How do you divide $\frac{x^{4} - 2 x^{3} - x + 2}{x^{3} - 1}$ $(x^4 - 2x^3 - x + 2 )/( x^3 - 1)$ ?

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Answer 1 · 2016-08-28T17:12:01

$The Quotient is (x - 2), &, Remainder = 0$ $"The Quotient is" (x-2), "&, Remainder"=0$ .

We see that the $Nr. = x^{4} - 2 x^{3} - x + 2$ $"Nr."=x^4-2x^3-x+2$

$=ul(x^4-x)-ul(2x^3+2)$

$=x(x^3-1)-2(x^3-1)$

$=(x^3-1)(x-2)$

$"Therefore"=(x^4-2x^3-x+2)/(x^3-1)$

$={cancel((x^3-1))(x-2)}/(cancel(x^3-1))$

$=x-2$ .

How do you divide (x^4 - 2x^3 - x + 2 )/( x^3 - 1)x4−2x3−x+2x3−1(x^4 - 2x^3 - x + 2 )/( x^3 - 1)?