Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Lösung 3.4:1c

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 1:
-
If we focus on the leading term
+
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,
-
<math>x^{3}</math>, we need to complement it with in order to get an expression that is divisible by the denominator ,
+
 +
{{Displayed math||<math>\frac{x^3+a^3}{x+a} = \frac{x^3+ax^2-ax^2+a^3}{x+a}\,\textrm{.}</math>}}
-
<math>\frac{x^{3}+a^{3}}{x+a}=\frac{x^{3}+ax^{2}-ax^{2}+a^{3}}{x+a}</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,
 +
{{Displayed math||<math>\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}</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
+
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>,
-
<math>-ax^{2}+a^{3}</math>
+
-
in the numerator:
+
 +
{{Displayed math||<math>\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}</math>}}
-
<math>\begin{align}
+
In the last quotient, the numerator has <math>x+a</math> as a factor, and we obtain a perfect division,
-
& \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}</math>
+
-
 
+
-
 
+
-
When we treat the new quotient, we add and take away
+
-
<math>-a^{2}x</math>
+
-
to/from
+
-
<math>-ax^{2}</math>
+
-
in order to get something divisible by
+
-
<math>x+a</math>
+
-
 
+
-
 
+
-
 
+
-
<math>\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}</math>
+
-
 
+
-
 
+
-
In the last quotient, the numerator has
+
-
<math>x+a</math>
+
-
as a factor, and we obtain a perfect division:
+
-
 
+
-
 
+
-
<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>
+
 +
{{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
 +
{{Displayed math||<math>\frac{x^3+a^3}{x+a} = x^2-ax+a^2</math>}}
-
<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>,
+
-
 
+
-
 
+
-
<math>x^{3}+a^{3}=\left( x^{2}-ax+a^{2} \right)\left( x+a \right)</math>
+
-
 
+
-
Then, expand the right-hand side and then we should get what is on the left-hand side:
+
{{Displayed math||<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,
-
<math>\begin{align}
+
{{Displayed math||<math>\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} \\
+
\text{RHS}
-
& =x^{3}+a^{3}=~~\text{LHS} \\
+
&= (x^2-ax+a^2)(x+a)\\[5pt]
-
\end{align}</math>
+
&= x^3+ax^2-ax^2-a^2x+a^2x+a^3\\[5pt]
 +
&= x^3+a^3\\[5pt]
 +
&= \text{LHS.}
 +
\end{align}</math>}}

Version vom 12:11, 31. Okt. 2008

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“