How do you divide $\frac{x^{2} + 3 x + 2}{1 - x}$ $(x^2 + 3x +2)/(1-x)$ ?

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How do you divide $\frac{x^{2} + 3 x + 2}{1 - x}$ $(x^2 + 3x +2)/(1-x)$ ?

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Answer 1 · 2016-02-26T10:57:27

$\frac{x^{2} + 3 x + 2}{1 - x} = - x - 4 + \frac{6}{1 - x}$ $(x^2+3x+2)/(1-x)=-x-4+6/(1-x)$

Explanation:

by long division

Our dividend = $x^{2} + 3 x + 2$ $x^2+3x+2$
our divisor $= - x + 1$ $=-x+1$

$" " " " " " " " " underline(-x-4)$
$-x+1" "|~x^2+3x+2$
$" " " " " " " " "underline(x^2-x)$
$" " " " " " " " " " " "4x+2$
$" " " " " " " " " " " "underline(4x-4)$
$" " " " " " " " " " " " " " "+6" " "larr$ the remainder

so we write the answer

$("dividend")/("divisor")="quotient"+("remainder")/("divisor")$

$(x^2+3x+2)/(1-x)=-x-4+6/(1-x)$

God bless...I hope the explanation is useful ..

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How do you divide $\frac{x^{2} + 3 x + 2}{1 - x}$ $(x^2 + 3x +2)/(1-x)$ ?

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