How do you divide $\frac{x^{3} - 12 x^{2} + 3 x - 4}{x - 2}$ $( x^3-12x^2+3x-4)/(x-2)$ ?

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Answer 1 · 2016-01-27T18:56:35

$x^{2} - 10 x - 17 - \frac{38}{x - 2}$ $x^2-10x-17-38/(x-2)$

$x^{3} - 12 x^{2} + 3 x - 4 ∣ (x - 2)$ $x^3-12x^2+3x-4 | (x-2)$

$- x^{3} + 2 x^{2} ... ... ... ... . ∣ x^{2} - 10 x - 17$ $-x^3 +2x^2 color(white)(.............)|x^2-10x-17$

$color(white)(....)-10x^2+3x-4$

$color(white)(....)+10x^2-20x$

$color(white)(................)-17x-4$

$color(white)(................)+17x-34$

$color(white)(...........................)-38$

So it is:

$x^2-10x-17-38/(x-2)$

How do you divide ( x^3-12x^2+3x-4)/(x-2)x3−12x2+3x−4x−2( x^3-12x^2+3x-4)/(x-2)?