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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 x2 in the numerator, so that, in combination with x3, we obtain the expression x3+x2=x2(x+1) which can be simplified with the denominator x+1,

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

The term −x2 in the remaining quotient needs to complemented with −x so that we get −x2−x=−x(x+1), which is divisible by x+1,

x2+x+1−x2+x+2=x2+x+1−x2−x+x+x+2=x2+x+1−x2−x+x+12x+2=x2+x+1−x(x+1)+x+12x+2=x2−x+x+12x+2.

The last quotient divides perfectly and we obtain

x2−x+x+12x+2=x2−x+2.

A quick check of whether

x+1x3+x+2=x2−x+2.

is the correct answer is to investigate whether

x3+x+2=(x2−x+2)(x+1)

holds. If we expand the right-hand side, we see that the relation really does hold

(x2−x+2)(x+1)=x3+x2−x2−x+2x+2=x3+x+2.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.4:1d“