Lösung 3.4:1d
Aus Online Mathematik Brückenkurs 2
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Version vom 10:51, 11. Mär. 2009
We start by adding and taking away \displaystyle x^2 in the numerator, so that, in combination with \displaystyle x^3, we obtain the expression \displaystyle x^3+x^2 = x^2(x+1) which can be simplified with the denominator \displaystyle x+1,
| \displaystyle \begin{align}
\frac{x^3+x+2}{x+1} &= \frac{x^3+x^2-x^2+x+2}{x+1}\\[5pt] &= \frac{x^3+x^2}{x+1} + \frac{-x^2+x+2}{x+1}\\[5pt] &= \frac{x^2(x+1)}{x+1} + \frac{-x^2+x+2}{x+1}\\[5pt] &= x^2 + \frac{-x^2+x+2}{x+1}\,\textrm{.} \end{align} |
The term \displaystyle -x^2 in the remaining quotient needs to complemented with \displaystyle -x so that we get \displaystyle -x^2-x = -x(x+1), which is divisible by \displaystyle x+1,
| \displaystyle \begin{align}
x^2 + \frac{-x^2+x+2}{x+1} &= x^2 + \frac{-x^2-x+x+x+2}{x+1}\\[5pt] &= x^2 + \frac{-x^2-x}{x+1} + \frac{2x+2}{x+1}\\[5pt] &= x^2 + \frac{-x(x+1)}{x+1} + \frac{2x+2}{x+1}\\[5pt] &= x^2 - x + \frac{2x+2}{x+1}\,\textrm{.} \end{align} |
The last quotient divides perfectly and we obtain
| \displaystyle x^2-x+\frac{2x+2}{x+1}=x^2-x+2\,\textrm{.} |
A quick check of whether
| \displaystyle \frac{x^3+x+2}{x+1} = x^2-x+2\,\textrm{.} |
is the correct answer is to investigate whether
| \displaystyle x^3+x+2 = (x^2-x+2)(x+1) |
holds. If we expand the right-hand side, we see that the relation really does hold
| \displaystyle \begin{align}
(x^2-x+2)(x+1) = x^3+x^2-x^2-x+2x+2 = x^3+x+2\,\textrm{.} \end{align} |
