2.1 Algebraic expressions - Revision history

Phil: Idiom, localisation of terminology and notation

2008-12-30T17:18:30Z

Idiom, localisation of terminology and notation

←Older revision		Revision as of 17:18, 30 December 2008
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	'''Contents:'''		'''Contents:'''
	* Distributive law		* Distributive law
-	* ~~Squaring rules~~	+	* Expansion and factorisation
	*Difference of two squares		*Difference of two squares
	* Rational expressions		* Rational expressions
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	*Simplify complicated algebraic expressions.		*Simplify complicated algebraic expressions.
-	*Factorise expressions ~~with the use of squaring rules~~ and the difference of two squares ~~rule~~.	+	*Factorise expressions, including perfect squares and the difference of two squares.
-	*Expand expressions ~~with the use of squaring rules~~ and the difference of two squares ~~rule~~.	+	*Expand expressions, including perfect squares and the difference of two squares.
	}}		}}

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-	== Squaring ~~rules~~ ==	+	== Squaring ==

	On occasions the distributive law has to be used repeatedly to deal with larger expressions.		On occasions the distributive law has to be used repeatedly to deal with larger expressions.
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	<div class="regel">		<div class="regel">
-	~~'''Squaring rules '''~~	+
	{{Displayed math\|\|<math>(a+b)^2 = a^2 +2ab + b^2</math>}}		{{Displayed math\|\|<math>(a+b)^2 = a^2 +2ab + b^2</math>}}
	{{Displayed math\|\|<math>(a-b)^2 = a^2 -2ab + b^2</math>}}		{{Displayed math\|\|<math>(a-b)^2 = a^2 -2ab + b^2</math>}}
	</div>		</div>
-
-	These formulas are called the first and second squaring rules.

	<div class="exempel">		<div class="exempel">
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	<ol type="a">		<ol type="a">
-	<li><math>(x+2)^2 = x^2 + 2\~~cdot~~ 2x+ 2^2 = x^2 +4x +4</math></li>	+	<li><math>(x+2)^2 = x^2 + 2\times 2x+ 2^2 = x^2 +4x +4</math></li>
-	<li><math>(-x+3)^2 = (-x)^2 + 2\~~cdot~~ 3(-x) + 3^2 = x^2 -6x +9</math> <br>	+	<li><math>(-x+3)^2 = (-x)^2 + 2\times 3(-x) + 3^2 = x^2 -6x +9</math> <br>
-	: where <math>(-x)^2 = ((-1)x)^2 = (-1)^2 x^2 = 1 \~~cdot~~ x^2 = x^2\,\mbox{.}</math></li>	+	: where <math>(-x)^2 = ((-1)x)^2 = (-1)^2 x^2 = 1 \times x^2 = x^2\,\mbox{.}</math></li>
-	<li><math>(x^2 -4)^2 = (x^2)^2 - 2 \~~cdot~~ 4x^2 + 4^2	+	<li><math>(x^2 -4)^2 = (x^2)^2 - 2 \times 4x^2 + 4^2
	= x^4 -8x^2 +16</math></li>		= x^4 -8x^2 +16</math></li>
	<li><math>(x+1)^2 - (x-1)^2 = (x^2 +2x +1)- (x^2-2x+1)</math><br/>		<li><math>(x+1)^2 - (x-1)^2 = (x^2 +2x +1)- (x^2-2x+1)</math><br/>
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	<math>\phantom{(2x+4)(x+2)}{}=2x^2 + 8x + 8</math></li>		<math>\phantom{(2x+4)(x+2)}{}=2x^2 + 8x + 8</math></li>
	<li><math>(x-2)^3 = (x-2)(x-2)^2 = (x-2)(x^2-4x+4)</math><br/>		<li><math>(x-2)^3 = (x-2)(x-2)^2 = (x-2)(x^2-4x+4)</math><br/>
-	<math>\phantom{(x-2)^3}{}=x \~~cdot~~ x^2 + x\~~cdot~~ (-4x) + x\~~cdot~~ 4	+	<math>\phantom{(x-2)^3}{}=x \times x^2 + x\times (-4x) + x\times 4
-	- 2\~~cdot~~ x^2 - 2 \~~cdot~~ (-4x)-2 \~~cdot~~ 4</math><br/>	+	- 2\times x^2 - 2 \times (-4x)-2 \times 4</math><br/>
	<math>\phantom{(x-2)^3}{}=x^3 -4x^2 + 4x-2x^2 +8x -8		<math>\phantom{(x-2)^3}{}=x^3 -4x^2 + 4x-2x^2 +8x -8
	= x^3-6x^2 + 12x -8</math></li>		= x^3-6x^2 + 12x -8</math></li>
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	</div>		</div>

-	~~The squaring~~ rules are also used in the reverse direction to factorise expressions.	+	These rules are also used in the reverse direction to factorise expressions.

