Lösung 3.4:1c

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If we focus on the leading term
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If we focus on the leading term <math>x^3</math>, we need to complement it with <math>ax^2</math> in order to get a sub expression that is divisible by the denominator,
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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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{{Displayed math||<math>\frac{x^3+a^3}{x+a} = \frac{x^3+ax^2-ax^2+a^3}{x+a}\,\textrm{.}</math>}}
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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 <math>-ax^2+a^3</math> in the numerator,
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{{Displayed math||<math>\begin{align}
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\frac{x^3+ax^2-ax^2+a^3}{x+a}
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&= \frac{x^3+ax^2}{x+a} + \frac{-ax^2+a^3}{x+a}\\[5pt]
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&= \frac{x^2(x+a)}{x+a} + \frac{-ax^2+a^3}{x+a}\\[5pt]
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&= x^2 + \frac{-ax^2+a^3}{x+a}\,\textrm{.}
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\end{align}</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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When we treat the new quotient, we add and take away <math>-a^2x</math> to/from <math>-ax^2</math> in order to get something divisible by <math>x+a</math>,
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<math>-ax^{2}+a^{3}</math>
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in the numerator:
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{{Displayed math||<math>\begin{align}
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x^2+\frac{-ax^2+a^3}{x+a}
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&= x^2 + \frac{-ax^2-a^2x+a^2x+a^3}{x+a}\\[5pt]
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&= x^2 + \frac{-ax^2-a^2x}{x+a} + \frac{a^2x+a^3}{x+a}\\[5pt]
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&= x^2 + \frac{-ax(x+a)}{x+a} + \frac{a^2x+a^3}{x+a}\\[5pt]
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&= x^2 - ax + \frac{a^2x+a^3}{x+a}\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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In the last quotient, the numerator has <math>x+a</math> as a factor, and we obtain a perfect division,
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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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{{Displayed math||<math>x^2 - ax + \frac{a^2x+a^3}{x+a} = x^2 - ax + \frac{a^2(x+a)}{x+a} = x^2-ax+a^2\,\textrm{.}</math>}}
If we have calculated correctly, we should have
If we have calculated correctly, we should have
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{{Displayed math||<math>\frac{x^3+a^3}{x+a} = x^2-ax+a^2</math>}}
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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 <math>x+a</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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{{Displayed math||<math>x^3+a^3 = (x^2-ax+a^2)(x+a)\,\textrm{.}</math>}}
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Then, expand the right-hand side and we should get what is on the left-hand side,
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<math>\begin{align}
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{{Displayed math||<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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\text{RHS}
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& =x^{3}+a^{3}=~~\text{LHS} \\
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&= (x^2-ax+a^2)(x+a)\\[5pt]
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\end{align}</math>
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&= x^3+ax^2-ax^2-a^2x+a^2x+a^3\\[5pt]
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&= x^3+a^3\\[5pt]
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&= \text{LHS.}
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\end{align}</math>}}

Version vom 12:11, 31. Okt. 2008

If we focus on the leading term \displaystyle x^3, we need to complement it with \displaystyle ax^2 in order to get a sub 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}\,\textrm{.}

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

When we treat the new quotient, we add and take away \displaystyle -a^2x 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^2x+a^2x+a^3}{x+a}\\[5pt] &= x^2 + \frac{-ax^2-a^2x}{x+a} + \frac{a^2x+a^3}{x+a}\\[5pt] &= x^2 + \frac{-ax(x+a)}{x+a} + \frac{a^2x+a^3}{x+a}\\[5pt] &= x^2 - ax + \frac{a^2x+a^3}{x+a}\,\textrm{.} \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^2x+a^3}{x+a} = x^2 - ax + \frac{a^2(x+a)}{x+a} = x^2-ax+a^2\,\textrm{.}

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 = (x^2-ax+a^2)(x+a)\,\textrm{.}

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

\displaystyle \begin{align}

\text{RHS} &= (x^2-ax+a^2)(x+a)\\[5pt] &= x^3+ax^2-ax^2-a^2x+a^2x+a^3\\[5pt] &= x^3+a^3\\[5pt] &= \text{LHS.} \end{align}

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