Aus Online Mathematik Brückenkurs 1

Version vom 08:37, 23. Sep. 2008 (bearbeiten) Tek (Diskussion | Beiträge) K ← Zum vorherigen Versionsunterschied		Version vom 08:23, 22. Okt. 2008 (bearbeiten) (rückgängig) Tekbot (Diskussion | Beiträge) K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) Zum nächsten Versionsunterschied →
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	Both terms contain ''x'', which can therefore be taken out as a factor (as can 2),		Both terms contain ''x'', which can therefore be taken out as a factor (as can 2),
-	{{Displayed math||<math>18x-2x^3=2x\cdot 9-2x \cdot x^2=2x(9-x^2)\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<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
-	{{Displayed math||<math> 2x(9-x^2)=2x(3^2-x^2)=2x(3+x)(3-x)\,,</math>}}	+	{{Abgesetzte Formel||<math> 2x(9-x^2)=2x(3^2-x^2)=2x(3+x)(3-x)\,,</math>}}
	which can also be written as <math>-2x(x+3)(x-3).</math>		which can also be written as <math>-2x(x+3)(x-3).</math>

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Version vom 08:23, 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).