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Lösung 3.4:1c

Aus Online Mathematik Brückenkurs 2

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Zeile 1: Zeile 1:
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,
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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{{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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{{Abgesetzte Formel||<math>\frac{x^3+a^3}{x+a} = \frac{x^3+ax^2-ax^2+a^3}{x+a}\,\textrm{.}</math>}}
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,
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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{{Abgesetzte Formel||<math>\begin{align}
\frac{x^3+ax^2-ax^2+a^3}{x+a}
\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^3+ax^2}{x+a} + \frac{-ax^2+a^3}{x+a}\\[5pt]
Zeile 14: Zeile 14:
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>,
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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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
x^2+\frac{-ax^2+a^3}{x+a}
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+a^2x+a^3}{x+a}\\[5pt]
Zeile 24: Zeile 24:
In the last quotient, the numerator has <math>x+a</math> as a factor, and we obtain a perfect division,
In the last quotient, the numerator has <math>x+a</math> as a factor, and we obtain a perfect division,
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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>}}
+
{{Abgesetzte Formel||<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>}}
+
{{Abgesetzte Formel||<math>\frac{x^3+a^3}{x+a} = x^2-ax+a^2</math>}}
and one way to check the answer is to multiply both sides by <math>x+a</math>,
and one way to check the answer is to multiply both sides by <math>x+a</math>,
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{{Displayed math||<math>x^3+a^3 = (x^2-ax+a^2)(x+a)\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>x^3+a^3 = (x^2-ax+a^2)(x+a)\,\textrm{.}</math>}}
Then, expand the right-hand side and we should get what is on the left-hand side,
Then, expand the right-hand side and we should get what is on the left-hand side,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\text{RHS}
\text{RHS}
&= (x^2-ax+a^2)(x+a)\\[5pt]
&= (x^2-ax+a^2)(x+a)\\[5pt]

Version vom 13:14, 10. Mär. 2009

If we focus on the leading term x3, we need to complement it with ax2 in order to get a sub expression that is divisible by the denominator,

x+ax3+a3=x+ax3+ax2−ax2+a3.

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 −ax2+a3 in the numerator,

x+ax3+ax2−ax2+a3=x+ax3+ax2+x+a−ax2+a3=x+ax2(x+a)+x+a−ax2+a3=x2+x+a−ax2+a3.

When we treat the new quotient, we add and take away −a2x to/from −ax2 in order to get something divisible by x+a,

x2+x+a−ax2+a3=x2+x+a−ax2−a2x+a2x+a3=x2+x+a−ax2−a2x+x+aa2x+a3=x2+x+a−ax(x+a)+x+aa2x+a3=x2−ax+x+aa2x+a3.

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

x2−ax+x+aa2x+a3=x2−ax+x+aa2(x+a)=x2−ax+a2.

If we have calculated correctly, we should have

x+ax3+a3=x2−ax+a2

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

x3+a3=(x2−ax+a2)(x+a).

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

RHS=(x2−ax+a2)(x+a)=x3+ax2−ax2−a2x+a2x+a3=x3+a3=LHS.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.4:1c“