Aus Online Mathematik Brückenkurs 2
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,
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| 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. |
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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,
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| 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. |
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The last quotient divides perfectly and we obtain
A quick check of whether
is the correct answer is to investigate whether
holds. If we expand the right-hand side, we see that the relation really does hold
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| (x2−x+2)(x+1)=x3+x2−x2−x+2x+2=x3+x+2. |
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