How do you divide $\frac{{(x - 2)}^{2}}{x^{2} - 5 x + 9}$ $(x-2)^2/(x^2-5x+9)$ ?

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How do you divide $\frac{{(x - 2)}^{2}}{x^{2} - 5 x + 9}$ $(x-2)^2/(x^2-5x+9)$ ?

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Answer 1 · 2016-06-14T16:20:29

$\frac{x^{2} - 4 x + 4}{x^{2} - 5 x + 9} = 1 + \frac{x - 5}{x^{2} - 5 x + 9}$ $(x^2-4x+4)/(x^2-5x+9)" "=" " 1 +(x-5)/(x^2-5x+9)$

Explanation:

Look at https://socratic.org/s/avjGBRiF as a further example of the method.

Write as $\frac{x^{2} - 4 x + 4}{x^{2} - 5 x + 9}$ $(color(blue)(x^2-4x+4))/(color(green)(x^2-5x+9))$

$Numerator \to x^{2} - 4 x + 4$ $"Numerator "->" "color(blue)( x^2-4x+4)$
$color(magenta)(1xx)color(green)((x^2-5x+9)) ->" "ul(x^2-5x+9)" "larr" Subtract"$
$"Remainder" ->" " 0+color(white)(.)x-5$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So $(x^2-4x+4)/(x^2-5x+9)" "=" "color(magenta)( 1) +(x-5)/(x^2-5x+9)$

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How do you divide $\frac{{(x - 2)}^{2}}{x^{2} - 5 x + 9}$ $(x-2)^2/(x^2-5x+9)$ ?

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Explanation:

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How do you divide (x-2)^2/(x^2-5x+9)(x−2)2x2−5x+9(x-2)^2/(x^2-5x+9)?

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How do you divide $\frac{{(x - 2)}^{2}}{x^{2} - 5 x + 9}$ $(x-2)^2/(x^2-5x+9)$ ?