	<div class="exempel">		<div class="exempel">
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	<ol type="a">		<ol type="a">
	<li><math>x^2 + 2x+ 1 = (x+1)^2</math></li>		<li><math>x^2 + 2x+ 1 = (x+1)^2</math></li>
-	<li><math>x^6-4x^3 +4 = (x^3)^2 - 2\~~cdot~~ 2x^3 +2^2 = (x^3-2)^2</math></li>	+	<li><math>x^6-4x^3 +4 = (x^3)^2 - 2\times 2x^3 +2^2 = (x^3-2)^2</math></li>
-	<li><math>x^2 +x + \frac{1}{4} = x^2 + 2\~~cdot~~\frac{1}{2}x	+	<li><math>x^2 +x + \frac{1}{4} = x^2 + 2\times\frac{1}{2}x
	+ \bigl(\frac{1}{2}\bigr)^2 = \bigl(x+\frac{1}{2}\bigr)^2</math></li>		+ \bigl(\frac{1}{2}\bigr)^2 = \bigl(x+\frac{1}{2}\bigr)^2</math></li>
	</ol>		</ol>
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	== Rational expressions==		== Rational expressions==

-	~~Calculations of~~ fractions containing algebraic expressions ~~are largely~~ similar to ordinary calculations with fractions.	+	Working with fractions containing algebraic expressions is very similar to carrying out ordinary calculations with fractions.

-	Multiplication and division of ~~fractions containing~~ algebraic ~~expressions~~ follow the same rules that apply to ordinary fractions,	+	Multiplication and division of algebraic fractions follow the same rules that apply to ordinary fractions,

	<div class="regel">		<div class="regel">
	{{Displayed math\|\|<math> \frac{a}{b} \cdot \frac{c}{d}		{{Displayed math\|\|<math> \frac{a}{b} \cdot \frac{c}{d}
-	= \frac{a\~~cdot~~ c}{b\~~cdot~~ d}	+	= \frac{a\, c}{b\, d}
	\quad \mbox{and} \quad		\quad \mbox{and} \quad
	\frac{\displaystyle\ \frac{a}{b}\ }{\displaystyle\frac{c}{d}}		\frac{\displaystyle\ \frac{a}{b}\ }{\displaystyle\frac{c}{d}}
-	= \frac{a\~~cdot~~ d}{b\~~cdot~~ c} \; \mbox{.}</math>}}	+	= \frac{a\, d}{b\, c} \; \mbox{.}</math>}}
	</div>		</div>

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	<ol type="a">		<ol type="a">
-	<li><math>\frac{3x}{x-y} \~~cdot~~ \frac{4x}{2x+y}	+	<li><math>\frac{3x}{x-y} \times \frac{4x}{2x+y}
-	= \frac{3x\~~cdot~~ 4x}{(x-y)\~~cdot~~(2x+y)}	+	= \frac{3x\times 4x}{(x-y)\times(2x+y)}
	= \frac{12x^2}{(x-y)(2x+y)}</math></li>		= \frac{12x^2}{(x-y)(2x+y)}</math></li>
	<li><math>\frac{\displaystyle \frac{a}{x}}{\displaystyle \frac{x+1}{a}}		<li><math>\frac{\displaystyle \frac{a}{x}}{\displaystyle \frac{x+1}{a}}
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	</div>		</div>

-	A fractional expression can have its numerator and denominator multiplied by the same factor	+	A fractional expression can have its numerator and denominator multiplied by the same factor

	{{Displayed math\|\|<math>\frac{x+2}{x+1}		{{Displayed math\|\|<math>\frac{x+2}{x+1}
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	= \dots</math>}}		= \dots</math>}}

-	~~The opposite of this is cancellation. Here we delete~~ factors that the numerator and denominator have in common	+	We can cancel factors that the numerator and denominator have in common
	{{Displayed math\|\|<math>\frac{(x+2)(x+3)(x+4)}{(x+1)(x+3)(x+4) }		{{Displayed math\|\|<math>\frac{(x+2)(x+3)(x+4)}{(x+1)(x+3)(x+4) }
	= \frac{(x+2)(x+4)}{(x+1)(x+4)}		= \frac{(x+2)(x+4)}{(x+1)(x+4)}
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	</div>		</div>

-	When fractional expressions are added or subtracted they may need to be ~~converted so that they have the same~~ denominator. ~~Only then can the numerators be combined together.~~	+	When fractional expressions are added or subtracted they may need to be placed over a common denominator.
-		+

