Solution 2.1:5d - Revision history

Tek at 11:09, 23 September 2008

2008-09-23T11:09:41Z

←Older revision		Revision as of 11:09, 23 September 2008
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	In this fraction, there is the possibility that the numerator and denominator contain common factors which can be eliminated and we therefore try to factorize all expressions to simplest possible form.		In this fraction, there is the possibility that the numerator and denominator contain common factors which can be eliminated and we therefore try to factorize all expressions to simplest possible form.

-	The factor	+	The factor <math>y^{2}+4y+4</math> can be rewritten as <math>y^{2}+2\cdot 2y+2^{2}</math>, which opens the way for using the squaring rule
-	<math>y^{2}+4y+4</math>	+
-	can be rewritten as	+
-	<math>y^{2}+2\~~centerdot~~ 2y+2^{2}</math>, which opens the way for using the squaring rule	+

		+	{{Displayed math\|\|<math>y^{2}+4y+4 = y^{2}+2\cdot 2y+2^{2} = (y+2)^{2}\textrm{.}</math>}}

-	<math>~~y^{~~2~~}+4y+~~4=~~y^{2}+~~2~~\centerdot 2y+2^{2}=~~\left( y+2 \right)~~^{2}~~</math>	+	The factor <math>2y-4</math> is already a first-order expression and can therefore not be divided up any further, other than by taking out a factor <math>2</math>, i.e. <math>2y-4=2\left( y-2 \right)\,</math>.


-	~~The factor~~	+	<math>y^{2}-4</math> can be factorized using the conjugate rule to give
-	<math>2y-4</math>	+
-	~~is already a first-order expression and~~ can ~~therefore not~~ be ~~divided up any further,~~	+
-	~~other than by taking out a factor~~	+
-	~~<math>2</math>~~	+
-	~~i.e.~~	+
-	~~<math>2y-4=2\left( y-2 \right)</math>~~	+
-	.	+

		+	{{Displayed math\|\|<math>y^{2}-4 = (y+2)(y-2)\,\textrm{.}</math>}}

-	~~<math>y^{2}+4</math>~~	+	On the other hand, <math>y^{2}+4</math> cannot be written as a product of first-order factors. If it were possible to write <math>y^{2}+4 = (y-a)(y-b)</math>, where ''a'' and ''b'' are some numbers, then <math>y=a</math> and
-	~~y2-4 can be factorized using the conjugate rule to give~~	+	<math>y=b</math> would be zeros of <math>y^{2}+4</math>, but because
-		+	<math>y^{2}+4</math> is the sum of a square, <math>y^{2}</math>, which cannot have a negative value and the number <math>4</math>, <math>y^{2}+4</math> is always greater than or equal <math>4</math> regardless of how <math>y</math> is chosen. Hence, <math>y^{2}+4</math> cannot be divided up into first-order factors.
-		+
-	~~<math>y^{2}-4=\left( y+2 \right)\left( y-2 \right)</math>~~	+
-		+
-		+
-	On the other hand,	+
-	<math>y^{2}+4</math>	+
-	cannot be written as a product of first-order factors. If it were possible to	+
-	write	+
-	<math>y^{2}+4=~~\left~~( y-a ~~\right~~)~~\left~~( y-b ~~\right~~)</math>, where	+
-	~~<math>~~a~~</math>~~	+
-	and	+
-	~~<math>~~b~~</math>~~	+
-	are some numbers, then	+
-	<math>y=a</math>	+
-	and	+
-	<math>y=b</math>	+
-	would be zeros of	+
-	<math>y^{2}+4</math>, but because	+
-	<math>y^{2}+4</math>	+
-	is the sum of a square,	+
-	<math>y^{2}</math>, which cannot have a negative value and the number	+
-	<math>4</math>,	+
-	<math>y^{2}+4</math>	+
-	is always greater than or equal	+
-	<math>4</math>	+
-	regardless of how	+
-	<math>y</math>	+
-	is chosen. Hence,	+
-	<math>y^{2}+4</math>	+
-	cannot be divided up into first-order factors.	+

