Lösung 3.4:1e

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Imagine for a moment taking away all the terms in the numerator apart from
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<center> [[Image:3_4_1e.gif]] </center>
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<math>x^{3}</math>. If we are to make
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<math>x^{3}</math>
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divisible by the denominator
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<math>x^{2}+3x+1</math>, we need to add and subtract
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<math>3x^{2}+x</math>
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in order to obtain the expression
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<math>x^{3}+3x^{2}+x=x\left( x^{2}+3x+1 \right)</math>,
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<math>\begin{align}
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& \frac{x^{3}+2x^{2}+1}{x^{2}+3x+1}=\frac{x^{3}+\underline{3x^{2}+x-3x^{2}-x}+2x^{2}+1}{x^{2}+3x+1} \\
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& =\frac{x^{3}+3x^{2}+x}{x^{2}+3x+1}+\frac{-3x^{2}-x+2x^{2}+1}{x^{2}+3x+1} \\
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& =\frac{x\left( x^{2}+3x+1 \right)}{x^{2}+3x+1}+\frac{-x^{2}-x+1}{x^{2}+3x+1} \\
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& =x+\frac{-x^{2}-x+1}{x^{2}+3x+1} \\
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\end{align}</math>
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Now, we carry out the same procedure with the new quotient. To the term
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<math>-x^{2}</math>, we add and subtract
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<math>-\text{3}x-\text{1}</math>
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and obtain
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<math>\begin{align}
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& x+\frac{-x^{2}-x+1}{x^{2}+3x+1}=x+\frac{-x^{2}\underline{-3x-1+3x+1}-x+1}{x^{2}+3x+1} \\
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& =x+\frac{-x^{2}-3x-1}{x^{2}+3x+1}+\frac{3x+1-x+1}{x^{2}+3x+1} \\
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& =x-1+\frac{2x+2}{x^{2}+3x+1} \\
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\end{align}</math>

Version vom 13:42, 26. Okt. 2008

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\left( x^{2}+3x+1 \right),


\displaystyle \begin{align} & \frac{x^{3}+2x^{2}+1}{x^{2}+3x+1}=\frac{x^{3}+\underline{3x^{2}+x-3x^{2}-x}+2x^{2}+1}{x^{2}+3x+1} \\ & =\frac{x^{3}+3x^{2}+x}{x^{2}+3x+1}+\frac{-3x^{2}-x+2x^{2}+1}{x^{2}+3x+1} \\ & =\frac{x\left( x^{2}+3x+1 \right)}{x^{2}+3x+1}+\frac{-x^{2}-x+1}{x^{2}+3x+1} \\ & =x+\frac{-x^{2}-x+1}{x^{2}+3x+1} \\ \end{align}


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


\displaystyle \begin{align} & x+\frac{-x^{2}-x+1}{x^{2}+3x+1}=x+\frac{-x^{2}\underline{-3x-1+3x+1}-x+1}{x^{2}+3x+1} \\ & =x+\frac{-x^{2}-3x-1}{x^{2}+3x+1}+\frac{3x+1-x+1}{x^{2}+3x+1} \\ & =x-1+\frac{2x+2}{x^{2}+3x+1} \\ \end{align}

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