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)))#