	{{Displayed math\|\|<math>\frac{1}{x} - \frac{1}{x-1}		{{Displayed math\|\|<math>\frac{1}{x} - \frac{1}{x-1}
-	= \frac{1}{x} \~~cdot~~ \frac{x-1}{x-1} - \frac{1}{x-1} \~~cdot~~ \frac{x}{x}	+	= \frac{1}{x} \times\frac{x-1}{x-1} - \frac{1}{x-1} \times\frac{x}{x}
	= \frac{x-1}{x(x-1)} - \frac{x}{x(x-1)}		= \frac{x-1}{x(x-1)} - \frac{x}{x(x-1)}
	= \frac{x-1-x}{x(x-1)}		= \frac{x-1-x}{x(x-1)}
	= \frac{-1}{x(x-1)} \; \mbox{.}</math>}}		= \frac{-1}{x(x-1)} \; \mbox{.}</math>}}

-	~~One normally tries~~ to ~~convert the fractions~~ by multiplying the ~~numerators and~~ denominators ~~by minimal factors~~ to ~~facilitate~~ the ~~calculations. The~~ lowest common denominator (LCD~~) is the common denominator which contains the least number of factors~~.	+	As with ordinary fractions, it's possible to find a common denominator by simply multiplying the two denominators, but it is better to find the smallest, simplest expression that both denominators go into: the lowest common denominator or LCD.

	<div class="exempel">		<div class="exempel">
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	</div>		</div>

-	To simplify large expressions it is often necessary to both cancel factors and multiply numerators and denominators by factors. ~~As cancellation implies that we have performed factorisations it is obvious~~ we should ~~try to~~ keep expressions ~~(such as the denominator)~~ factorised and not expand ~~something that~~ we will later need to factorise.	+	To simplify large expressions it is often necessary both to cancel factors and to multiply numerators and denominators by factors. Where possible, therefore, we should keep expressions in a factorised form, and not expand expressions we will later need to factorise again.
-		+

	<div class="exempel">		<div class="exempel">

Phil: Partial edit: more to do

2008-12-30T16:28:27Z

Partial edit: more to do

←Older revision		Revision as of 16:28, 30 December 2008
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	After this section you will have learned how to:		After this section you will have learned how to:

-	*Simplify ~~complex~~ algebraic expressions.	+	*Simplify complicated algebraic expressions.
	*Factorise expressions with the use of squaring rules and the difference of two squares rule.		*Factorise expressions with the use of squaring rules and the difference of two squares rule.
	*Expand expressions with the use of squaring rules and the difference of two squares rule.		*Expand expressions with the use of squaring rules and the difference of two squares rule.
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	<li><math>2(a-b) = 2a -2b</math></li>		<li><math>2(a-b) = 2a -2b</math></li>
	<li><math>x \left(\frac{1}{x} + \frac{1}{x^2} \right)		<li><math>x \left(\frac{1}{x} + \frac{1}{x^2} \right)
-	= x\~~cdot~~ \frac{1}{x} + x \~~cdot~~ \frac{1}{x^2}	+	= x\times \frac{1}{x} + x \times \frac{1}{x^2}
	= \frac{\not{x}}{\not{x}} + \frac{\not{x}}{x^{\not{2}}}		= \frac{\not{x}}{\not{x}} + \frac{\not{x}}{x^{\not{2}}}
	= 1 + \frac{1}{x}</math></li>		= 1 + \frac{1}{x}</math></li>
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	Using the distributive law we can also see how to tackle		Using the distributive law we can also see how to tackle
	a minus sign in front of a bracketed expression.		a minus sign in front of a bracketed expression.
-	The rule says that a minus sign ~~infront~~ of a bracket can be	+	The rule says that a minus sign in front of a bracket can be
	eliminated if all the terms inside the brackets switch signs.		eliminated if all the terms inside the brackets switch signs.

