How do you divide $\frac{5 x^{2} - 2 x + 4}{x - 1}$ $(5x^2-2x+4)/(x-1)$ ?

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How do you divide $\frac{5 x^{2} - 2 x + 4}{x - 1}$ $(5x^2-2x+4)/(x-1)$ ?

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Answer 1 · 2018-04-06T17:51:56

$5 x + 3 + \frac{7}{x - 1}$ $5x+3+7/(x-1)$

Explanation:

$\frac{5 x^{2} - 2 x + 4}{x - 1}$ $(5x^2-2x+4)/(x-1)$

$color(white)(....,,......)5x+3$
$x-1|overline(5x^2-2x+4)$
$color(white)(............)ul(5x^2-5x)$
$color(white)(.....................)3x+4$
$color(white)(..........,,......)ul(3x-3)$
$color(white)(............................)7$

$(5x^2-2x+4) / (x-1) = 5x+3+7/(x-1)$

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How do you divide $\frac{5 x^{2} - 2 x + 4}{x - 1}$ $(5x^2-2x+4)/(x-1)$ ?

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How do you divide (5x^2-2x+4)/(x-1)5x2−2x+4x−1(5x^2-2x+4)/(x-1)?

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How do you divide $\frac{5 x^{2} - 2 x + 4}{x - 1}$ $(5x^2-2x+4)/(x-1)$ ?