Question

How do you evaluate `(3x^3+2x^2-x+3) / (x-3)`?

Answer 1

`(3x^3+2x^2-x+3)/(x-3) = 3x^2+11x+32+99/(x-3)`

Explanation:

The way that I would evaluate it goes as follows (effectively long dividing, but written in a more compact form):

Write down:

`3x^3+2x^2-x+3 = (x-3)(...`

then complete the line one term at a time.

The first term we need will be `3x^2` to get `3x^3` when multiplied by the `x` in `(x-3)`...

`3x^3+2x^2-x+3 = (x-3)(3x^2...`

Notice that when `3x^2` is multiplied by `-3` it will give us `-9x^2`, but we want `+2x^2`, so we need another `11x^2` which means the next term must be `+11x`...

`3x^3+2x^2-x+3 = (x-3)(3x^2+11x...`

Then `-3 * 11x = -33x`, but we want just `-x`, so the next term we want is `+32`...

`3x^3+2x^2-x+3 = (x-3)(3x^2+11x+32)...`

Finally note that `-3 * 32 = -96`, but we want `+3`, so we need to add `99`...

`3x^3+2x^2-x+3 = (x-3)(3x^2+11x+32)+99`

So:

`(3x^3+2x^2-x+3)/(x-3) = 3x^2+11x+32" "` with remainder `99`

or:

`(3x^3+2x^2-x+3)/(x-3) = 3x^2+11x+32+99/(x-3)`