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I know I posted this question yesterday but I really nead help on it urgently

 

 

Thank You

Guest Dec 2, 2017
 #1
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Here you go:

 

\(\frac{p(x)}{x-2}=(x+1)q(x)+\frac{Ax+B}{x-2}\\ \frac{p(x)}{x-2}=(x+1)q(x)+A+\frac{B+2A}{x-2}\qquad \text{Note: I used algebraic division here}\\ where\;\;B+2A=3\qquad (1)\\~\\ \frac{p(x)}{x+1}=(x-2)q(x)+\frac{Ax+B}{x+1}\\ \frac{p(x)}{x+1}=(x-2)q(x)+A+\frac{B-A}{x+1}\\ where\;\;B-A=9 \qquad (2)\)

 

Solve (1) and (2) simultaneously and you get A=-2 and B=7

 

so

\(p(x)=(x-2)(x+1)q(x)-2x+7\\ \frac{p(x)}{(x-2)(x+1)}=q(x)\;\;Remainder\;-2x+7\\\)

Melody  Dec 3, 2017

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