Lösung 3.4:1e

Aus Online Mathematik Brückenkurs 2

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Imagine for a moment taking away all the terms in the numerator apart from
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Imagine for a moment taking away all the terms in the numerator apart from <math>x^3</math>. If we are to make <math>x^3</math> divisible by the denominator
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<math>x^{3}</math>. If we are to make
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<math>x^2+3x+1</math>, we need to add and subtract <math>3x^2+x</math> in order to obtain the expression <math>x^3+3x^2+x=x(x^2+3x+1)</math>,
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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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{{Displayed math||<math>\begin{align}
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\frac{x^3+2x^2+1}{x^2+3x+1}
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&= \frac{x^3\bbox[#FFEEAA;,1.5pt]{{}+3x^2+x-3x^2-x}+2x^2+1}{x^2+3x+1}\\[5pt]
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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}\\[5pt]
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&= \frac{x(x^2+3x+1)}{x^2+3x+1} + \frac{-x^2-x+1}{x^2+3x+1}\\[5pt]
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&= x+\frac{-x^2-x+1}{x^2+3x+1}\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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Now, we carry out the same procedure with the new quotient. To the term <math>-x^2</math>, we add and subtract <math>-3x-1</math> and obtain
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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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{{Displayed math||<math>\begin{align}
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Now, we carry out the same procedure with the new quotient. To the term
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x + \frac{-x^2-x+1}{x^2+3x+1}
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<math>-x^{2}</math>, we add and subtract
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&= x + \frac{-x^2\bbox[#FFEEAA;,1.5pt]{{}-3x-1+3x+1}-x+1}{x^2+3x+1}\\[5pt]
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<math>-\text{3}x-\text{1}</math>
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&= x + \frac{-x^2-3x-1}{x^2+3x+1} + \frac{3x+1-x+1}{x^2+3x+1}\\[5pt]
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and obtain
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&= x - 1 + \frac{2x+2}{x^2+3x+1}\,\textrm{.}
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\end{align}</math>}}
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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>
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Version vom 12:40, 31. 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(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}

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