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	<ol type="a">		<ol type="a">
-	<li><math>-(x+y) = (-1) \~~cdot~~ (x+y) = (-1)x + (-1)y = -x-y</math></li>	+	<li><math>-(x+y) = (-1) \times (x+y) = (-1)x + (-1)y = -x-y</math></li>
-	<li><math>-(x^2-x) = (-1) \~~cdot~~ (x^2-x) = (-1)x^2 -(-1)x	+	<li><math>-(x^2-x) = (-1) \times (x^2-x) = (-1)x^2 -(-1)x
	= -x^2 +x</math><br/>		= -x^2 +x</math><br/>
-	where we have in the final step used <math>-(-1)x = (-1)(-1)x = 1\~~cdot~~ x = x\,\mbox{.}</math></li>	+	where we have in the final step used <math>-(-1)x = (-1)(-1)x = 1\times x = x\,\mbox{.}</math></li>
-	<li><math>-(x+y-y^3) = (-1)\~~cdot~~ (x+y-y^3) = (-1)\~~cdot~~ x	+	<li><math>-(x+y-y^3) = (-1)\times (x+y-y^3) = (-1)\times x
-	+ (-1) \~~cdot~~ y -(-1)\~~cdot~~ y^3</math><br/>	+	+ (-1) \times y -(-1)\times y^3</math><br/>
	<math>\phantom{-(x+y-y^3)}{} = -x-y+y^3</math></li>		<math>\phantom{-(x+y-y^3)}{} = -x-y+y^3</math></li>
	<li><math>x^2 - 2x -(3x+2) = x^2 -2x -3x-2 = x^2 -(2+3)x -2</math><br/>		<li><math>x^2 - 2x -(3x+2) = x^2 -2x -3x-2 = x^2 -(2+3)x -2</math><br/>
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	</div>		</div>

-	If the distributive law is applied in reverse we say we “factorise” the expression.	+	Applying the distributive law this way round - converting a product of factors into a sum of terms - is called ''expanding''. If the distributive law is applied in reverse we say we “factorise” the expression. We usually want to factorise as thoroughly as possible, by
-	~~One would often like~~ to factorise ~~out the largest~~ possible ~~numerical~~ factor.	+	identifying the highet factor shared by all the terms.

	<div class="exempel">		<div class="exempel">
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	<ol type="a">		<ol type="a">
-	<li><math>3x +9y = 3x + 3\~~cdot~~ 3y = 3(x+3y)</math></li>	+	<li><math>3x +9y = 3x + 3\times 3y = 3(x+3y)</math></li>
-	<li><math>xy + y^2 = xy + y\~~cdot~~ y = y(x+y)</math></li>	+	<li><math>xy + y^2 = xy + y\times y = y(x+y)</math></li>
-	<li><math>2x^2 -4x = 2x\~~cdot~~ x - 2\~~cdot~~ 2\~~cdot~~ x = 2x(x-2)</math></li>	+	<li><math>2x^2 -4x = 2x\times x - 2\times 2\times x = 2x(x-2)</math></li>
	<li><math>\frac{y-x}{x-y} = \frac{-(x-y)}{x-y} = \frac{-1}{1} = -1</math></li>		<li><math>\frac{y-x}{x-y} = \frac{-(x-y)}{x-y} = \frac{-1}{1} = -1</math></li>
	</ol>		</ol>
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	<ol type="a">		<ol type="a">
-	<li><math>(x+1)(x-2) = x\~~cdot~~ x + x \~~cdot~~ (-2) + 1 \~~cdot~~ x + 1 \~~cdot~~ (-2)	+	<li><math>(x+1)(x-2) = x\times x + x \times (-2) + 1 \times x + 1 \times (-2)
	= x^2 -2x+x-2</math><br/>		= x^2 -2x+x-2</math><br/>
	<math>\phantom{(x+1)(x-2)}{}=x^2 -x-2</math></li>		<math>\phantom{(x+1)(x-2)}{}=x^2 -x-2</math></li>
-	<li><math>3(x-y)(2x+1) = 3(x\~~cdot~~ 2x + x\~~cdot~~ 1 - y \~~cdot~~ 2x - y \~~cdot~~ 1)	+	<li><math>3(x-y)(2x+1) = 3(x\times 2x + x\times 1 - y \times 2x - y \times 1)
	= 3(2x^2 +x-2xy-y)</math><br/>		= 3(2x^2 +x-2xy-y)</math><br/>
	<math>\phantom{3(x-y)(2x+1)}{}=6x^2 +3x-6xy-3y</math></li>		<math>\phantom{3(x-y)(2x+1)}{}=6x^2 +3x-6xy-3y</math></li>
-	<li><math>(1-x)(2-x) = 1\~~cdot~~ 2 + 1 \~~cdot~~ (-x) -x\~~cdot~~ 2 - x\~~cdot~~ (-x)	+	<li><math>(1-x)(2-x) = 1\times 2 + 1 \times (-x) -x\times 2 - x\times (-x)
	= 2-x-2x+x^2</math><br/>		= 2-x-2x+x^2</math><br/>
	<math>\phantom{(1-x)(2-x)}{}=2-3x+x^2</math>		<math>\phantom{(1-x)(2-x)}{}=2-3x+x^2</math>
-	where we have used <math>-x\~~cdot~~ (-x) = (-1)x \~~cdot~~ (-1)x = (-1)^2 x^2 = 1\~~cdot~~ x^2 = x^2</math>.	+	where we have used <math>-x\times (-x) = (-1)x \times (-1)x = (-1)^2 x^2 = 1\times x^2 = x^2</math>.
	</ol>		</ol>
	</div>		</div>

