How do you divide #(6x^4+12x+4)/(3x^2+2x+5)#?

1 Answer
Tony B
Nov 2, 2017

Taken you to a point where you can continue from. Method is demonstrated.

Explanation:

I am using no value place keepers. Example #0x^3#

#color(white)("ddddddddd.dddddddd.dd")6x^4+0x^3+0x^2+12x+4#
#color(magenta)(+2x^2)(3x^2+2x+5) ->color(white)("d") ul(6x^4+4x^3+10x^2 larr" Subtract")#
# color(white)("dddddddddddddddddddddd")0-4x^3 -10x^2+12x+4#
#color(magenta)(-4/3x)(3x^2+2x+5) ->color(white)("dd.dd")ul(-4x^3-8/3x^2-20/3 x larr" Sub." )#
#color(white)("dddddddddddddddddddddddddd") 0 -22/3x^2+56/3x+4#

This should be enough to demonstrate the method. I will let you take over from this point.

SO FAR WE HAVE AS AN ANSWER

#color(white)("d")color(magenta)(2x^2-4/3x + ? + ("remainder")/(3x^2+2x+5)#

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