How do you divide `(5x^4+2x^3-9x+12)/(x^2-3x+4)`?

We perform a long division

`color(white)(aa)``x^2-3x+4``color(white)(a)``|``5x^4+2x^3+0x^2-9x+12``|``5x^2+17x+31`

`color(white)(aaaaaaaaaaaaaa)``5x^4-15x^3+20x^2`

`color(white)(aaaaaaaaaaaaaa)``0+17x^3-20x^2-9x`

`color(white)(aaaaaaaaaaaaaaaa)``+17x^3-51x^2+68x`

`color(white)(aaaaaaaaaaaaaaaaaa)``+0+31x^2-77x+12`

`color(white)(aaaaaaaaaaaaaaaaaaaaaa)``+31x^2-93x+124`

`color(white)(aaaaaaaaaaaaaaaaaaaaaaaa)``+0+16x-112`

The remainder is `=16x-112` and the quotient is `=5x^2+17x+31`

`(5x^4+2x^3+0x^2-9x+12)/(x^2-3x+4)=5x^2+17x+31+(16x-112)/(x^2-3x+4)`

`(5x^4+2x^3-9x+12)/(x^2-3x+4)`

Let's start by getting rid of our `5x^4` term. We can do this by pulling out `5x^2*(x^2-3x+4)`.

`5x^2*(x^2-3x+4) = 5x^4 - 15x^3 + 20x^2`

So we can re-write the polynomial as:

`(5x^2(x^2-3x+4))/(x^2-3x+4)+(5x^4+2x^3-9x+12)/(x^2-3x+4) - (5x^4-15x^3+20x^2)/(x^2-3x+4)`

`= 5x^2 + (17x^3-20x^2-9x+12)/(x^2-3x+4)`

Now, let's divide the remaining part again. This time, to get rid of the `17x^3`, we need to pull out `17x*(x^2-3x+4)`.

`(17x*(x^2-3x+4))/(x^2-3x+4)+ (17x^3-20x^2-9x+12)/(x^2-3x+4)-(17x^3-51x^2+68x)/(x^2-3x+4)`

`=17x + (31x^2-77x+12)/(x^2-3x+4)`

Finally, let's divide the remaining part one last time. This time, to get rid of the `31x^2`, we need to pull out `31*(x^2-3x+4)`.

`(31(x^2-3x+4))/(x^2-3x+4)+(31x^2-77x+12)/(x^2-3x+4) - (31x^2-93x+124)/(x^2-3x+4)`

`= 31 + (16x-112)/(x^2-3x+4)`

We can't divide any more since the numerator is smaller than the denominator. So, our final answer is:

`5x^2+17x+31+(16x-112)/(x^2-3x+4)`