Eakc07: Minor grammatical changes

2008-10-31T16:44:07Z

Minor grammatical changes

←Older revision		Revision as of 16:44, 31 October 2008
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	'''Learning outcomes:'''		'''Learning outcomes:'''

-	After this section, you will have learned how to:	+	After this section you will have learned how to:

	*Simplify complex algebraic expressions.		*Simplify complex algebraic expressions.
-	*Factorise expressions ~~using~~ squaring rules and the difference of two squares rule.	+	*Factorise expressions with the use of squaring rules and the difference of two squares rule.
-	*Expand expressions ~~using~~ squaring rules and the difference of two squares rule.	+	*Expand expressions with the use of squaring rules and the difference of two squares rule.
	}}		}}

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	</div>		</div>

-	Using the distributive law, we can also see how to tackle	+	Using the distributive law we can also see how to tackle
	a minus sign in front of a bracketed expression.		a minus sign in front of a bracketed expression.
-	The rule says that ~~a bracket with~~ a minus sign ~~in front~~ can be	+	The rule says that a minus sign infront of a bracket can be
-	eliminated if all the terms inside the brackets, switch signs.	+	eliminated if all the terms inside the brackets switch signs.

	<div class="exempel">		<div class="exempel">
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	</div>		</div>

-	If the distributive law is applied in reverse we say we ~~“factor”~~ the expression.	+	If the distributive law is applied in reverse we say we “factorise” the expression.
-	One ~~often~~ would like to to factorise out ~~as large a~~ numerical factor ~~as possible~~.	+	One would often like to factorise out the largest possible numerical factor.

	<div class="exempel">		<div class="exempel">
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	== Squaring rules ==		== Squaring rules ==

-	~~The~~ distributive law ~~occasionally~~ has to be used repeatedly to deal with larger expressions.	+	On occasions the distributive law has to be used repeatedly to deal with larger expressions.
	If we consider		If we consider

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	</div>		</div>

-	These formulas are called the first and second squaring rules	+	These formulas are called the first and second squaring rules.

	<div class="exempel">		<div class="exempel">
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	= \dots</math>}}		= \dots</math>}}

-	The opposite of this, is cancellation~~, where~~ we delete factors that the numerator and denominator have in common	+	The opposite of this is cancellation. Here we delete factors that the numerator and denominator have in common
	{{Displayed math\|\|<math>\frac{(x+2)(x+3)(x+4)}{(x+1)(x+3)(x+4) }		{{Displayed math\|\|<math>\frac{(x+2)(x+3)(x+4)}{(x+1)(x+3)(x+4) }
	= \frac{(x+2)(x+4)}{(x+1)(x+4)}		= \frac{(x+2)(x+4)}{(x+1)(x+4)}
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	</div>		</div>

-	When fractional expressions are added or subtracted, they may need to be converted so that they have the same denominator ~~before~~ the numerators ~~can~~ be combined together,	+	When fractional expressions are added or subtracted they may need to be converted so that they have the same denominator. Only then can the numerators be combined together.


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	= \frac{-1}{x(x-1)} \; \mbox{.}</math>}}		= \frac{-1}{x(x-1)} \; \mbox{.}</math>}}

-	One normally tries to convert the fractions by multiplying the numerators and denominators by minimal factors to facilitate the calculations. The lowest common denominator (LCD) is the common denominator which contains the least number of factors.	+	One normally tries to convert the fractions by multiplying the numerators and denominators by minimal factors to facilitate the calculations. The lowest common denominator (LCD) is the common denominator which contains the least number of factors.

	<div class="exempel">		<div class="exempel">
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	</div>		</div>

-	To simplify large expressions, it is often necessary to both cancel factors and multiply numerators and denominators by factors. As cancellation implies that we have performed factorisations, it is obvious we should try to keep expressions (such as the denominator) factorised and not expand something that we will later need to factorise.	+	To simplify large expressions it is often necessary to both cancel factors and multiply numerators and denominators by factors. As cancellation implies that we have performed factorisations it is obvious we should try to keep expressions (such as the denominator) factorised and not expand something that we will later need to factorise.

Jdinglis at 13:58, 26 October 2008

2008-10-26T13:58:36Z

←Older revision		Revision as of 13:58, 26 October 2008
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-	'''Keep in mind that: '''	+	'''Keep in mind that...'''