	Thus,		Thus,

-		+	{{Displayed math\|\|<math>\frac{(y^{2}+4y+4)(2y-4)}{(y^{2}+4)(y^{2}-4)} = \frac{(y+2)^{2}\cdot 2(y-2)}{(y^{2}+4)(y+2)(y-2)} = \frac{2(y+2)}{(y^{2}+4)}\,\textrm{.}</math>}}
-	<math>\frac{~~\left~~( y^{2}+4y+4 ~~\right~~)~~\left~~( 2y-4 ~~\right~~)}{~~\left~~( y^{2}+4 ~~\right~~)~~\left~~( y^{2}-4 ~~\right~~)}=\frac{~~\left~~( y+2 ~~\right~~)^{2}\~~centerdot~~ 2~~\left~~( y-2 ~~\right~~)}{~~\left~~( y^{2}+4 ~~\right~~)~~\left~~( y+2 ~~\right~~)~~\left~~( y-2 ~~\right~~)}=\frac{2~~\left~~( y+2 ~~\right~~)}{~~\left~~( y^{2}+4 ~~\right~~)}</math>	+

Ian at 09:55, 16 September 2008

2008-09-16T09:55:58Z

←Older revision		Revision as of 09:55, 16 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	In this fraction, there is the possibility that the numerator and denominator contain common factors which can be eliminated and we therefore try to factorize all expressions to simplest possible form.
-	<~~center~~> ~~[[Image:2_1_5d~~-1(2).~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	The factor
-	{{~~NAVCONTENT_START~~}}	+	<math>y^{2}+4y+4</math>
-	<~~center~~> ~~[[Image:2_1_5d-2(~~2).~~gif]]~~ </~~center~~>	+	can be rewritten as
-	{{~~NAVCONTENT_STOP~~}}	+	<math>y^{2}+2\centerdot 2y+2^{2}</math>, which opens the way for using the squaring rule
		+
		+
		+	<math>y^{2}+4y+4=y^{2}+2\centerdot 2y+2^{2}=\left( y+2 \right)^{2}</math>
		+
		+
		+	The factor
		+	<math>2y-4</math>
		+	is already a first-order expression and can therefore not be divided up any further,
		+	other than by taking out a factor
		+	<math>2</math>
		+	i.e.
		+	<math>2y-4=2\left( y-2 \right)</math>
		+	.
		+
		+
		+	<math>y^{2}+4</math>
		+	y2-4 can be factorized using the conjugate rule to give
		+
		+
		+	<math>y^{2}-4=\left( y+2 \right)\left( y-2 \right)</math>
		+
		+
		+	On the other hand,
		+	<math>y^{2}+4</math>
		+	cannot be written as a product of first-order factors. If it were possible to
		+	write
		+	<math>y^{2}+4=\left( y-a \right)\left( y-b \right)</math>, where
		+	<math>a</math>
		+	and
		+	<math>b</math>
		+	are some numbers, then
		+	<math>y=a</math>
		+	and
		+	<math>y=b</math>
		+	would be zeros of
		+	<math>y^{2}+4</math>, but because
		+	<math>y^{2}+4</math>
		+	is the sum of a square,
		+	<math>y^{2}</math>, which cannot have a negative value and the number
		+	<math>4</math>,
		+	<math>y^{2}+4</math>
		+	is always greater than or equal
		+	<math>4</math>
		+	regardless of how
		+	<math>y</math>
		+	is chosen. Hence,
		+	<math>y^{2}+4</math>
		+	cannot be divided up into first-order factors.
		+
		+	Thus,
		+
		+
		+	<math>\frac{\left( y^{2}+4y+4 \right)\left( 2y-4 \right)}{\left( y^{2}+4 \right)\left( y^{2}-4 \right)}=\frac{\left( y+2 \right)^{2}\centerdot 2\left( y-2 \right)}{\left( y^{2}+4 \right)\left( y+2 \right)\left( y-2 \right)}=\frac{2\left( y+2 \right)}{\left( y^{2}+4 \right)}</math>

Tekbot: Lösning 2.1:5d moved to Solution 2.1:5d: Robot: moved page

2008-09-09T08:38:43Z

Lösning 2.1:5d moved to Solution 2.1:5d: Robot: moved page

←Older revision

Revision as of 08:38, 9 September 2008

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

2008-08-21T06:36:29Z

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

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Lina at 11:11, 31 March 2008

2008-03-31T11:11:27Z

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Lina at 11:11, 31 March 2008

2008-03-31T11:11:00Z

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Lina: Ny sida: {{NAVCONTENT_START}} Bild:2_1_5d-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:2_1_5d-1(2).gif {{NAVCONTENT_STOP}}

2008-03-31T11:08:55Z

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