Lösung 3.4:1c

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If we focus on the leading term
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<center> [[Image:3_4_1c.gif]] </center>
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<math>x^{3}</math>, we need to complement it with in order to get an expression that is divisible by the denominator ,
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<math>\frac{x^{3}+a^{3}}{x+a}=\frac{x^{3}+ax^{2}-ax^{2}+a^{3}}{x+a}</math>
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With this form on the right-hand side, we can separate away the first two terms in the numerator and have left a polynomial quotient with
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<math>-ax^{2}+a^{3}</math>
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in the numerator:
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<math>\begin{align}
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& \frac{x^{3}+ax^{2}-ax^{2}+a^{3}}{x+a}=\frac{x^{3}+ax^{2}}{x+a}+\frac{-ax^{2}+a^{3}}{x+a} \\
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& =\frac{x^{2}\left( x+a \right)}{x+a}+\frac{-ax^{2}+a^{3}}{x+a} \\
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& =x^{2}+\frac{-ax^{2}+a^{3}}{x+a} \\
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\end{align}</math>
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When we treat the new quotient, we add and take away
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<math>-a^{2}x</math>
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to/from
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<math>-ax^{2}</math>
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in order to get something divisible by
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<math>x+a</math>
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<math>\begin{align}
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& x^{2}+\frac{-ax^{2}+a^{3}}{x+a}=x^{2}+\frac{-ax^{2}-a^{2}x+a^{2}x+a^{3}}{x+a} \\
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& =x^{2}+\frac{-ax^{2}-a^{2}x}{x+a}+\frac{a^{2}x+a^{3}}{x+a} \\
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& =x^{2}+\frac{-ax\left( x+a \right)}{x+a}+\frac{a^{2}x+a^{3}}{x+a} \\
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& =x^{2}-ax+\frac{a^{2}x+a^{3}}{x+a} \\
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\end{align}</math>
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In the last quotient, the numerator has
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<math>x+a</math>
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as a factor, and we obtain a perfect division:
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<math>x^{2}-ax+\frac{a^{2}x+a^{3}}{x+a}=x^{2}-ax+\frac{a^{2}\left( x+a \right)}{x+a}=x^{2}-ax+a^{2}</math>
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If we have calculated correctly, we should have
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<math>\frac{x^{3}+a^{3}}{x+a}=x^{2}-ax+a^{2}</math>
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and one way to check the answer is to multiply both sides by
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<math>x+a</math>,
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<math>x^{3}+a^{3}=\left( x^{2}-ax+a^{2} \right)\left( x+a \right)</math>
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Then, expand the right-hand side and then we should get what is on the left-hand side:
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<math>\begin{align}
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& \text{RHS}=\left( x^{2}-ax+a^{2} \right)\left( x+a \right)=x^{3}+ax^{2}-ax^{2}-a^{2}x+a^{2}x+a^{3} \\
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& =x^{3}+a^{3}=~~\text{LHS} \\
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\end{align}</math>

Version vom 13:04, 26. Okt. 2008

If we focus on the leading term \displaystyle x^{3}, we need to complement it with in order to get an expression that is divisible by the denominator ,


\displaystyle \frac{x^{3}+a^{3}}{x+a}=\frac{x^{3}+ax^{2}-ax^{2}+a^{3}}{x+a}


With this form on the right-hand side, we can separate away the first two terms in the numerator and have left a polynomial quotient with \displaystyle -ax^{2}+a^{3} in the numerator:


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


When we treat the new quotient, we add and take away \displaystyle -a^{2}x to/from \displaystyle -ax^{2} in order to get something divisible by \displaystyle x+a


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


In the last quotient, the numerator has \displaystyle x+a as a factor, and we obtain a perfect division:


\displaystyle x^{2}-ax+\frac{a^{2}x+a^{3}}{x+a}=x^{2}-ax+\frac{a^{2}\left( x+a \right)}{x+a}=x^{2}-ax+a^{2}


If we have calculated correctly, we should have


\displaystyle \frac{x^{3}+a^{3}}{x+a}=x^{2}-ax+a^{2}


and one way to check the answer is to multiply both sides by \displaystyle x+a,


\displaystyle x^{3}+a^{3}=\left( x^{2}-ax+a^{2} \right)\left( x+a \right)


Then, expand the right-hand side and then we should get what is on the left-hand side:


\displaystyle \begin{align} & \text{RHS}=\left( x^{2}-ax+a^{2} \right)\left( x+a \right)=x^{3}+ax^{2}-ax^{2}-a^{2}x+a^{2}x+a^{3} \\ & =x^{3}+a^{3}=~~\text{LHS} \\ \end{align}

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