-	~~Be careful.~~ If you make a mistake somewhere the rest of the calculation will be wrong.	+	If you make a mistake somewhere the rest of the calculation will be wrong, so be careful!

	Use many intermediate steps. If you are unsure of a calculation do it in many small steps rather than one big step.		Use many intermediate steps. If you are unsure of a calculation do it in many small steps rather than one big step.

Tekbot: Robot: Automated text replacement (-{{:2.1 - Figur - Distributiva lagen två gånger}} +{{:2.1 - Figure - The distributive law twice}})

2008-09-12T12:19:33Z

Robot: Automated text replacement (-{{:2.1 - Figur - Distributiva lagen två gånger}} +{{:2.1 - Figure - The distributive law twice}})

←Older revision		Revision as of 12:19, 12 September 2008
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	A mnemonic for this formula is:		A mnemonic for this formula is:

-	<center>{{:2.1 - ~~Figur~~ - ~~Distributiva lagen två gånger~~}}</center>	+	<center>{{:2.1 - Figure - The distributive law twice}}</center>

	<div class="exempel">		<div class="exempel">

Tekbot: Robot: Automated text replacement (-{{:2.1 - Figur - Distributiva lagen}} +{{:2.1 - Figure - The distributive law}})

2008-09-12T12:18:27Z

Robot: Automated text replacement (-{{:2.1 - Figur - Distributiva lagen}} +{{:2.1 - Figure - The distributive law}})

←Older revision		Revision as of 12:18, 12 September 2008
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	The distributive law specifies how to multiply a bracketed expression by a factor.		The distributive law specifies how to multiply a bracketed expression by a factor.

-	<center>{{:2.1 - ~~Figur~~ - ~~Distributiva lagen~~}}</center>	+	<center>{{:2.1 - Figure - The distributive law}}</center>

	<div class="exempel">		<div class="exempel">

Tekbot: Robot: Automated text replacement (-{{Fristående formel +{{Displayed math)

2008-09-10T13:54:19Z

Robot: Automated text replacement (-{{Fristående formel +{{Displayed math)

←Older revision		Revision as of 13:54, 10 September 2008
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	If we consider		If we consider

-	{{~~Fristående formel~~\|\|<math>(a+b)(c+d)</math>}}	+	{{Displayed math\|\|<math>(a+b)(c+d)</math>}}

	and regard <math>a+b</math> as a factor that multiplies the bracketed expression <math>(c+d)</math> we get		and regard <math>a+b</math> as a factor that multiplies the bracketed expression <math>(c+d)</math> we get

-	{{~~Fristående formel~~\|\|<math>\eqalign{	+	{{Displayed math\|\|<math>\eqalign{
	\bbox[#AAEEFF,0pt]{\phantom{(a+b)}}\,(c+d)		\bbox[#AAEEFF,0pt]{\phantom{(a+b)}}\,(c+d)
	&= \bbox[#AAEEFF,0pt]{\phantom{(a+b)}}\,c		&= \bbox[#AAEEFF,0pt]{\phantom{(a+b)}}\,c
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	Then the <math>c</math> and the <math>d</math> are multiplied into their respective brackets,		Then the <math>c</math> and the <math>d</math> are multiplied into their respective brackets,

-	{{~~Fristående formel~~\|\|<math>(a+b)c + (a+b)d = ac + bc + ad + bd \, \mbox{.}</math>}}	+	{{Displayed math\|\|<math>(a+b)c + (a+b)d = ac + bc + ad + bd \, \mbox{.}</math>}}

	A mnemonic for this formula is:		A mnemonic for this formula is:
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	<div class="regel">		<div class="regel">
	'''Squaring rules '''		'''Squaring rules '''
-	{{~~Fristående formel~~\|\|<math>(a+b)^2 = a^2 +2ab + b^2</math>}}	+	{{Displayed math\|\|<math>(a+b)^2 = a^2 +2ab + b^2</math>}}
-	{{~~Fristående formel~~\|\|<math>(a-b)^2 = a^2 -2ab + b^2</math>}}	+	{{Displayed math\|\|<math>(a-b)^2 = a^2 -2ab + b^2</math>}}
	</div>		</div>

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	<div class="regel">		<div class="regel">
	'''Difference of two squares:'''		'''Difference of two squares:'''
-	{{~~Fristående formel~~\|\|<math>(a+b)(a-b) = a^2 -b^2</math>}}	+	{{Displayed math\|\|<math>(a+b)(a-b) = a^2 -b^2</math>}}
	</div>		</div>

	This formula can be obtained directly by expanding the left hand side		This formula can be obtained directly by expanding the left hand side

