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__NOTOC__
{| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
| style="border-bottom:1px solid #797979" width="5px" | &nbsp;
{{Vald flik|[[2.1algebraic expressions|Theory]]}}
{{Ej vald flik|[[2.1 Övningar|Exercises]]}}
| style="border-bottom:1px solid #797979"  width="100%"| &nbsp;
|}

{{Info|
'''Content:'''
* Distributive law
* Squaring rules
*Difference of two squares 
* Rational expression 
}}

{{Info|
'''Learning outcomes:'''

After this section, you will have learned how to:

*Simplify  complex algebraic expression. 
*Factorise expressions  using squaring rules and  and the difference of two squares rule.
*Expand expressions using squaring rules and  and the difference of two squares rule.
}}

== Distributive Law ==

[[Bild:miniräknare_skämt.gif|right]]
The distributive law specifies how to multiply  a bracketed expression by a factor. 

<center>{{:2.1 - Figur - Distributiva lagen}}</center> 

<div class="exempel">
''' Example 1'''

<ol type="a">
  <li><math>4(x+y) = 4x + 4y</math></li>
  <li><math>2(a-b) = 2a -2b</math></li>
  <li><math>x \left(\frac{1}{x} + \frac{1}{x^2} \right)
    = x\cdot \frac{1}{x} + x \cdot \frac{1}{x^2}
    = \frac{\not{x}}{\not{x}} + \frac{\not{x}}{x^{\not{2}}}
    = 1 + \frac{1}{x}</math></li>
  <li><math>a(x+y+z) = ax + ay + az</math></li>
</ol>

</div>

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

<div class="exempel">
''' Example 2'''

<ol type="a">
  <li><math>-(x+y) = (-1) \cdot (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
    = -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>
  <li><math>-(x+y-y^3) = (-1)\cdot (x+y-y^3) = (-1)\cdot x
    + (-1) \cdot y -(-1)\cdot y^3</math><br/>
<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/>
<math>\phantom{x^2-2x-(3x+2)}{} = x^2 -5x -2</math></li>
</ol>
</div>

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

<div class="exempel">
''' Example 3'''

<ol type="a">
  <li><math>3x +9y = 3x + 3\cdot 3y = 3(x+3y)</math></li>
  <li><math>xy + y^2 = xy + y\cdot 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>\frac{y-x}{x-y} = \frac{-(x-y)}{x-y} = \frac{-1}{1} = -1</math></li>
</ol>
</div>


== Squaring rules  ==

The distributive law occasionally has to be used repeatedly to deal with larger expression. 
If we consider 

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

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

{{Fristående formel||<math>\eqalign{
  \bbox[#AAEEFF,0pt]{\phantom{(a+b)}}\,(c+d)
    &= \bbox[#AAEEFF,0pt]{\phantom{(a+b)}}\,c
       + \bbox[#AAEEFF,0pt]{\phantom{(a+b)}}\,d\mbox{,}\cr
  (a+b)\,(c+d)
    &= (a+b)\,c + (a+b)\,d\mbox{.}}</math>}}

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>}}

A mnemonic for this  formula is: 

<center>{{:2.1 - Figur - Distributiva lagen två gånger}}</center>

<div class="exempel">
''' Example 4'''

<ol type="a">
  <li><math>(x+1)(x-2) = x\cdot x + x \cdot (-2) + 1 \cdot x + 1 \cdot (-2)
    = x^2 -2x+x-2</math><br/>
<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)
    = 3(2x^2 +x-2xy-y)</math><br/>
<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)
    = 2-x-2x+x^2</math><br/>
<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>. 
</ol>
</div>

Two important special cases of the above formula are when <math>a+b</math> and <math>c+d</math>  are the same expression

<div class="regel">
'''Squaring rules '''
{{Fristående formel||<math>(a+b)^2 = a^2 +2ab + b^2</math>}}
{{Fristående formel||<math>(a-b)^2 = a^2 -2ab + b^2</math>}}
</div>

These formulas are called  the first and second squaring rules 

<div class="exempel">
''' Example 5'''

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

The squaring rules  are also used in the reverse direction to factorise expressions. 

