How can I change $f (x) = \frac{2 x - 1}{x + 2}$ $f(x) = (2x-1)/(x+2)$ into the form $A + \frac{B}{x + 2}$ $A + B/(x+2)$ where A and B are both integers?

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How can I change $f (x) = \frac{2 x - 1}{x + 2}$ $f(x) = (2x-1)/(x+2)$ into the form $A + \frac{B}{x + 2}$ $A + B/(x+2)$ where A and B are both integers?

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Answer 1 · 2017-08-20T10:25:26

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

Explanation:

$using (x + 2) as a factor in the numerator$ $"using "(x+2)" as a factor in the numerator"$

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

$2 (x + 2) - 4 - 1$ $2(x+2)color(magenta)(-4)-1$

$= 2 (x + 2) - 5$ $=2(x+2)-5$

$\Rightarrow \frac{2 (x + 2) - 5}{x + 2} = 2 - \frac{5}{x + 2}$ $rArr(2(x+2)-5)/(x+2)=2-5/(x+2)$

$As a check$ $color(blue)"As a check"$

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

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

$= \frac{2 x + 4 - 5}{x + 2} = \frac{2 x - 1}{x + 2} \leftarrow True$ $=(2x+4-5)/(x+2)=(2x-1)/(x+2)larr" True"$

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How can I change $f (x) = \frac{2 x - 1}{x + 2}$ $f(x) = (2x-1)/(x+2)$ into the form $A + \frac{B}{x + 2}$ $A + B/(x+2)$ where A and B are both integers?

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How can I change $f (x) = \frac{2 x - 1}{x + 2}$ $f(x) = (2x-1)/(x+2)$ into the form $A + \frac{B}{x + 2}$ $A + B/(x+2)$ where A and B are both integers?