-	{{~~Fristående formel~~\|\|<math>(a+b)(a-b)	+	{{Displayed math\|\|<math>(a+b)(a-b)
	= a \cdot a + a\cdot (-b) + b\cdot a + b \cdot (-b)		= a \cdot a + a\cdot (-b) + b\cdot a + b \cdot (-b)
	= a^2 -ab+ab-b^2		= a^2 -ab+ab-b^2
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	<div class="regel">		<div class="regel">
-	{{~~Fristående formel~~\|\|<math> \frac{a}{b} \cdot \frac{c}{d}	+	{{Displayed math\|\|<math> \frac{a}{b} \cdot \frac{c}{d}
	= \frac{a\cdot c}{b\cdot d}		= \frac{a\cdot c}{b\cdot d}
	\quad \mbox{and} \quad		\quad \mbox{and} \quad
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	A fractional expression can have its numerator and denominator multiplied by the same factor		A fractional expression can have its numerator and denominator multiplied by the same factor

-	{{~~Fristående formel~~\|\|<math>\frac{x+2}{x+1}	+	{{Displayed math\|\|<math>\frac{x+2}{x+1}
	= \frac{(x+2)(x+3)}{(x+1)(x+3)}		= \frac{(x+2)(x+3)}{(x+1)(x+3)}
	= \frac{(x+2)(x+3)(x+4)}{(x+1)(x+3)(x+4)}		= \frac{(x+2)(x+3)(x+4)}{(x+1)(x+3)(x+4)}
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	The opposite of this, is cancellation, where we delete factors that the numerator and denominator have in common		The opposite of this, is cancellation, where we delete factors that the numerator and denominator have in common
-	{{~~Fristående formel~~\|\|<math>\frac{(x+2)(x+3)(x+4)}{(x+1)(x+3)(x+4) }	+	{{Displayed math\|\|<math>\frac{(x+2)(x+3)(x+4)}{(x+1)(x+3)(x+4) }
	= \frac{(x+2)(x+4)}{(x+1)(x+4)}		= \frac{(x+2)(x+4)}{(x+1)(x+4)}
	= \frac{x+2}{x+1} \mbox{.}</math>}}		= \frac{x+2}{x+1} \mbox{.}</math>}}
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-	{{~~Fristående formel~~\|\|<math>\frac{1}{x} - \frac{1}{x-1}	+	{{Displayed math\|\|<math>\frac{1}{x} - \frac{1}{x-1}
	= \frac{1}{x} \cdot \frac{x-1}{x-1} - \frac{1}{x-1} \cdot \frac{x}{x}		= \frac{1}{x} \cdot \frac{x-1}{x-1} - \frac{1}{x-1} \cdot \frac{x}{x}
	= \frac{x-1}{x(x-1)} - \frac{x}{x(x-1)}		= \frac{x-1}{x(x-1)} - \frac{x}{x(x-1)}
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	= (x+1)(x+2)</math> <br><br>		= (x+1)(x+2)</math> <br><br>
	Convert the first term using <math>(x+2)</math> and the second term using <math>(x+1)</math>		Convert the first term using <math>(x+2)</math> and the second term using <math>(x+1)</math>
-	{{~~Fristående formel~~\|\|<math>\begin{align*}	+	{{Displayed math\|\|<math>\begin{align*}
	\frac{1}{x+1} + \frac{1}{x+2}		\frac{1}{x+1} + \frac{1}{x+2}
	&= \frac{x+2}{(x+1)(x+2)} + \frac{x+1}{(x+2)(x+1)}\\[4pt]		&= \frac{x+2}{(x+1)(x+2)} + \frac{x+1}{(x+2)(x+1)}\\[4pt]
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	= x^2</math><br><br>		= x^2</math><br><br>
	We only need to convert the first term to get a common denominator		We only need to convert the first term to get a common denominator
-	{{~~Fristående formel~~\|\|<math>\frac{1}{x} + \frac{1}{x^2}	+	{{Displayed math\|\|<math>\frac{1}{x} + \frac{1}{x^2}
	= \frac{x}{x^2} + \frac{1}{x^2}		= \frac{x}{x^2} + \frac{1}{x^2}
	= \frac{x+1}{x^2}\,\mbox{.}</math>}}</li>		= \frac{x+1}{x^2}\,\mbox{.}</math>}}</li>
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	\text{LCD}= x^2(x+1)^2(x+2)</math><br><br>		\text{LCD}= x^2(x+1)^2(x+2)</math><br><br>
	The first term is converted using <math>x(x+2)</math> while the other term is converted using <math>(x+1)^2</math>		The first term is converted using <math>x(x+2)</math> while the other term is converted using <math>(x+1)^2</math>
-	{{~~Fristående formel~~\|\|<math>\begin{align*}	+	{{Displayed math\|\|<math>\begin{align*}
	\frac{1}{x(x+1)^2} - \frac{1}{x^2(x+2)}		\frac{1}{x(x+1)^2} - \frac{1}{x^2(x+2)}
	&= \frac{x(x+2)}{x^2(x+1)^2(x+2)}		&= \frac{x(x+2)}{x^2(x+1)^2(x+2)}
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	\text{LCD}=x(x-1)(x+1)</math><br><br>		\text{LCD}=x(x-1)(x+1)</math><br><br>
	We must convert all the terms so that they have the common denominator <math>x(x-1)(x+1)</math>		We must convert all the terms so that they have the common denominator <math>x(x-1)(x+1)</math>
-	{{~~Fristående formel~~\|\|<math>\begin{align*}	+	{{Displayed math\|\|<math>\begin{align*}
	\frac{x}{x+1} - \frac{1}{x(x-1)} -1		\frac{x}{x+1} - \frac{1}{x(x-1)} -1
	&= \frac{x^2(x-1)}{x(x-1)(x+1)} - \frac{x+1}{x(x-1)(x+1)}		&= \frac{x^2(x-1)}{x(x-1)(x+1)} - \frac{x+1}{x(x-1)(x+1)}

