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What is the quotient of #d-2# divided by #d^4-6d^3+d+17#?

Answer 1

Hazel Anglin

The quotient is #=(d^3-4d^2-8d-15)#

Now let's divide that long way.

#d-2##color(white)(aaaa)##|##d^4-6d^3+0d^2+d+17##color(white)(aa)##|##d^3-4d^2-8d-15#

#color(white)(aaaaaaaaaa)##d^4-2d^3#

#color(white)(aaaaaaaaaaa)##0-4d^3+0d^2#

#color(white)(aaaaaaaaaaaaa)##-4d^3+8d^2#

#color(white)(aaaaaaaaaaaaaa)##-0-8d^2+d#

#color(white)(aaaaaaaaaaaaaaaaa)##-8d^2+16d#

#color(white)(aaaaaaaaaaaaaaaaaaa)##-0-15d+17#

#color(white)(aaaaaaaaaaaaaaaaaaaaaaa)##-15d+30#

#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaa)##-0-13#

Consequently,

#(d^4-6d^3+d+17)/(d-2)=d^3-4d^2-8d-15-13/(d-2)#

The remainder is #=-13# and the quotient is #=(d^3-4d^2-8d-15)#

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Answer 2

Genesis Allen

The quotient of (d-2) divided by (d^4-6d^3+d+17) is 1/(d^3-4d^2+8d-15).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.