How do you divide #(6x^4-44x^3+48x^2-2)-:(6x-8)# using long division?

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Answer 1 · 2017-02-04T16:44:52

The answer is #=x^3-6x^2-1/(3x-4)#

Let's perform the long division

#color(white)(aaaa)##6x^4-44x^3+48x^2##color(white)(aaaa)##-2##color(white)(aaaa)##|##6x-8#

#color(white)(aaaa)##6x^4-8x^3##color(white)(aaaaaaaaaaaaaa)####color(white)(aaaa)##|##x^3-6x^2#

#color(white)(aaaaaa)##0-36x^3+48x^2##color(white)(aaaaaaaaaaaaaa)####color(white)(aaaa)#

#color(white)(aaaaaaaa)##-36x^3+48x^2##color(white)(aaaaaaaaaaaaaa)####color(white)(aaaa)#

#color(white)(aaaaaaaaaaa)##-0+0##color(white)(aaaaaaa)##-2##color(white)(aaaa)#

The quotient is #=x^3-6x^2#

The remainder is #=-2#

Therefore,

#(6x^4-44x^3+48x^2-2)/(6x-8)=x^3-6x^2-2/(6x-8)#

#=x^3-6x^2-1/(3x-4)#