Lösung 3.4:7a - Versionsgeschichte

Sekretariat1 um 11:03, 4. Sep. 2009

Sekretariat1 — Fri, 04 Sep 2009 11:03:21 GMT

← Nächstältere Version		Version vom 11:03, 4. Sep. 2009
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	{{Abgesetzte Formel\|\|<math>(z-1)(z-2)(z-4) = z^3-7z^2+14z-8\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>(z-1)(z-2)(z-4) = z^3-7z^2+14z-8\,\textrm{.}</math>}}

-	Hinweis: Es ist möglich, das Polynom mit einer Konstante zu multiplizieren und dieselben Nullstellen zu ~~behalten~~.	+	Hinweis: Es ist möglich, das Polynom mit einer Konstante zu multiplizieren und dieselben Nullstellen zu erhalten.

Sekretariat1 um 10:59, 4. Sep. 2009

Sekretariat1 — Fri, 04 Sep 2009 10:59:39 GMT

← Nächstältere Version		Version vom 10:59, 4. Sep. 2009
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-	Ein Polynom mit der Nullstelle <math>z=a</math>, enthält den Faktor <math>(z-a)</math>.	+	Ein Polynom mit der Nullstelle <math>z=a</math> enthält den Faktor <math>(z-a)</math>.

-	Ein Polynom mit den Nullstellen <math>1</math>, <math>2</math> und <math>4</math>, enthält daher die Faktoren <math>(z-1)</math>, <math>(z-2)</math> und <math>(z-4)</math>, zum Beispiel	+	Ein Polynom mit den Nullstellen <math>1</math>, <math>2</math> und <math>4</math> enthält daher die Faktoren <math>(z-1)</math>, <math>(z-2)</math> und <math>(z-4)</math>, zum Beispiel

	{{Abgesetzte Formel\|\|<math>(z-1)(z-2)(z-4) = z^3-7z^2+14z-8\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>(z-1)(z-2)(z-4) = z^3-7z^2+14z-8\,\textrm{.}</math>}}

-	Hinweis: Es ist möglich das Polynom mit einer Konstante zu multiplizieren, und dieselben Nullstellen zu behalten.	+	Hinweis: Es ist möglich, das Polynom mit einer Konstante zu multiplizieren und dieselben Nullstellen zu behalten.

Martin um 14:31, 21. Mai 2009

Martin — Thu, 21 May 2009 14:31:35 GMT

← Nächstältere Version		Version vom 14:31, 21. Mai 2009
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-	~~There exists a simple relation between a zero and the polynomial's factorization:~~ <math>z=a</math> ~~is a zero if and only if the polynomial contains the factor~~ <math>(z-a)</math>. ~~(This is the meaning of the factor theorem.)~~	+	Ein Polynom mit der Nullstelle <math>z=a</math>, enthält den Faktor <math>(z-a)</math>.

-	~~If we are to have a polynomial with zeros at~~ <math>1</math>, <math>2</math> ~~and~~ <math>4</math>, ~~the polynomial must therefore contain the factors~~ <math>(z-1)</math>, <math>(z-2)</math> ~~and~~ <math>(z-4)</math>~~. For example~~,	+	Ein Polynom mit den Nullstellen <math>1</math>, <math>2</math> und <math>4</math>, enthält daher die Faktoren <math>(z-1)</math>, <math>(z-2)</math> und <math>(z-4)</math>, zum Beispiel

	{{Abgesetzte Formel\|\|<math>(z-1)(z-2)(z-4) = z^3-7z^2+14z-8\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>(z-1)(z-2)(z-4) = z^3-7z^2+14z-8\,\textrm{.}</math>}}

-		+	Hinweis: Es ist möglich das Polynom mit einer Konstante zu multiplizieren, und dieselben Nullstellen zu behalten.
-	~~Note~~: ~~It is possible to multiply the polynomial above by a non-zero constant and get another third-degree polynomial with the same roots~~.	+

Tekbot: Solution 3.4:7a moved to Lösung 3.4:7a: Robot: moved page

Tekbot — Wed, 11 Mar 2009 10:53:35 GMT

Solution 3.4:7a moved to Lösung 3.4:7a: Robot: moved page

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Version vom 10:53, 11. Mär. 2009

