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

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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 $=-13$

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

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