Lösung 3.4:1e

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Imagine for a moment taking away all the terms in the numerator apart from \displaystyle x^3. If we are to make \displaystyle x^3 divisible by the denominator \displaystyle x^2+3x+1, we need to add and subtract \displaystyle 3x^2+x in order to obtain the expression \displaystyle x^3+3x^2+x=x(x^2+3x+1),

\displaystyle \begin{align}

\frac{x^3+2x^2+1}{x^2+3x+1} &= \frac{x^3\bbox[#FFEEAA;,1.5pt]{{}+3x^2+x-3x^2-x}+2x^2+1}{x^2+3x+1}\\[5pt] &= \frac{x^3+3x^2+x}{x^2+3x+1} + \frac{-3x^2-x+2x^2+1}{x^2+3x+1}\\[5pt] &= \frac{x(x^2+3x+1)}{x^2+3x+1} + \frac{-x^2-x+1}{x^2+3x+1}\\[5pt] &= x+\frac{-x^2-x+1}{x^2+3x+1}\,\textrm{.} \end{align}

Now, we carry out the same procedure with the new quotient. To the term \displaystyle -x^2, we add and subtract \displaystyle -3x-1 and obtain

\displaystyle \begin{align}

x + \frac{-x^2-x+1}{x^2+3x+1} &= x + \frac{-x^2\bbox[#FFEEAA;,1.5pt]{{}-3x-1+3x+1}-x+1}{x^2+3x+1}\\[5pt] &= x + \frac{-x^2-3x-1}{x^2+3x+1} + \frac{3x+1-x+1}{x^2+3x+1}\\[5pt] &= x - 1 + \frac{2x+2}{x^2+3x+1}\,\textrm{.} \end{align}