Lösung 3.4:1e
Aus Online Mathematik Brückenkurs 2
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<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^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>, | ||
- | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\frac{x^3+2x^2+1}{x^2+3x+1} | \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\bbox[#FFEEAA;,1.5pt]{{}+3x^2+x-3x^2-x}+2x^2+1}{x^2+3x+1}\\[5pt] | ||
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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 | 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 | ||
- | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
x + \frac{-x^2-x+1}{x^2+3x+1} | x + \frac{-x^2-x+1}{x^2+3x+1} | ||
&= x + \frac{-x^2\bbox[#FFEEAA;,1.5pt]{{}-3x-1+3x+1}-x+1}{x^2+3x+1}\\[5pt] | &= x + \frac{-x^2\bbox[#FFEEAA;,1.5pt]{{}-3x-1+3x+1}-x+1}{x^2+3x+1}\\[5pt] |
Version vom 13:15, 10. Mär. 2009
Imagine for a moment taking away all the terms in the numerator apart from \displaystyle x^3. If we are to make \displaystyle x^3 divisible by the denominator \displaystyle x^2+3x+1, we need to add and subtract \displaystyle 3x^2+x in order to obtain the expression \displaystyle x^3+3x^2+x=x(x^2+3x+1),
\displaystyle \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} |
Now, we carry out the same procedure with the new quotient. To the term \displaystyle -x^2, we add and subtract \displaystyle -3x-1 and obtain
\displaystyle \begin{align}
x + \frac{-x^2-x+1}{x^2+3x+1} &= x + \frac{-x^2\bbox[#FFEEAA;,1.5pt]{{}-3x-1+3x+1}-x+1}{x^2+3x+1}\\[5pt] &= x + \frac{-x^2-3x-1}{x^2+3x+1} + \frac{3x+1-x+1}{x^2+3x+1}\\[5pt] &= x - 1 + \frac{2x+2}{x^2+3x+1}\,\textrm{.} \end{align} |