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

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

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Answer 1 · 2017-07-03T13:56:49

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

Explanation:

Let's perform the long division

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

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

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

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

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

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

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

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

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

Therefore,

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

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

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

1 Answer

Explanation:

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