2.1 Algebraic expressions
From Förberedande kurs i matematik 1
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- | {{ | + | {{Selected tab|[[2.1 Algebraic expressions|Theory]]}} |
- | {{ | + | {{Not selected tab|[[2.1 Exercises|Exercises]]}} |
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{{Info| | {{Info| | ||
- | ''' | + | '''Contents:''' |
- | * | + | * Distributive law |
- | * | + | * Squaring rules |
- | * | + | *Difference of two squares |
- | * | + | * Rational expressions |
}} | }} | ||
{{Info| | {{Info| | ||
- | ''' | + | '''Learning outcomes:''' |
- | + | After this section you will have learned how to: | |
- | * | + | *Simplify complex algebraic expressions. |
- | * | + | *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. |
}} | }} | ||
- | == | + | == Distributive Law == |
- | + | The distributive law specifies how to multiply a bracketed expression by a factor. | |
- | + | ||
- | <center>{{:2.1 - | + | <center>{{:2.1 - Figure - The distributive law}}</center> |
<div class="exempel"> | <div class="exempel"> | ||
- | ''' | + | ''' Example 1''' |
<ol type="a"> | <ol type="a"> | ||
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</div> | </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 a minus sign infront of a bracket can be | ||
+ | eliminated if all the terms inside the brackets switch signs. | ||
<div class="exempel"> | <div class="exempel"> | ||
- | ''' | + | ''' Example 2''' |
<ol type="a"> | <ol type="a"> | ||
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<li><math>-(x^2-x) = (-1) \cdot (x^2-x) = (-1)x^2 -(-1)x | <li><math>-(x^2-x) = (-1) \cdot (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> | |
<li><math>-(x+y-y^3) = (-1)\cdot (x+y-y^3) = (-1)\cdot x | <li><math>-(x+y-y^3) = (-1)\cdot (x+y-y^3) = (-1)\cdot x | ||
+ (-1) \cdot y -(-1)\cdot y^3</math><br/> | + (-1) \cdot y -(-1)\cdot y^3</math><br/> | ||
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</div> | </div> | ||
- | + | If the distributive law is applied in reverse we say we “factorise” the expression. | |
+ | One would often like to factorise out the largest possible numerical factor. | ||
<div class="exempel"> | <div class="exempel"> | ||
- | ''' | + | ''' Example 3''' |
<ol type="a"> | <ol type="a"> | ||
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- | == | + | == Squaring rules == |
- | + | On occasions the distributive law has to be used repeatedly to deal with larger expressions. | |
+ | If we consider | ||
- | {{ | + | {{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 | |
- | {{ | + | {{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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&= (a+b)\,c + (a+b)\,d\mbox{.}}</math>}} | &= (a+b)\,c + (a+b)\,d\mbox{.}}</math>}} | ||
- | + | Then the <math>c</math> and the <math>d</math> are multiplied into their respective brackets, | |
- | {{ | + | {{Displayed math||<math>(a+b)c + (a+b)d = ac + bc + ad + bd \, \mbox{.}</math>}} |
- | + | A mnemonic for this formula is: | |
- | <center>{{:2.1 - | + | <center>{{:2.1 - Figure - The distributive law twice}}</center> |
<div class="exempel"> | <div class="exempel"> | ||
- | ''' | + | ''' Example 4''' |
<ol type="a"> | <ol type="a"> | ||
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= 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>. | |
</ol> | </ol> | ||
</div> | </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"> | <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>}} |
</div> | </div> | ||
- | + | These formulas are called the first and second squaring rules. | |
<div class="exempel"> | <div class="exempel"> | ||
- | ''' | + | ''' Example 5''' |
<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\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> | <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 | <li><math>(x^2 -4)^2 = (x^2)^2 - 2 \cdot 4x^2 + 4^2 | ||
= x^4 -8x^2 +16</math></li> | = x^4 -8x^2 +16</math></li> | ||
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</div> | </div> | ||
- | + | The squaring rules are also used in the reverse direction to factorise expressions. | |
<div class="exempel"> | <div class="exempel"> | ||
- | ''' | + | ''' Example 6''' |
<ol type="a"> | <ol type="a"> | ||
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- | == | + | == 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"> | <div class="regel"> | ||
- | ''' | + | '''Difference of two squares:''' |
- | {{ | + | {{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 | |
- | {{ | + | {{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="exempel"> | <div class="exempel"> | ||
- | ''' | + | ''' Example 7''' |
<ol type="a"> | <ol type="a"> | ||
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- | == | + | == 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"> | <div class="regel"> | ||
- | {{ | + | {{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{ | + | \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\cdot d}{b\cdot c} \; \mbox{.}</math>}} | ||
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<div class="exempel"> | <div class="exempel"> | ||
- | ''' | + | ''' Example 8''' |
<ol type="a"> | <ol type="a"> | ||
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</div> | </div> | ||
- | + | A fractional expression can have its numerator and denominator multiplied by the same factor | |
- | {{ | + | {{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. 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) } | |
- | {{ | + | |
= \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>}} | ||
