How do you divide $\frac{2 x^{3} - 4 x^{2} + 16 x - 8}{x - 2}$ $(2x^3-4x^2+16x-8) / (x-2)$ ?

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How do you divide $\frac{2 x^{3} - 4 x^{2} + 16 x - 8}{x - 2}$ $(2x^3-4x^2+16x-8) / (x-2)$ ?

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Answer 1 · 2018-04-08T12:59:47

$2 x^{2} + 16 + \frac{24}{x - 2}$ $2x^2+16+24/(x-2)$

Explanation:

$one way is to use the divisor as a factor in the numerator$ $"one way is to use the divisor as a factor in the numerator"$

$consider the numerator$ $"consider the numerator"$

$color(red)(2x^2)(x-2)cancel(color(magenta)(+4x^2))cancel(-4x^2)+16x-8$

$=color(red)(2x^2)(x-2)color(red)(+16)(x-2)color(magenta)(+32)-8$

$=color(red)(2x^2)(x-2)color(red)(+16)(x-2)+24$

$"quotient "=color(red)(2x^2+16)," remainder "=24$

$rArr(2x^3-4x^2+16x-8)/(x-2)=2x^2+16x+24/(x-2)$

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How do you divide $\frac{2 x^{3} - 4 x^{2} + 16 x - 8}{x - 2}$ $(2x^3-4x^2+16x-8) / (x-2)$ ?

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Explanation:

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How do you divide (2x^3-4x^2+16x-8) / (x-2)2x3−4x2+16x−8x−2(2x^3-4x^2+16x-8) / (x-2)?

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How do you divide $\frac{2 x^{3} - 4 x^{2} + 16 x - 8}{x - 2}$ $(2x^3-4x^2+16x-8) / (x-2)$ ?