Lösung 3.4:1d

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We start by adding and taking away
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We start by adding and taking away <math>x^2</math> in the numerator, so that, in combination with <math>x^3</math>, we obtain the expression <math>x^3+x^2 = x^2(x+1)</math> which can be simplified with the denominator <math>x+1</math>,
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<math>x^{2}</math>
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in the numerator, so that, in combination with
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<math>x^{3}</math>, we obtain the expression
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<math>x^{3}+x^{2}=x^{2}\left( x+1 \right)</math>
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which can be simplified with the denominator
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<math>x+1</math>,
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{{Displayed math||<math>\begin{align}
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\frac{x^3+x+2}{x+1}
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&= \frac{x^3+x^2-x^2+x+2}{x+1}\\[5pt]
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&= \frac{x^3+x^2}{x+1} + \frac{-x^2+x+2}{x+1}\\[5pt]
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&= \frac{x^2(x+1)}{x+1} + \frac{-x^2+x+2}{x+1}\\[5pt]
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&= x^2 + \frac{-x^2+x+2}{x+1}\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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The term <math>-x^2</math> in the remaining quotient needs to complemented with
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& \frac{x^{3}+x+2}{x+1}=\frac{x^{3}+x^{2}-x^{2}+x+2}{x+1} \\
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<math>-x</math> so that we get <math>-x^2-x = -x(x+1)</math>, which is divisible by
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& =\frac{x^{3}+x^{2}}{x+1}+\frac{-x^{2}+x+2}{x+1}=\frac{x^{2}\left( x+1 \right)}{x+1}+\frac{-x^{2}+x+2}{x+1} \\
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& =x^{2}+\frac{-x^{2}+x+2}{x+1} \\
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\end{align}</math>
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The term
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<math>-x^{2}</math>
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in the remaining quotient needs to complemented with
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<math>-x</math>
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so that we get
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<math>-x^{2}-x=-x\left( x+1 \right)</math>, which is divisible by
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<math>x+1</math>,
<math>x+1</math>,
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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x^2 + \frac{-x^2+x+2}{x+1}
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& x^{2}+\frac{-x^{2}+x+2}{x+1}=x^{2}+\frac{-x^{2}-x+x+x+2}{x+1} \\
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&= x^2 + \frac{-x^2-x+x+x+2}{x+1}\\[5pt]
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& =x^{2}+\frac{-x^{2}-x}{x+1}+\frac{2x+2}{x+1} \\
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&= x^2 + \frac{-x^2-x}{x+1} + \frac{2x+2}{x+1}\\[5pt]
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& =x^{2}+\frac{-x\left( x+1 \right)}{x+1}+\frac{2x+2}{x+1} \\
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&= x^2 + \frac{-x(x+1)}{x+1} + \frac{2x+2}{x+1}\\[5pt]
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& =x^{2}-x+\frac{2x+2}{x+1} \\
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&= x^2 - x + \frac{2x+2}{x+1}\,\textrm{.}
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\end{align}</math>
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\end{align}</math>}}
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The last quotient divides perfectly and we obtain
The last quotient divides perfectly and we obtain
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{{Displayed math||<math>x^2-x+\frac{2x+2}{x+1}=x^2-x+2\,\textrm{.}</math>}}
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<math>x^{2}-x+\frac{2x+2}{x+1}=x^{2}-x+2.</math>
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A quick check of whether
A quick check of whether
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{{Displayed math||<math>\frac{x^3+x+2}{x+1} = x^2-x+2\,\textrm{.}</math>}}
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<math>\frac{x^{3}+x+2}{x+1}=x^{2}-x+2.</math>
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is the correct answer is to investigate whether
is the correct answer is to investigate whether
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{{Displayed math||<math>x^3+x+2 = (x^2-x+2)(x+1)</math>}}
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<math>x^{3}+x+2=\left( x^{2}-x+2 \right)\left( x+1 \right)</math>
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holds. If we expand the right-hand side, we see that the relation really does hold
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holds. If we expand the right-hand side, we see that the relation really does hold:
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& \left( x^{2}-x+2 \right)\left( x+1 \right)=x^{3}+x^{2}-x^{2}-x+2x+2 \\
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(x^2-x+2)(x+1) = x^3+x^2-x^2-x+2x+2 = x^3+x+2\,\textrm{.}
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& =x^{3}+x+2 \\
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\end{align}</math>}}
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\end{align}</math>
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Version vom 12:29, 31. Okt. 2008

We start by adding and taking away \displaystyle x^2 in the numerator, so that, in combination with \displaystyle x^3, we obtain the expression \displaystyle x^3+x^2 = x^2(x+1) which can be simplified with the denominator \displaystyle x+1,

\displaystyle \begin{align}

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

The term \displaystyle -x^2 in the remaining quotient needs to complemented with \displaystyle -x so that we get \displaystyle -x^2-x = -x(x+1), which is divisible by \displaystyle x+1,

\displaystyle \begin{align}

x^2 + \frac{-x^2+x+2}{x+1} &= x^2 + \frac{-x^2-x+x+x+2}{x+1}\\[5pt] &= x^2 + \frac{-x^2-x}{x+1} + \frac{2x+2}{x+1}\\[5pt] &= x^2 + \frac{-x(x+1)}{x+1} + \frac{2x+2}{x+1}\\[5pt] &= x^2 - x + \frac{2x+2}{x+1}\,\textrm{.} \end{align}

The last quotient divides perfectly and we obtain

\displaystyle x^2-x+\frac{2x+2}{x+1}=x^2-x+2\,\textrm{.}

A quick check of whether

\displaystyle \frac{x^3+x+2}{x+1} = x^2-x+2\,\textrm{.}

is the correct answer is to investigate whether

\displaystyle x^3+x+2 = (x^2-x+2)(x+1)

holds. If we expand the right-hand side, we see that the relation really does hold

\displaystyle \begin{align}

(x^2-x+2)(x+1) = x^3+x^2-x^2-x+2x+2 = x^3+x+2\,\textrm{.} \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.4:1d“