Question

How do you divide `(x^4-2x^3-2x^2+9x+3)/(x^2-3)`?

Answer 1

`(x^4-2x^3-2x^2+9x+3)/(x^2-3) = (x^2-2x+1) + (3x+6)/(x^2-3)`

Explanation:

I like to long divide the coefficients, not forgetting to include `0`'s for any missing powers of `x`. In our example, that means the missing `x` term in the divisor, which is therefore repesented as `1, 0, -3`.

The process is similar to long division of numbers.

We find a quotient `1, -2, 1`, meaning `x^2-2x+1` and remainder `3, 6`, meaning `3x+6`

So: `(x^4-2x^3-2x^2+9x+3)/(x^2-3) = (x^2-2x+1) + (3x+6)/(x^2-3)`