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

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Answer 1 · 2017-07-03T13:56:49

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

Let's perform the long division

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

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

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

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

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

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

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

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

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

Therefore,

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

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

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