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p(x)/x+1=(2x^2-3x+1) + -2/x+1, find p(x)
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multiply both sides by (x+1)
See what that gets you
\[\frac{ P(x) }{ x+1 }= (2x^2-3x+1)+ (\frac{ -2 }{ x+1 })\] Find the common denominator of the second side. \[\frac{ P(x) }{ x+1 }=\frac{ (x+1)(2x^2 -3x+1)-2}{ x+1}\] and now x+1 can just cancel out and you are left with p(x)= (x+1)(2x^2-3x+1)-2
thanks
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