Question

How do you divide `( -3x^3+12x^2-7x-6 )/((x + 1)(4x+12) )`?

Answer 1

`-3x/4+6-(47x+39)/(2(x^2+4x+3))`

Explanation:

Expand `(x+1)(4x+12)` into `4x^2+16x+12`

Now you can long divide `(-3x^3+12x^2-7x-6)/(4x^2+16x+12)`.

First, divide the leading coefficients, that is `-3x^3` and `4x^2` into `-3x/4`. Now multiply `(4x^2+16x+12)` by `-3x/4` to get `-3x^3-12x^2-9x`. Subtract `-3x^3-12x^2-9x` from `-3x^3+12x^2-7x-6` to get the remainder, that is `24x^2+2x-6`.

Therefore,
`(-3x^3+12x^2-7x-6)/(4x^2+16x+12)=-3x/4+(24x^2+2x-6)/(4x^2+16x+12)`.

Now, divide `(24x^2+2x-6)/(4x^2+16x+12)` with the same steps above. First, divide the leading coefficients to get `6`. Multiply `4x^2+16x+12` by `6` to get `24x^2+96x+72`. Subtract `24x^2+96x+72` from `24x^2+2x-6` to get the second remainder, `-94x-78`.

Therefore,

`(24x^2+2x-6)/(4x^2+16x+12)=6+(-94x-78)/(4x^2+16x+12)`.

Adding together gives us

`-3x/4+6+(-94x-78)/(4x^2+16x+12)`

which simplifies into

`-3x/4+6-(47x+39)/(2(x^2+4x+3))`