Tekbot: Robot: Automated text replacement (-{{Vald flik +{{Selected tab)

2008-09-10T13:49:25Z

Robot: Automated text replacement (-{{Vald flik +{{Selected tab)

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-	{{~~Vald flik~~\|[[2.1 Algebraic expressions\|Theory]]}}	+	{{Selected tab\|[[2.1 Algebraic expressions\|Theory]]}}
	{{Not selected tab\|[[2.1 Exercises\|Exercises]]}}		{{Not selected tab\|[[2.1 Exercises\|Exercises]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|

Tekbot: Robot: Automated text replacement (-{{Ej vald flik +{{Not selected tab)

2008-09-10T13:44:02Z

Robot: Automated text replacement (-{{Ej vald flik +{{Not selected tab)

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	\| style="border-bottom:1px solid #797979" width="5px" \|		\| style="border-bottom:1px solid #797979" width="5px" \|
	{{Vald flik\|[[2.1 Algebraic expressions\|Theory]]}}		{{Vald flik\|[[2.1 Algebraic expressions\|Theory]]}}
-	{{~~Ej vald flik~~\|[[2.1 Exercises\|Exercises]]}}	+	{{Not selected tab\|[[2.1 Exercises\|Exercises]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
	\|}		\|}

Tekbot: Robot: Automated text replacement (-2.1 Algebraiska uttryck +2.1 Algebraic expressions)

2008-09-10T13:13:51Z

Robot: Automated text replacement (-2.1 Algebraiska uttryck +2.1 Algebraic expressions)

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	\| style="border-bottom:1px solid #797979" width="5px" \|		\| style="border-bottom:1px solid #797979" width="5px" \|
-	{{Vald flik\|[[2.1 ~~Algebraiska uttryck~~\|Theory]]}}	+	{{Vald flik\|[[2.1 Algebraic expressions\|Theory]]}}
	{{Ej vald flik\|[[2.1 Exercises\|Exercises]]}}		{{Ej vald flik\|[[2.1 Exercises\|Exercises]]}}
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	\| style="border-bottom:1px solid #797979" width="5px" \|		\| style="border-bottom:1px solid #797979" width="5px" \|
-	{{~~Vald flik~~\|[[2.1 Algebraic expressions\|Theory]]}}	+	{{Selected tab\|[[2.1 Algebraic expressions\|Theory]]}}
	{{Not selected tab\|[[2.1 Exercises\|Exercises]]}}		{{Not selected tab\|[[2.1 Exercises\|Exercises]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|

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	{{Vald flik\|[[2.1 Algebraic expressions\|Theory]]}}		{{Vald flik\|[[2.1 Algebraic expressions\|Theory]]}}
-	{{~~Ej vald flik~~\|[[2.1 Exercises\|Exercises]]}}	+	{{Not selected tab\|[[2.1 Exercises\|Exercises]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
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	\| style="border-bottom:1px solid #797979" width="5px" \|		\| style="border-bottom:1px solid #797979" width="5px" \|
-	{{Vald flik\|[[2.1 ~~Algebraiska uttryck~~\|Theory]]}}	+	{{Vald flik\|[[2.1 Algebraic expressions\|Theory]]}}
	{{Ej vald flik\|[[2.1 Exercises\|Exercises]]}}		{{Ej vald flik\|[[2.1 Exercises\|Exercises]]}}
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