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

2 Answers
Nov 3, 2017

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.

Nov 3, 2017

-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)))