2.1 Algebraic expressions

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{| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
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| style="border-bottom:1px solid #797979" width="5px" |  
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{{Vald flik|[[2.1 Algebraiska uttryck|Theory]]}}
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{{Selected tab|[[2.1 Algebraic expressions|Theory]]}}
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{{Ej vald flik|[[2.1 Övningar|Exercises]]}}
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{{Not selected tab|[[2.1 Exercises|Exercises]]}}
| style="border-bottom:1px solid #797979" width="100%"|  
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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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'''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 complicated algebraic expressions.
-
*Factorise expressions using squaring rules and the difference of two squares rule.
+
*Factorise expressions, including perfect squares and the difference of two squares.
-
*Expand expressions using squaring rules and the difference of two squares rule.
+
*Expand expressions, including perfect squares and the difference of two squares.
}}
}}
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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">
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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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</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 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.
<div class="exempel">
<div class="exempel">
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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 “factor” 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 often would like to to factorise out as large a numerical factor as possible.
+
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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-
== Squaring rules ==
+
== Squaring ==
-
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
-
{{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:
-
<center>{{:2.1 - Figur - Distributiva lagen två gånger}}</center>
+
<center>{{:2.1 - Figure - The distributive law twice}}</center>
<div class="exempel">
<div class="exempel">
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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>
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<div class="regel">
<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
+
{{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 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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<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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== 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">
-
{{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\, 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
-
{{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)}
= \dots</math>}}
= \dots</math>}}
-
The opposite of this, is cancellation, where we delete factors that the numerator and denominator have in common
+
We can cancel 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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</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 placed over a common denominator.
-
 
+
{{Displayed math||<math>\frac{1}{x} - \frac{1}{x-1}
-
{{Fristående formel||<math>\frac{1}{x} - \frac{1}{x-1}
+
= \frac{1}{x} \times\frac{x-1}{x-1} - \frac{1}{x-1} \times\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)}
= \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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= (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)}
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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">
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-
[[2.1 Övningar|Exercises]]
+
[[2.1 Exercises|Exercises]]
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-
'''Keep in mind that: '''
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'''Keep in mind that...'''
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Be careful. If you make a mistake somewhere the rest of the calculation will be wrong.
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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.

Current revision

       Theory          Exercises      

Contents:

  • Distributive law
  • Expansion and factorisation
  • Difference of two squares
  • Rational expressions

Learning outcomes:

After this section you will have learned how to:

  • Simplify complicated algebraic expressions.
  • Factorise expressions, including perfect squares and the difference of two squares.
  • Expand expressions, including perfect squares and the difference of two squares.

Distributive Law

The distributive law specifies how to multiply a bracketed expression by a factor.

[Image]

Example 1

  1. \displaystyle 4(x+y) = 4x + 4y
  2. \displaystyle 2(a-b) = 2a -2b
  3. \displaystyle x \left(\frac{1}{x} + \frac{1}{x^2} \right) = x\times \frac{1}{x} + x \times \frac{1}{x^2} = \frac{\not{x}}{\not{x}} + \frac{\not{x}}{x^{\not{2}}} = 1 + \frac{1}{x}
  4. \displaystyle a(x+y+z) = ax + ay + az

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

Example 2

  1. \displaystyle -(x+y) = (-1) \times (x+y) = (-1)x + (-1)y = -x-y
  2. \displaystyle -(x^2-x) = (-1) \times (x^2-x) = (-1)x^2 -(-1)x = -x^2 +x
    where we have in the final step used \displaystyle -(-1)x = (-1)(-1)x = 1\times x = x\,\mbox{.}
  3. \displaystyle -(x+y-y^3) = (-1)\times (x+y-y^3) = (-1)\times x + (-1) \times y -(-1)\times y^3
    \displaystyle \phantom{-(x+y-y^3)}{} = -x-y+y^3
  4. \displaystyle x^2 - 2x -(3x+2) = x^2 -2x -3x-2 = x^2 -(2+3)x -2
    \displaystyle \phantom{x^2-2x-(3x+2)}{} = x^2 -5x -2

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 identifying the highet factor shared by all the terms.

