Lösung 2.1:3e
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Version vom 13:46, 22. Okt. 2008
Both terms contain 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^2)\,\textrm{.} |
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).