Tekbot: Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

Tekbot — Tue, 10 Mar 2009 13:16:14 GMT

Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

← Nächstältere Version		Version vom 13:16, 10. Mär. 2009
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	If we are to have a polynomial with zeros at <math>1</math>, <math>2</math> and <math>4</math>, the polynomial must therefore contain the factors <math>(z-1)</math>, <math>(z-2)</math> and <math>(z-4)</math>. For example,		If we are to have a polynomial with zeros at <math>1</math>, <math>2</math> and <math>4</math>, the polynomial must therefore contain the factors <math>(z-1)</math>, <math>(z-2)</math> and <math>(z-4)</math>. For example,

-	{{~~Displayed math~~\|\|<math>(z-1)(z-2)(z-4) = z^3-7z^2+14z-8\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>(z-1)(z-2)(z-4) = z^3-7z^2+14z-8\,\textrm{.}</math>}}


	Note: It is possible to multiply the polynomial above by a non-zero constant and get another third-degree polynomial with the same roots.		Note: It is possible to multiply the polynomial above by a non-zero constant and get another third-degree polynomial with the same roots.

Tek um 14:37, 31. Okt. 2008

Tek — Fri, 31 Oct 2008 14:37:21 GMT

← Nächstältere Version		Version vom 14:37, 31. Okt. 2008
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-	There exists a simple relation between a zero and the polynomial's factorization:	+	There exists a simple relation between a zero and the polynomial's factorization: <math>z=a</math> is a zero if and only if the polynomial contains the factor <math>(z-a)</math>. (This is the meaning of the factor theorem.)
-	<math>z=a~~\text{ }~~</math>	+
-	is a zero if and only if the polynomial contains the factor	+
-	<math>~~\left~~( z-a ~~\right~~)</math>. (This is the meaning of the factor theorem.)	+

-	If we are to have a polynomial with zeros at	+	If we are to have a polynomial with zeros at <math>1</math>, <math>2</math> and <math>4</math>, the polynomial must therefore contain the factors <math>(z-1)</math>, <math>(z-2)</math> and <math>(z-4)</math>. For example,
-	<math>1,\ 2</math>	+
-	and	+
-	<math>~~\text{~~4}</math>, the polynomial must therefore contain the factors	+
-	<math>~~\left~~( z-1 ~~\right~~),~~\ \left~~( z-2 ~~\right~~)</math>	+
-	and	+
-	<math>~~\left~~( z-4 ~~\right~~)</math>. For example,	+

		+	{{Displayed math\|\|<math>(z-1)(z-2)(z-4) = z^3-7z^2+14z-8\,\textrm{.}</math>}}

-	<math>\left( z-1 \right)\left( z-2 \right)\left( z-4 \right)=z^{3}-7z^{2}+14z-8</math>

-		+	Note: It is possible to multiply the polynomial above by a non-zero constant and get another third-degree polynomial with the same roots.
-	~~NOTE~~: it is possible to multiply the polynomial above by a non-zero constant and get another third-degree polynomial with the same roots.	+

Ian um 15:44, 28. Okt. 2008

Ian — Tue, 28 Oct 2008 15:44:45 GMT

← Nächstältere Version		Version vom 15:44, 28. Okt. 2008
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-	{~~{NAVCONTENT_START}~~}	+	There exists a simple relation between a zero and the polynomial's factorization:
-	<~~center~~> ~~[[Image:3_4_7a~~.~~gif]]~~ </~~center~~>	+	<math>z=a\text{ }</math>
-	{{~~NAVCONTENT_STOP~~}}	+	is a zero if and only if the polynomial contains the factor
		+	<math>\left( z-a \right)</math>. (This is the meaning of the factor theorem.)
		+
		+	If we are to have a polynomial with zeros at
		+	<math>1,\ 2</math>
		+	and
		+	<math>\text{4}</math>, the polynomial must therefore contain the factors
		+	<math>\left( z-1 \right),\ \left( z-2 \right)</math>
		+	and
		+	<math>\left( z-4 \right)</math>. For example,
		+
		+
		+	<math>\left( z-1 \right)\left( z-2 \right)\left( z-4 \right)=z^{3}-7z^{2}+14z-8</math>
		+
		+
		+	NOTE: it is possible to multiply the polynomial above by a non-zero constant and get another third-degree polynomial with the same roots.

Tekbot: Lösning 3.4:7a moved to Solution 3.4:7a: Robot: moved page

Tekbot — Tue, 16 Sep 2008 15:14:01 GMT

Lösning 3.4:7a moved to Solution 3.4:7a: Robot: moved page

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Version vom 15:14, 16. Sep. 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

Tekbot — Thu, 21 Aug 2008 08:16:44 GMT

Robot: Automated text replacement (-[[Bild: +[[Image:)

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Oskar — Mon, 07 Apr 2008 15:10:41 GMT

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