<div class="exempel">
''' Example 6'''

<ol type="a">
  <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^2 +x + \frac{1}{4} = x^2 + 2\cdot\frac{1}{2}x
    + \bigl(\frac{1}{2}\bigr)^2 = \bigl(x+\frac{1}{2}\bigr)^2</math></li>
</ol>
</div>


== Difference of two squares  ==

A  third special case of the first formula in the last section is  the difference of two squares rule.

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

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

{{Fristående formel||<math>(a+b)(a-b)
  = a \cdot a + a\cdot (-b) + b\cdot a + b \cdot (-b)
  = a^2 -ab+ab-b^2
  = a^2 -b^2\mbox{.}</math>}}

<div class="exempel">
''' Example 7'''

<ol type="a">
  <li><math>(x-4y)(x+4y) = x^2 -(4y)^2 = x^2 -16y^2</math></li>
  <li><math>(x^2+2x)(x^2-2x)= (x^2)^2 - (2x)^2 = x^4 -4x^2</math></li>
  <li><math>(y+3)(3-y)= (3+y)(3-y) = 3^2 -y^2 = 9-y^2</math></li>
  <li><math>x^4 -16 = (x^2)^2 -4^2 = (x^2+4)(x^2-4)
    = (x^2+4)(x^2-2^2)</math><br/> 
<math>\phantom{x^4-16}{}=(x^2+4)(x+2)(x-2)</math></li>
</ol>
</div>


== Rational expressions==

Calculations of fractions containing algebraic expressions are largely similar to ordinary calculations with fractions. 

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

<div class="regel">
{{Fristående formel||<math> \frac{a}{b} \cdot \frac{c}{d}
  = \frac{a\cdot c}{b\cdot d}
  \quad \mbox{och} \quad
  \frac{\displaystyle\ \frac{a}{b}\ }{\displaystyle\frac{c}{d}}
  = \frac{a\cdot d}{b\cdot c} \; \mbox{.}</math>}}
</div>

<div class="exempel">
''' Example 8'''

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

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

{{Fristående formel||<math>\frac{x+2}{x+1}
  = \frac{(x+2)(x+3)}{(x+1)(x+3)}
  = \frac{(x+2)(x+3)(x+4)}{(x+1)(x+3)(x+4)}
  = \dots</math>}}

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) }
  = \frac{(x+2)(x+4)}{(x+1)(x+4)}
  = \frac{x+2}{x+1} \mbox{.}</math>}}

<div class="exempel">
''' Example 9'''

<ol type="a">
  <li><math>\frac{x}{x+1} = \frac{x}{x+1} \cdot \frac{x+2}{x+2}
    = \frac{x(x+2)}{(x+1)(x+2)}</math></li>
  <li><math>\frac{x^2 -1}{x(x^2-1)}= \frac{1}{x}</math></li>
  <li><math>\frac{(x^2-y^2)(x-2)}{(x^2-4)(x+y)}
    = \left\{\,\text{Difference of two squares}\,\right\}
    = \frac{(x+y)(x-y)(x-2)}{(x+2)(x-2)(x+y)}
    = \frac{x-y}{x+2}</math></li>
</ol>
</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, 


{{Fristående formel||<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{x-1}{x(x-1)} - \frac{x}{x(x-1)}
  = \frac{x-1-x}{x(x-1)}
  = \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. 

<div class="exempel">
''' Example 10'''

