^3-2x^2`

`(x-1)(x^4+x^3+2x^2+2x color(white)(xxxx)` gives `x^5+x^3-2x`

`(x-1)(x^4+x^3+2x^2+2x+2) color(white)(x)` gives `x^5+x^3-2`

`(x-1)(x^4+x^3+2x^2+2x+2) -7 color(white)(x)` gives `x^5+x^3-9`