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

Lösung 3.4:1e

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 1:
-
Imagine for a moment taking away all the terms in the numerator apart from
+
Imagine for a moment taking away all the terms in the numerator apart from <math>x^3</math>. If we are to make <math>x^3</math> divisible by the denominator
-
<math>x^{3}</math>. If we are to make
+
<math>x^2+3x+1</math>, we need to add and subtract <math>3x^2+x</math> in order to obtain the expression <math>x^3+3x^2+x=x(x^2+3x+1)</math>,
-
<math>x^{3}</math>
+
-
divisible by the denominator
+
-
<math>x^{2}+3x+1</math>, we need to add and subtract
+
-
<math>3x^{2}+x</math>
+
-
in order to obtain the expression
+
-
<math>x^{3}+3x^{2}+x=x\left( x^{2}+3x+1 \right)</math>,
+
 +
{{Displayed math||<math>\begin{align}
 +
\frac{x^3+2x^2+1}{x^2+3x+1}
 +
&= \frac{x^3\bbox[#FFEEAA;,1.5pt]{{}+3x^2+x-3x^2-x}+2x^2+1}{x^2+3x+1}\\[5pt]
 +
&= \frac{x^3+3x^2+x}{x^2+3x+1} + \frac{-3x^2-x+2x^2+1}{x^2+3x+1}\\[5pt]
 +
&= \frac{x(x^2+3x+1)}{x^2+3x+1} + \frac{-x^2-x+1}{x^2+3x+1}\\[5pt]
 +
&= x+\frac{-x^2-x+1}{x^2+3x+1}\,\textrm{.}
 +
\end{align}</math>}}
-
<math>\begin{align}
+
Now, we carry out the same procedure with the new quotient. To the term <math>-x^2</math>, we add and subtract <math>-3x-1</math> and obtain
-
& \frac{x^{3}+2x^{2}+1}{x^{2}+3x+1}=\frac{x^{3}+\underline{3x^{2}+x-3x^{2}-x}+2x^{2}+1}{x^{2}+3x+1} \\
+
-
& =\frac{x^{3}+3x^{2}+x}{x^{2}+3x+1}+\frac{-3x^{2}-x+2x^{2}+1}{x^{2}+3x+1} \\
+
-
& =\frac{x\left( x^{2}+3x+1 \right)}{x^{2}+3x+1}+\frac{-x^{2}-x+1}{x^{2}+3x+1} \\
+
-
& =x+\frac{-x^{2}-x+1}{x^{2}+3x+1} \\
+
-
\end{align}</math>
+
-
 
+
{{Displayed math||<math>\begin{align}
-
Now, we carry out the same procedure with the new quotient. To the term
+
x + \frac{-x^2-x+1}{x^2+3x+1}
-
<math>-x^{2}</math>, we add and subtract
+
&= x + \frac{-x^2\bbox[#FFEEAA;,1.5pt]{{}-3x-1+3x+1}-x+1}{x^2+3x+1}\\[5pt]
-
<math>-\text{3}x-\text{1}</math>
+
&= x + \frac{-x^2-3x-1}{x^2+3x+1} + \frac{3x+1-x+1}{x^2+3x+1}\\[5pt]
-
and obtain
+
&= x - 1 + \frac{2x+2}{x^2+3x+1}\,\textrm{.}
-
 
+
\end{align}</math>}}
-
 
+
-
<math>\begin{align}
+
-
& x+\frac{-x^{2}-x+1}{x^{2}+3x+1}=x+\frac{-x^{2}\underline{-3x-1+3x+1}-x+1}{x^{2}+3x+1} \\
+
-
& =x+\frac{-x^{2}-3x-1}{x^{2}+3x+1}+\frac{3x+1-x+1}{x^{2}+3x+1} \\
+
-
& =x-1+\frac{2x+2}{x^{2}+3x+1} \\
+
-
\end{align}</math>
+

Version vom 12:40, 31. Okt. 2008

Imagine for a moment taking away all the terms in the numerator apart from x3. If we are to make x3 divisible by the denominator x2+3x+1, we need to add and subtract 3x2+x in order to obtain the expression x3+3x2+x=x(x2+3x+1),

x2+3x+1x3+2x2+1=x2+3x+1x3+3x2+x−3x2−x+2x2+1=x2+3x+1x3+3x2+x+x2+3x+1−3x2−x+2x2+1=x2+3x+1x(x2+3x+1)+x2+3x+1−x2−x+1=x+x2+3x+1−x2−x+1.

Now, we carry out the same procedure with the new quotient. To the term −x2, we add and subtract −3x−1 and obtain

x+x2+3x+1−x2−x+1=x+x2+3x+1−x2−3x−1+3x+1−x+1=x+x2+3x+1−x2−3x−1+x2+3x+13x+1−x+1=x−1+2x+2x2+3x+1.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.4:1e“