Example 3

  1. \displaystyle 3x +9y = 3x + 3\times 3y = 3(x+3y)
  2. \displaystyle xy + y^2 = xy + y\times y = y(x+y)
  3. \displaystyle 2x^2 -4x = 2x\times x - 2\times 2\times x = 2x(x-2)
  4. \displaystyle \frac{y-x}{x-y} = \frac{-(x-y)}{x-y} = \frac{-1}{1} = -1


Squaring

On occasions the distributive law has to be used repeatedly to deal with larger expressions. If we consider

\displaystyle (a+b)(c+d)

and regard \displaystyle a+b as a factor that multiplies the bracketed expression \displaystyle (c+d) we get

\displaystyle \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{.}}

Then the \displaystyle c and the \displaystyle d are multiplied into their respective brackets,

\displaystyle (a+b)c + (a+b)d = ac + bc + ad + bd \, \mbox{.}

A mnemonic for this formula is:

[Image]

Example 4

  1. \displaystyle (x+1)(x-2) = x\times x + x \times (-2) + 1 \times x + 1 \times (-2) = x^2 -2x+x-2
    \displaystyle \phantom{(x+1)(x-2)}{}=x^2 -x-2
  2. \displaystyle 3(x-y)(2x+1) = 3(x\times 2x + x\times 1 - y \times 2x - y \times 1) = 3(2x^2 +x-2xy-y)
    \displaystyle \phantom{3(x-y)(2x+1)}{}=6x^2 +3x-6xy-3y
  3. \displaystyle (1-x)(2-x) = 1\times 2 + 1 \times (-x) -x\times 2 - x\times (-x) = 2-x-2x+x^2
    \displaystyle \phantom{(1-x)(2-x)}{}=2-3x+x^2 where we have used \displaystyle -x\times (-x) = (-1)x \times (-1)x = (-1)^2 x^2 = 1\times x^2 = x^2.

Two important special cases of the above formula are when \displaystyle a+b and \displaystyle c+d are the same expression

\displaystyle (a+b)^2 = a^2 +2ab + b^2
\displaystyle (a-b)^2 = a^2 -2ab + b^2

Example 5

  1. \displaystyle (x+2)^2 = x^2 + 2\times 2x+ 2^2 = x^2 +4x +4
  2. \displaystyle (-x+3)^2 = (-x)^2 + 2\times 3(-x) + 3^2 = x^2 -6x +9
    where \displaystyle (-x)^2 = ((-1)x)^2 = (-1)^2 x^2 = 1 \times x^2 = x^2\,\mbox{.}
  3. \displaystyle (x^2 -4)^2 = (x^2)^2 - 2 \times 4x^2 + 4^2 = x^4 -8x^2 +16
  4. \displaystyle (x+1)^2 - (x-1)^2 = (x^2 +2x +1)- (x^2-2x+1)
    \displaystyle \phantom{(x+1)^2-(x-1)^2}{}= x^2 +2x +1 -x^2 + 2x-1
    \displaystyle \phantom{(x+1)^2-(x-1)^2}{} = 2x+2x = 4x
  5. \displaystyle (2x+4)(x+2) = 2(x+2)(x+2) = 2(x+2)^2 = 2(x^2 + 4x+ 4)
    \displaystyle \phantom{(2x+4)(x+2)}{}=2x^2 + 8x + 8
  6. \displaystyle (x-2)^3 = (x-2)(x-2)^2 = (x-2)(x^2-4x+4)
    \displaystyle \phantom{(x-2)^3}{}=x \times x^2 + x\times (-4x) + x\times 4 - 2\times x^2 - 2 \times (-4x)-2 \times 4
    \displaystyle \phantom{(x-2)^3}{}=x^3 -4x^2 + 4x-2x^2 +8x -8 = x^3-6x^2 + 12x -8

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

Example 6

  1. \displaystyle x^2 + 2x+ 1 = (x+1)^2
  2. \displaystyle x^6-4x^3 +4 = (x^3)^2 - 2\times 2x^3 +2^2 = (x^3-2)^2
  3. \displaystyle 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


Difference of two squares

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

Difference of two squares:

