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Lösung 3.4:1d

Aus Online Mathematik Brückenkurs 2

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K (Lösning 3.4:1d moved to Solution 3.4:1d: Robot: moved page)
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We start by adding and taking away
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<center> [[Image:3_4_1d.gif]] </center>
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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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<math>\begin{align}
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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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& =\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>,
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<math>\begin{align}
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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+1}+\frac{2x+2}{x+1} \\
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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}-x+\frac{2x+2}{x+1} \\
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\end{align}</math>
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The last quotient divides perfectly and we obtain
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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
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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
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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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<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^{3}+x+2 \\
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\end{align}</math>

Version vom 13:22, 26. Okt. 2008

We start by adding and taking away x2 in the numerator, so that, in combination with x3, we obtain the expression x3+x2=x2x+1  which can be simplified with the denominator x+1,


x+1x3+x+2=x+1x3+x2−x2+x+2=x+1x3+x2+x+1−x2+x+2=x+1x2x+1+x+1−x2+x+2=x2+x+1−x2+x+2


The term −x2 in the remaining quotient needs to complemented with −x so that we get −x2−x=−xx+1 , which is divisible by x+1,


x2+x+1−x2+x+2=x2+x+1−x2−x+x+x+2=x2+x+1−x2−x+x+12x+2=x2+x+1−xx+1+x+12x+2=x2−x+x+12x+2


The last quotient divides perfectly and we obtain


x2−x+x+12x+2=x2−x+2


A quick check of whether


x+1x3+x+2=x2−x+2


is the correct answer is to investigate whether


x3+x+2=x2−x+2x+1 

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


x2−x+2x+1=x3+x2−x2−x+2x+2=x3+x+2 

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