Lösung 2.1:3e
Aus Online Mathematik Brückenkurs 1
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Version vom 08:36, 9. Sep. 2008
Both terms contain \displaystyle x, which can therefore be taken out as a factor (as can 2).
- \displaystyle 18x-2x^3=2x\cdot 9-2x \cdot x^2=2x(9-x).
The remaining second-degree factor \displaystyle 9-x^2 can then be factorized using the conjugate rule
- \displaystyle 2x(9-x^2)=2x(3^2-x^2)=2x(3+x)(3-x),
which can also be written as \displaystyle -2x(x+3)(x-3).