Solution 2.1:3e

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Both terms contain ''x'', which can therefore be taken out as a factor (as can 2),
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<!--center> [[Bild:2_1_3e.gif]] </center-->
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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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{{Displayed math||<math>18x-2x^3=2x\cdot 9-2x \cdot x^2=2x(9-x^2)\,\textrm{.}</math>}}
The remaining second-degree factor <math> 9-x^2 </math> can then be factorized using the conjugate rule
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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{{Displayed math||<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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which can also be written as <math>-2x(x+3)(x-3).</math>
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Current revision

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).

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