How do you divide $(x^{4} - 3 x^{3} + 5 x - 6) \div (x + 2)$ $(x^4-3x^3+5x-6)div (x+2)$ ?

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Answer 1 · 2016-05-14T16:37:28

$Quotient = (x^{3} - 5 x^{2} + 10 x - 15)$ $"Quotient"=(x^3-5x^2+10x-15)$
$Remainder = 24$ $"Remainder"=24$

$(x^{4} - 3 x^{3} + 5 x - 6) \div (x + 2)$ $(x^4-3x^3+5x-6)div (x+2)$

$= \frac{x^{4} - 3 x^{3} + 5 x - 6}{x + 2}$ $=(x^4-3x^3+5x-6)/ (x+2)$

$= \frac{x^{3} (x + 2) - 2 x^{3} - 3 x^{3} + 5 x - 6}{x + 2}$ $=(x^3(x+2)-2x^3-3x^3+5x-6)/ (x+2)$

$= \frac{x^{3} (x + 2) - 5 x^{3} + 5 x - 6}{x + 2}$ $=(x^3(x+2)-5x^3+5x-6)/ (x+2)$

$= \frac{x^{3} (x + 2) - 5 x^{2} (x + 2) + 10 x^{2} + 5 x - 6}{x + 2}$ $=(x^3(x+2)-5x^2(x+2)+10x^2+5x-6)/ (x+2)$

$= \frac{x^{3} (x + 2) - 5 x^{2} (x + 2) + 10 x (x + 2) - 20 x + 5 x - 6}{x + 2}$ $=(x^3(x+2)-5x^2(x+2)+10x(x+2)-20x+5x-6)/ (x+2)$

$= \frac{x^{3} (x + 2) - 5 x^{2} (x + 2) + 10 x (x + 2) - 15 x - 6}{x + 2}$ $=(x^3(x+2)-5x^2(x+2)+10x(x+2)-15x-6)/ (x+2)$

$= \frac{x^{3} (x + 2) - 5 x^{2} (x + 2) + 10 x (x + 2) - 15 (x + 2) + 30 - 6}{x + 2}$ $=(x^3(x+2)-5x^2(x+2)+10x(x+2)-15(x+2)+30-6)/ (x+2)$

$= \frac{x^{3} (x + 2) - 5 x^{2} (x + 2) + 10 x (x + 2) - 15 (x + 2) + 24}{x + 2}$ $=(x^3(x+2)-5x^2(x+2)+10x(x+2)-15(x+2)+24)/ (x+2)$

$=(cancel((x+2))(x^3-5x^2+10x-15))/cancel((x+2))+24/ (x+2)$

$=(x^3-5x^2+10x-15)+24/ (x+2)$

Result
$"Quotient"=(x^3-5x^2+10x-15)$
$"Remainder"=24$

How do you divide (x^4-3x^3+5x-6)div (x+2)(x4−3x3+5x−6)÷(x+2)(x^4-3x^3+5x-6)div (x+2)?