Lösung 2.1:3e

Aus Online Mathematik Brückenkurs 1

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Both terms contain <math>x</math>, which can therefore be taken out as a factor (as can 2).
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:<math>18x-2x^3=2x\cdot 9-2x \cdot x^2=2x(9-x). </math>
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The remaining second-degree factor <math> 9-x^2 </math> can then be factorized using the conjugate rule
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:<math> 2x(9-x^2)=2x(3^2-x^2)=2x(3+x)(3-x), </math>
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which can also be written as <math> -2x(x+3)(x-3).</math>
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Version vom 13:15, 13. Aug. 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).

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:3e“