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

Question

1786 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-09T08:04:52

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

Perform a long division to obtain the quotient

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

#color(white)(aaaa)##d^4-2d^3##color(white)(aaaaaaaaaaaaaaaaa)##|##d^3-4d^2-8d-15#

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

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

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

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

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

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

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

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

The remainder is #=-13#

#(d^4-6d^3+0d^2+d+17)/(d-2)=d^3-4d^2-8d-15-13/(d-2)#