Question

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

Answer 1

The result is `(-x^2-3x+1)` remainder `5x-10`.

Explanation:

Looking at the orders of the numerator and the denominator, we can see the result should be of order 2.
We can deduce the first term will be `-x^2` in order for the first term of the product to be `-x^4`.
Looking at the constants in the numerator it seems there must be a remainder, since 3 is not a factor of 7.
We write the numerator as a product of two quadratics (the known denominator and the unknown quotient) plus a linear term (the unknown remainder):
`(-x^4-3x^3-2x^2-4x-7)`
`=(x^2+3)(-x^2+Ax+B)+Cx+D`
Comparing coefficients of `x^3`, we see that: `A=-3`
Comparing coefficients of `x^2`, we see that: `B-3=-2 rArr B=1`
Comparing coefficients of `x` ,we see that: `3A+C=-4 rArr C=5`
Comparing constants, we see that: `3B+D=-7 rArr D=-10`
Hence the result is `(-x^2-3x+1)` remainder `5x-10`.

Answer 2

`-x^2-3x+1+(5x-10)/(x^2+3)`

Explanation:

`(-x^4-3x^3-2x^2-4x-7 )/ (color(green)(x^2+3))`

Note that I use place keepers such as `0x^3` for eas of alignment.

`color(white)("dddddddddddddd") -x^4-3x^3-2x^2-4x-7`
`color(magenta)(-x^2) color(green)((x^2+3)) ->color(white)("d") ul(-x^4+0x^3-3x^2larr" Subtract")`
# color(white)("dddddddddddddddd")0-3x^3 +color(white)("d")x^2-4x-7#
`color(magenta)(-3x)color(green)((x^2+3))->color(white)("ddd.d") ul(-3x^3+0x^2-9xlarr" Subtract")`
`color(white)("dddddddddddddddddddd")0color(white)("d")+color(white)("d")x^2+5x-7`
`color(white)(1)color(magenta)(+1)color(green)((x^2+3)) ->color(white)("ddddddddddd.d")ul( x^2+0x+3larr" Subtract")`
`color(white)("dddddddddddddddddddddddddd")0+color(magenta)(5x-10)larr" Stop"`

Putting all this together we have:

`color(magenta)(-x^2-3x+1+(5x-10)/(color(black)(color(green)(x^2+3)))`