Lösung 3.4:1b

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In our example, we write first
In our example, we write first
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{{Displayed math||<math>\frac{x^2}{x+1}=\frac{x^2+x-x}{x+1}\,\textrm{.}</math>}}
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<math>\frac{x^{2}}{x+1}=\frac{x^{2}+x-x}{x+1}</math>
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Note that we have added <math>x</math> and then taken away the same <math>x</math> in the numerator. The reason why we do it in this way is that we can now eliminate terms in <math>x^2</math>,
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{{Displayed math||<math>\begin{align}
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\frac{x^2+x-x}{x+1}
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&= \frac{x^2+x}{x+1}-\frac{x}{x+1}
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= \frac{x(x+1)}{x+1}-\frac{x}{x+1}
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= x-\frac{x}{x+1}\,\textrm{.}
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\end{align}</math>}}
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Note that we have added
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Thus, we added <math>x</math> which is precisely what is necessary in order that the numerator <math>x^2</math> should contain a sub expression which has <math>x+1</math> as a factor.
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<math>x</math>
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and then taken away the same
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<math>x</math>
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in the numerator. The reason why we do it in this way is that we can now eliminate terms in
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<math>x^{2}</math>:
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We can treat the second term <math>x/(x+1)</math> in a similar way. Add and take away <math>1</math> in order to get <math>x+1</math>, which can then be divided away,
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& \frac{x^{2}+x-x}{x+1}=\frac{x^{2}+x}{x+1}-\frac{x}{x+1}=\frac{x\left( x+1 \right)}{x+1}-\frac{x}{x+1} \\
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x-\frac{x}{x+1} = x-\frac{x+1-1}{x+1} = x-\frac{x+1}{x+1}+\frac{1}{x+1}
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& =x-\frac{x}{x+1} \\
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= x-1+\frac{1}{x+1}\,\textrm{.}
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\end{align}</math>
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\end{align}</math>}}
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In the remaining polynomial fraction, the numerator is of lower degree than the denominator and we cannot go any further with the division.
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Thus, we added <math>x</math>
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which is precisely what is necessary in order that the numerator
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<math>x^{2}+x</math>
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should contain a sub expression which has
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<math>x+1</math>
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as a factor.
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We can treat the second term
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<math>{x}/{\left( x+1 \right)}\;</math> in a similar way. Add and take away
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<math>\text{1}</math>
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in order to get
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<math>x+\text{1}</math>, which can then be divided away:
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<math>\begin{align}
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& x-\frac{x}{x+1}=x-\frac{x+1-1}{x+1}=x-\frac{x+1}{x+1}+\frac{1}{x+1} \\
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& =x-1+\frac{1}{x+1} \\
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\end{align}</math>
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In the remaining polynomial fraction, the numerator is of higher degree than the denominator and we cannot go any further with the division.
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Version vom 11:51, 31. Okt. 2008

The trick is to complement the numerator with terms that are missing so that the division of the leading term (the term of highest degree) becomes possible, and then successively work down in degree until the numerator has lower degree than the denominator.

In our example, we write first

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

Note that we have added \displaystyle x and then taken away the same \displaystyle x in the numerator. The reason why we do it in this way is that we can now eliminate terms in \displaystyle x^2,

\displaystyle \begin{align}

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

Thus, we added \displaystyle x which is precisely what is necessary in order that the numerator \displaystyle x^2 should contain a sub expression which has \displaystyle x+1 as a factor.

We can treat the second term \displaystyle x/(x+1) in a similar way. Add and take away \displaystyle 1 in order to get \displaystyle x+1, which can then be divided away,

\displaystyle \begin{align}

x-\frac{x}{x+1} = x-\frac{x+1-1}{x+1} = x-\frac{x+1}{x+1}+\frac{1}{x+1} = x-1+\frac{1}{x+1}\,\textrm{.} \end{align}

In the remaining polynomial fraction, the numerator is of lower degree than the denominator and we cannot go any further with the division.

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