How do you find the quotient of #(t^4+6t^3-3t^2-t+8)/(t+1)# using long division?

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Answer 1 · 2016-12-31T07:14:47

The quotient is #=t^3+5t^2-8t+7# and the remainder is #=1#

Let's do the lomg division

#color(white)(aaaa)##t^4+6t^3-3t^2-t+8##color(white)(aaaa)##∣##t+1#

#color(white)(aaaa)##t^4+t^3##color(white)(aaaaaaaaaaaaaaaa)##∣##t^3+5t^2-8t+7#

#color(white)(aaaa)##0+5t^3-3t^2#

#color(white)(aaaaaa)##+5t^3+5t^2#

#color(white)(aaaaaaa)##+0-8t^2-t#

#color(white)(aaaaaaaaaaa)##-8t^2-8t#

#color(white)(aaaaaaaaaaaa)##-0+7t+8#

#color(white)(aaaaaaaaaaaaaaaa)##+7t+7#

#color(white)(aaaaaaaaaaaaaaaaa)##+0+1#

The quotient is #=t^3+5t^2-8t+7# and the remainder is #=1#