Solution 2.1:4c - Revision history

Tek at 10:26, 23 September 2008

2008-09-23T10:26:40Z

←Older revision		Revision as of 10:26, 23 September 2008
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-	Instead of multiplying together the whole expression, and then reading off the coefficients, we investigate which terms from the three brackets together give terms in	+	Instead of multiplying together the whole expression, and then reading off the coefficients, we investigate which terms from the three brackets together give terms in ''x''¹ and ''x''².
-	~~<math>~~x~~^{1}</math>~~	+
-	and	+
-	~~<math>~~x~~^{2}</math>~~.	+

-	If we start with the term in	+	If we start with the term in ''x'', we see that there is only one combination of a term from each bracket which, when multiplied, gives ''x''¹,
-	~~<math>~~x~~</math>~~, we see that there is only one combination of a term from each bracket	+
-	which, when multiplied, gives	+
-	~~<math>~~x~~^{1}</math>~~,	+

		+	{{Displayed math\|\|<math>(\underline{x}-x^{3}+x^{5})(\underline{1}+3x+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 1\cdot 2} + \cdots</math>}}

-	<math>~~\left( x-x^{3}+x^{5} \right)\left(~~ 1~~+3x+5x^{2}~~ \~~right)\left(~~ 2~~-7x^{~~2~~}-x^{4}~~ \~~right)=...+x\centerdot 1\centerdot 2+...~~</math>	+	so, the coefficient in front of ''x'' is <math>1\cdot 2 = 2\,</math>.

-	~~so, the coefficient in front of~~	+	As for ''x''², we also have only one possible combination
-	~~<math>~~x~~</math>~~	+
-	is	+
-	~~<math>1\centerdot 2=2</math>.~~	+

-	~~As for~~	+	{{Displayed math\|\|<math>(\underline{x}-x^{3}+x^{5})(1+\underline{3x}+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 3x\cdot 2} + \cdots</math>}}
-	<math>x^{2}</math>~~, we also have only one possible combination:~~	+

-		+	The coefficient in front of ''x''² is <math>3\cdot 2 = 6\,</math>.
-	~~<math>\left( x-x^{3}+x^{5} \right)\left( 1+3x+5x^{2} \right)\left( 2-7x^{2}-x^{4} \right)=...+x\centerdot 3x\centerdot 2+...</math>~~	+
-		+
-		+
-	The coefficient in front of	+
-	~~<math>~~x~~^{2}</math>~~	+
-	is	+
-	<math>3\~~centerdot~~ 2=6</math>	+

Ian at 14:36, 15 September 2008

2008-09-15T14:36:02Z

←Older revision		Revision as of 14:36, 15 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	Instead of multiplying together the whole expression, and then reading off the coefficients, we investigate which terms from the three brackets together give terms in
-	<~~center~~> ~~[[Image:2_1_4c~~.~~gif]]~~ </~~center~~>	+	<math>x^{1}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and
		+	<math>x^{2}</math>.
		+
		+	If we start with the term in
		+	<math>x</math>, we see that there is only one combination of a term from each bracket
		+	which, when multiplied, gives
		+	<math>x^{1}</math>,
		+
		+
		+	<math>\left( x-x^{3}+x^{5} \right)\left( 1+3x+5x^{2} \right)\left( 2-7x^{2}-x^{4} \right)=...+x\centerdot 1\centerdot 2+...</math>
		+
		+	so, the coefficient in front of
		+	<math>x</math>
		+	is
		+	<math>1\centerdot 2=2</math>.
		+
		+	As for
		+	<math>x^{2}</math>, we also have only one possible combination:
		+
		+
		+	<math>\left( x-x^{3}+x^{5} \right)\left( 1+3x+5x^{2} \right)\left( 2-7x^{2}-x^{4} \right)=...+x\centerdot 3x\centerdot 2+...</math>
		+
		+
		+	The coefficient in front of
		+	<math>x^{2}</math>
		+	is
		+	<math>3\centerdot 2=6</math>

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

2008-09-09T08:37:23Z

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

←Older revision

Revision as of 08:37, 9 September 2008

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

2008-08-21T06:35:49Z

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

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2008-03-31T10:01:53Z

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