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Lösung 3.4:1e

Aus Online Mathematik Brückenkurs 2

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Version vom 10:51, 11. Mär. 2009

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“