Solution 2.1:4a - Revision history

Tek at 08:54, 23 September 2008

Tek — Tue, 23 Sep 2008 08:54:29 GMT

←Older revision		Revision as of 08:54, 23 September 2008
Line 1:		Line 1:
-	First, we multiply the second bracket by	+	First, we multiply the second bracket by ''x'' from the first bracket,
-	~~<math>~~x~~</math>~~	+
-	from the first bracket,	+

		+	{{Displayed math\|\|<math>(\bbox[#FFEEAA;,1.5pt]{\strut x}+\bbox[#FFFFFF;,1.5pt]{\strut 2})(3x^{2}-x+5) = \bbox[#FFEEAA;,1.5pt]{\strut x\cdot 3x^{2}-x\cdot x+x\cdot 5}+{}\rlap{\cdots}\phantom{\bbox[#FFEEAA;,1.5pt]{\strut 2\cdot 3x^{2}-2\cdot x+2\cdot 5}\,\textrm{.}}</math>}}

-	~~<math>\left( x+~~2 ~~\right)\left( 3x^{2}-x+5 \right)=x\centerdot 3x^{2}-x\centerdot x+x\centerdot 5+...</math>~~	+	Then, do the same for 2 from the first bracket

		+	{{Displayed math\|\|
		+	<math>(\bbox[#FFFFFF;,1.5pt]x+\bbox[#FFEEAA;,1.5pt]{\strut 2})(3x^{2}-x+5) = \secondcbox{#FFFFFF;}{\strut x\cdot 3x^{2}-x\cdot x+x\cdot 5}{3x^{3}-x^{2}+5x}+\bbox[#FFEEAA;,1.5pt]{\strut 2\cdot 3x^{2}-2\cdot x+2\cdot 5}\,\textrm{.}</math>}}

-	~~Then~~, ~~do the same for~~	+	Now, collect together ''x''³-, ''x''²-, ''x''- and the constant terms
-	~~<math>2</math>~~	+
-	~~from~~ the ~~first bracket:~~	+

		+	{{Displayed math\|\|
		+	<math>3x^{3}+(-1+6)x^{2}+(5-2)x+10=3x^{3}+5x^{2}+3x+10\,\textrm{.}</math>}}

-	~~<math>\left( x+2 \right)\left( 3x^{2}-x+5 \right)=3x^{3}-x^{2}+5x+2\centerdot 3x^{2}-2\centerdot x+2\centerdot 5</math>~~	+	The coefficient in front of ''x''² is 5 and the coefficient in front of ''x'' is 3.
-		+
-		+
-	~~Now, collect together~~	+
-	~~<math>x^{3}</math>-,~~	+
-	~~<math>x^{2}</math>-,~~	+
-	~~<math>x</math>- and the constant terms:~~	+
-		+
-		+
-	~~<math>3x^{3}+\left( -1+6 \right)x^{2}+\left( 5-2 \right)x+10=3x^{3}+5x^{2}+3x+10</math>~~	+
-		+
-		+
-	The coefficient in front of	+
-	~~<math>~~x~~^{2}</math>~~	+
-	is	+
-	~~<math>~~5~~</math>~~	+
-	and the coefficient in front of x is	+
-	~~<math>~~3~~</math>~~.	+

Ian at 13:42, 15 September 2008

Ian — Mon, 15 Sep 2008 13:42:35 GMT

←Older revision		Revision as of 13:42, 15 September 2008
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-	{{~~NAVCONTENT_START}~~}	+	First, we multiply the second bracket by
-	<~~center~~> ~~[[Image~~:~~2_1_4a.gif]]~~ </~~center~~>	+	<math>x</math>
-	{{~~NAVCONTENT_STOP~~}}	+	from the first bracket,
		+
		+
		+	<math>\left( x+2 \right)\left( 3x^{2}-x+5 \right)=x\centerdot 3x^{2}-x\centerdot x+x\centerdot 5+...</math>
		+
		+
		+	Then, do the same for
		+	<math>2</math>
		+	from the first bracket:
		+
		+
		+	<math>\left( x+2 \right)\left( 3x^{2}-x+5 \right)=3x^{3}-x^{2}+5x+2\centerdot 3x^{2}-2\centerdot x+2\centerdot 5</math>
		+
		+
		+	Now, collect together
		+	<math>x^{3}</math>-,
		+	<math>x^{2}</math>-,
		+	<math>x</math>- and the constant terms:
		+
		+
		+	<math>3x^{3}+\left( -1+6 \right)x^{2}+\left( 5-2 \right)x+10=3x^{3}+5x^{2}+3x+10</math>
		+
		+
		+	The coefficient in front of
		+	<math>x^{2}</math>
		+	is
		+	<math>5</math>
		+	and the coefficient in front of x is
		+	<math>3</math>.

Tekbot: Lösning 2.1:4a moved to Solution 2.1:4a: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:36:43 GMT

Lösning 2.1:4a moved to Solution 2.1:4a: Robot: moved page

←Older revision

Revision as of 08:36, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 06:35:29 GMT

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

←Older revision		Revision as of 06:35, 21 August 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_1_4a.gif]] </center>	+	<center> [[Image:2_1_4a.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Lina: Ny sida: {{NAVCONTENT_START}} Bild:2_1_4a.gif {{NAVCONTENT_STOP}}

Lina — Mon, 31 Mar 2008 09:55:56 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:2_1_4a.gif </center> {{NAVCONTENT_STOP}}

New page

{{NAVCONTENT_START}}
<center> [[Bild:2_1_4a.gif]] </center>
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