Question

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

Answer 1

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

Explanation:

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)`

Answer 2

Pull out factors of `(x^2-3x+4)` until there aren't any left. Start by removing the biggest degree term, and work your way down until the degree of the numerator is less than the degree of the denominator.

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

Explanation:

`(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)`