Lösung 3.4:1c
Aus Online Mathematik Brückenkurs 2
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Version vom 10:50, 11. Mär. 2009
If we focus on the leading term \displaystyle x^3, we need to complement it with \displaystyle ax^2 in order to get a sub expression that is divisible by the denominator,
| \displaystyle \frac{x^3+a^3}{x+a} = \frac{x^3+ax^2-ax^2+a^3}{x+a}\,\textrm{.} |
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 \displaystyle -ax^2+a^3 in the numerator,
| \displaystyle \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} |
When we treat the new quotient, we add and take away \displaystyle -a^2x to/from \displaystyle -ax^2 in order to get something divisible by \displaystyle x+a,
| \displaystyle \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} |
In the last quotient, the numerator has \displaystyle x+a as a factor, and we obtain a perfect division,
| \displaystyle 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{.} |
If we have calculated correctly, we should have
| \displaystyle \frac{x^3+a^3}{x+a} = x^2-ax+a^2 |
and one way to check the answer is to multiply both sides by \displaystyle x+a,
| \displaystyle x^3+a^3 = (x^2-ax+a^2)(x+a)\,\textrm{.} |
Then, expand the right-hand side and we should get what is on the left-hand side,
| \displaystyle \begin{align}
\text{RHS} &= (x^2-ax+a^2)(x+a)\\[5pt] &= x^3+ax^2-ax^2-a^2x+a^2x+a^3\\[5pt] &= x^3+a^3\\[5pt] &= \text{LHS.} \end{align} |