\displaystyle (a+b)(a-b) = a^2 -b^2

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

\displaystyle (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{.}

Example 7

  1. \displaystyle (x-4y)(x+4y) = x^2 -(4y)^2 = x^2 -16y^2
  2. \displaystyle (x^2+2x)(x^2-2x)= (x^2)^2 - (2x)^2 = x^4 -4x^2
  3. \displaystyle (y+3)(3-y)= (3+y)(3-y) = 3^2 -y^2 = 9-y^2
  4. \displaystyle x^4 -16 = (x^2)^2 -4^2 = (x^2+4)(x^2-4) = (x^2+4)(x^2-2^2)
    \displaystyle \phantom{x^4-16}{}=(x^2+4)(x+2)(x-2)


Rational expressions

Working with fractions containing algebraic expressions is very similar to carrying out ordinary calculations with fractions.

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

\displaystyle \frac{a}{b} \cdot \frac{c}{d} 
 = \frac{a\, c}{b\, d}
 \quad \mbox{and} \quad
 \frac{\displaystyle\ \frac{a}{b}\ }{\displaystyle\frac{c}{d}}
 = \frac{a\, d}{b\, c} \; \mbox{.}

Example 8

  1. \displaystyle \frac{3x}{x-y} \times \frac{4x}{2x+y} = \frac{3x\times 4x}{(x-y)\times(2x+y)} = \frac{12x^2}{(x-y)(2x+y)}
  2. \displaystyle \frac{\displaystyle \frac{a}{x}}{\displaystyle \frac{x+1}{a}} = \frac{a^2}{x(x+1)}
  3. \displaystyle \frac{\displaystyle \frac{x}{(x+1)^2}}{\displaystyle \frac{x-2}{x-1}} = \frac{x(x-1)}{(x-2)(x+1)^2}

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

\displaystyle \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

We can cancel factors that the numerator and denominator have in common

\displaystyle \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{.}

Example 9

  1. \displaystyle \frac{x}{x+1} = \frac{x}{x+1} \cdot \frac{x+2}{x+2} = \frac{x(x+2)}{(x+1)(x+2)}
  2. \displaystyle \frac{x^2 -1}{x(x^2-1)}= \frac{1}{x}
  3. \displaystyle \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}

When fractional expressions are added or subtracted they may need to be placed over a common denominator.

\displaystyle \frac{1}{x} - \frac{1}{x-1} 
 = \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(x-1)}
 = \frac{-1}{x(x-1)} \; \mbox{.}

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.

Example 10

  1. \displaystyle \frac{1}{x+1} + \frac{1}{x+2}\quad has \displaystyle \ \text{LCD} = (x+1)(x+2)

    Convert the first term using \displaystyle (x+2) and the second term using \displaystyle (x+1)
    \displaystyle \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*}
  2. \displaystyle \frac{1}{x} + \frac{1}{x^2}\quad has \displaystyle \ \text{LCD} = x^2

    We only need to convert the first term to get a common denominator
    \displaystyle \frac{1}{x} + \frac{1}{x^2} 
       = \frac{x}{x^2} + \frac{1}{x^2}
   = \frac{x+1}{x^2}\,\mbox{.}
  3. \displaystyle \frac{1}{x(x+1)^2} - \frac{1}{x^2(x+2)}\quad has \displaystyle \ \text{LCD}= x^2(x+1)^2(x+2)

    The first term is converted using \displaystyle x(x+2) while the other term is converted using \displaystyle (x+1)^2
    \displaystyle \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*}
  4. \displaystyle \frac{x}{x+1} - \frac{1}{x(x-1)} -1 \quad has \displaystyle \ \text{LCD}=x(x-1)(x+1)

    We must convert all the terms so that they have the common denominator \displaystyle x(x-1)(x+1)
    \displaystyle \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*}

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.

Example 11

  1. \displaystyle \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\}

    \displaystyle \phantom{\frac{1}{x-2} - \frac{4}{x^2-4}}{} = \frac{x+2}{(x+2)(x-2)} - \frac{4}{(x+2)(x-2)}

    \displaystyle \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}
  2. \displaystyle \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}
  3. \displaystyle \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)}

    \displaystyle \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}


Exercises


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...

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.

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


Reviews

Learn more about algebra in the English Wikipedia

Understanding Algebra - English text on the Web


Useful web sites

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