Solution 2.1:2a - Revision history

Tek at 08:10, 23 September 2008

2008-09-23T08:10:13Z

←Older revision		Revision as of 08:10, 23 September 2008
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-	<!--center> [[Image:2_1_2a.gif]] </center-->
	First, multiply the brackets together. In the first product, every term in the first bracket is multiplied by every term in the second bracket,		First, multiply the brackets together. In the first product, every term in the first bracket is multiplied by every term in the second bracket,

-	<math>	+	{{Displayed math\|\|<math>\begin{align}
-	~~\qquad~~	+
-	\begin{align}	+
	(x-4)(x-5)-3x(2x-3)&= x\cdot x-x\cdot 5- 4\cdot x-4\cdot (-5)-(3x \cdot 2x-3x\cdot 3)\\		(x-4)(x-5)-3x(2x-3)&= x\cdot x-x\cdot 5- 4\cdot x-4\cdot (-5)-(3x \cdot 2x-3x\cdot 3)\\
	&= x^2-5x-4x+20-(6x^2-9x)\\		&= x^2-5x-4x+20-(6x^2-9x)\\
-	&=x^2-5x-4x+20-6x^2+9x	+	&=x^2-5x-4x+20-6x^2+9x\,\textrm{.}
-	\end{align}	+	\end{align}</math>}}
-	</math>	+

-	Then, gather together ~~<math>~~x^2-, \, x- ~~</math>~~ and the constant terms and simplify	+	Then, gather together ''x''²-, ''x''- and the constant terms and simplify

-	<math>	+	{{Displayed math\|\|<math>\begin{align}
-	~~\qquad~~	+
-	\begin{align}	+
	\phantom{(x-4)(x-5)-3x(2x-3)}&= (x^2-6x^2)+(-5x-4x+9x)+20 \\		\phantom{(x-4)(x-5)-3x(2x-3)}&= (x^2-6x^2)+(-5x-4x+9x)+20 \\
	&= -5x^2+0+20\\		&= -5x^2+0+20\\
-	&= -5x^2+20	+	&= \rlap{-5x^2+20\,\textrm{.}}\phantom{x\cdot x-x\cdot 5- 4\cdot x-4\cdot (-5)-(3x \cdot 2x-3x\cdot 3)}
-	\end{align}	+	\end{align}</math>}}
-	</math>	+
-		+
-	~~{{NAVCONTENT_STOP~~}}	+

2008-09-09T08:33:03Z

←Older revision

Revision as of 08:33, 9 September 2008

2008-08-21T06:33:39Z

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

←Older revision		Revision as of 06:33, 21 August 2008
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-	<!--center> [[~~Bild~~:2_1_2a.gif]] </center-->	+	<!--center> [[Image:2_1_2a.gif]] </center-->
	First, multiply the brackets together. In the first product, every term in the first bracket is multiplied by every term in the second bracket,		First, multiply the brackets together. In the first product, every term in the first bracket is multiplied by every term in the second bracket,

2008-08-13T10:01:59Z

←Older revision		Revision as of 10:01, 13 August 2008
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-	<center> [[Bild:2_1_2a.gif]] </center>	+	<!--center> [[Bild:2_1_2a.gif]] </center-->
	First, multiply the brackets together. In the first product, every term in the first bracket is multiplied by every term in the second bracket,		First, multiply the brackets together. In the first product, every term in the first bracket is multiplied by every term in the second bracket,

2008-08-13T10:01:44Z

←Older revision		Revision as of 10:01, 13 August 2008
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	<center> [[Bild:2_1_2a.gif]] </center>		<center> [[Bild:2_1_2a.gif]] </center>
		+	First, multiply the brackets together. In the first product, every term in the first bracket is multiplied by every term in the second bracket,
		+
		+	<math>
		+	\qquad
		+	\begin{align}
		+	(x-4)(x-5)-3x(2x-3)&= x\cdot x-x\cdot 5- 4\cdot x-4\cdot (-5)-(3x \cdot 2x-3x\cdot 3)\\
		+	&= x^2-5x-4x+20-(6x^2-9x)\\
		+	&=x^2-5x-4x+20-6x^2+9x
		+	\end{align}
		+	</math>
		+
		+	Then, gather together <math>x^2-, \, x- </math> and the constant terms and simplify
		+
		+	<math>
		+	\qquad
		+	\begin{align}
		+	\phantom{(x-4)(x-5)-3x(2x-3)}&= (x^2-6x^2)+(-5x-4x+9x)+20 \\
		+	&= -5x^2+0+20\\
		+	&= -5x^2+20
		+	\end{align}
		+	</math>
		+
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2008-03-31T09:43:38Z

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