<ol type="a">
  <li><math>\frac{1}{x+1} + \frac{1}{x+2}\quad</math> has <math>\ \text{LCD}
    = (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>
{{Fristående formel||<math>\begin{align*}
    \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+1)(x+2)}
       = \frac{2x+3}{(x+1)(x+2)}\:\mbox{.}
    \end{align*}</math>}}</li>
  <li><math>\frac{1}{x} + \frac{1}{x^2}\quad</math> has <math>\ \text{LCD}
    = x^2</math><br><br>
We only need to convert the first term to get a common denominator 
{{Fristående formel||<math>\frac{1}{x} + \frac{1}{x^2}
    = \frac{x}{x^2} + \frac{1}{x^2}
    = \frac{x+1}{x^2}\,\mbox{.}</math>}}</li>
  <li><math>\frac{1}{x(x+1)^2} - \frac{1}{x^2(x+2)}\quad</math> has <math>\ 
    \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>
{{Fristående formel||<math>\begin{align*}
    \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+1)^2}{x^2(x+1)^2(x+2)}\\[4pt]
      &= \frac{x^2+2x}{x^2(x+1)^2(x+2)} - \frac{x^2+2x+1}{x^2(x+1)^2(x+2)}\\[4pt]
      &= \frac{x^2+2x-(x^2+2x+1)}{x^2(x+1)^2(x+2)}\\[4pt]
      &= \frac{x^2+2x-x^2-2x-1}{x^2(x+1)^2(x+2)}\\[4pt]
      &= \frac{-1}{x^2(x+1)^2(x+2)}\,\mbox{.}
    \end{align*}</math>}}</li>
  <li><math>\frac{x}{x+1} - \frac{1}{x(x-1)} -1 \quad</math> has <math>\ 
    \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>
{{Fristående formel||<math>\begin{align*}
    \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(x-1)(x+1)}{x(x-1)(x+1)}\\[4pt]
      &= \frac{x^3-x^2}{x(x-1)(x+1)} - \frac{x+1}{x(x-1)(x+1)}
         - \frac{x^3 -x}{x(x-1)(x+1)}\\[4pt]
      &= \frac{x^3-x^2 -(x+1) -(x^3-x)}{x(x-1)(x+1)}\\[4pt]
      &= \frac{x^3-x^2 -x-1 -x^3+x}{x(x-1)(x+1)}\\[4pt]
      &= \frac{-x^2-1}{x(x-1)(x+1)}\,\mbox{.}
    \end{align*}</math>}}</li>
</ol>
</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. 


<div class="exempel">
''' Example 11'''

<ol type="a">
  <li><math>\frac{1}{x-2} - \frac{4}{x^2-4}
    = \frac{1}{x-2} - \frac{4}{(x+2)(x-2)}
    =  \left\{\,\mbox{MGN}
    = (x+2)(x-2)\,\right\}</math><br/><br/>
<math>\phantom{\frac{1}{x-2} - \frac{4}{x^2-4}}{}
    = \frac{x+2}{(x+2)(x-2)} - \frac{4}{(x+2)(x-2)}</math><br/><br/>
<math>\phantom{\frac{1}{x-2} - \frac{4}{x^2-4}}{}
    = \frac{x+2 -4}{(x+2)(x-2)}
    = \frac{x-2}{(x+2)(x-2)}
    = \frac{1}{x+2}</math></li>
  <li><math>\frac{x + \displaystyle \frac{1}{x}}{x^2+1}
    = \frac{\displaystyle \frac{x^2}{x} + \frac{1}{x}}{x^2+1}
    = \frac{\displaystyle \frac{x^2+1}{x}}{x^2+1}
    = \frac{x^2+1}{x(x^2+1)}
    = \frac{1}{x}</math></li>
  <li><math>\frac{\displaystyle \frac{1}{x^2} - \frac{1}{y^2}}{x+y}
    = \frac{\displaystyle \frac{y^2}{x^2y^2} - \frac{x^2}{x^2y^2}}{x+y}
    = \frac{\displaystyle \frac{y^2-x^2}{x^2y^2}}{x+y}
    = \frac{y^2-x^2}{x^2y^2(x+y)}</math> <br><br>
<math>\phantom{\smash{\frac{\displaystyle \frac{1}{x^2}
        - \frac{1}{y^2}}{x+y}}}{}
    = \frac{(y+x)(y-x)}{x^2y^2(x+y)}
    = \frac{y-x}{x^2y^2}</math></li>
</ol>
</div>


[[2.1 Övningar|Exercises]]



<div class="inforuta" style="width:580px;">
'''Study advice'''

'''The basic and final tests'''

After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge. 


'''Keep in mind that: '''

Be careful. If you make a mistake somewhere the rest of the calculation will be wrong. 

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

Do not expand unnecessarily. You later may  be forced to factorise what you earlier expanded. 


'''Reviews
'''

[http://en.wikipedia.org/wiki/Algebra  Learn more about algebra  in the English Wikipedia ]

[http://www.jamesbrennan.org/algebra/  Understanding Algebra - English text on the Web ]


'''Useful web sites'''

</div>
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