<div class="exempel"> | <div class="exempel"> | ||
- | ''' | + | ''' Example 9''' |
<ol type="a"> | <ol type="a"> | ||
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<li><math>\frac{x^2 -1}{x(x^2-1)}= \frac{1}{x}</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)} | <li><math>\frac{(x^2-y^2)(x-2)}{(x^2-4)(x+y)} | ||
- | = \left\{\,\text{ | + | = \left\{\,\text{Difference of two squares}\,\right\} |
= \frac{(x+y)(x-y)(x-2)}{(x+2)(x-2)(x+y)} | = \frac{(x+y)(x-y)(x-2)}{(x+2)(x-2)(x+y)} | ||
= \frac{x-y}{x+2}</math></li> | = \frac{x-y}{x+2}</math></li> | ||
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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. | |
- | {{ | + | |
+ | {{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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= \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. | |
<div class="exempel"> | <div class="exempel"> | ||
- | ''' | + | ''' Example 10''' |
<ol type="a"> | <ol type="a"> | ||
- | <li><math>\frac{1}{x+1} + \frac{1}{x+2}\quad</math> | + | <li><math>\frac{1}{x+1} + \frac{1}{x+2}\quad</math> has <math>\ \text{LCD} |
= (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> | |
- | {{ | + | {{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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= \frac{2x+3}{(x+1)(x+2)}\:\mbox{.} | = \frac{2x+3}{(x+1)(x+2)}\:\mbox{.} | ||
\end{align*}</math>}}</li> | \end{align*}</math>}}</li> | ||
- | <li><math>\frac{1}{x} + \frac{1}{x^2}\quad</math> | + | <li><math>\frac{1}{x} + \frac{1}{x^2}\quad</math> has <math>\ \text{LCD} |
= x^2</math><br><br> | = x^2</math><br><br> | ||
- | + | We only need to convert the first term to get a common denominator | |
- | {{ | + | {{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}</math>}}</li> | + | = \frac{x+1}{x^2}\,\mbox{.}</math>}}</li> |
- | <li><math>\frac{1}{x(x+1)^2} - \frac{1}{x^2(x+2)}\quad</math> | + | <li><math>\frac{1}{x(x+1)^2} - \frac{1}{x^2(x+2)}\quad</math> has <math>\ |
- | \text{ | + | \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> | |
- | {{ | + | {{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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&= \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{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)} | + | &= \frac{-1}{x^2(x+1)^2(x+2)}\,\mbox{.} |
\end{align*}</math>}}</li> | \end{align*}</math>}}</li> | ||
- | <li><math>\frac{x}{x+1} - \frac{1}{x(x-1)} -1 \quad</math> | + | <li><math>\frac{x}{x+1} - \frac{1}{x(x-1)} -1 \quad</math> has <math>\ |
- | \text{ | + | \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> | |
- | {{ | + | {{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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&= \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^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)} | + | &= \frac{-x^2-1}{x(x-1)(x+1)}\,\mbox{.} |
\end{align*}</math>}}</li> | \end{align*}</math>}}</li> | ||
</ol> | </ol> | ||
</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. | |
+ | |||
<div class="exempel"> | <div class="exempel"> | ||
- | ''' | + | ''' Example 11''' |
<ol type="a"> | <ol type="a"> | ||
<li><math>\frac{1}{x-2} - \frac{4}{x^2-4} | <li><math>\frac{1}{x-2} - \frac{4}{x^2-4} | ||
= \frac{1}{x-2} - \frac{4}{(x+2)(x-2)} | = \frac{1}{x-2} - \frac{4}{(x+2)(x-2)} | ||
- | = \left\{\,\mbox{ | + | = \left\{\,\mbox{LCD} |
= (x+2)(x-2)\,\right\}</math><br/><br/> | = (x+2)(x-2)\,\right\}</math><br/><br/> | ||
<math>\phantom{\frac{1}{x-2} - \frac{4}{x^2-4}}{} | <math>\phantom{\frac{1}{x-2} - \frac{4}{x^2-4}}{} | ||
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- | [[2.1 | + | [[2.1 Exercises|Exercises]] |
<div class="inforuta" style="width:580px;"> | <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...''' |
- | + | 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 |
+ | ''' | ||
- | [http://en.wikipedia.org/wiki/Algebra | + | [http://en.wikipedia.org/wiki/Algebra Learn more about algebra in the English Wikipedia ] |
- | [http://www.jamesbrennan.org/algebra/ Understanding Algebra - | + | [http://www.jamesbrennan.org/algebra/ Understanding Algebra - English text on the Web ] |
- | ''' | + | '''Useful web sites''' |
</div> | </div> |
Current revision
Theory | Exercises |
Contents:
- Distributive law
- Squaring rules
- Difference of two squares
- Rational expressions
Learning outcomes:
After this section you will have learned how to:
- Simplify complex algebraic expressions.
- 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.
Distributive Law
The distributive law specifies how to multiply a bracketed expression by a factor.
Example 1
- \displaystyle 4(x+y) = 4x + 4y
- \displaystyle 2(a-b) = 2a -2b
- \displaystyle 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}
- \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 infront of a bracket can be eliminated if all the terms inside the brackets switch signs.
Example 2
- \displaystyle -(x+y) = (-1) \cdot (x+y) = (-1)x + (-1)y = -x-y
- \displaystyle -(x^2-x) = (-1) \cdot (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\cdot x = x\,\mbox{.} - \displaystyle -(x+y-y^3) = (-1)\cdot (x+y-y^3) = (-1)\cdot x
+ (-1) \cdot y -(-1)\cdot y^3
\displaystyle \phantom{-(x+y-y^3)}{} = -x-y+y^3 - \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
If the distributive law is applied in reverse we say we “factorise” the expression. One would often like to factorise out the largest possible numerical factor.
Example 3
- \displaystyle 3x +9y = 3x + 3\cdot 3y = 3(x+3y)
- \displaystyle xy + y^2 = xy + y\cdot y = y(x+y)
- \displaystyle 2x^2 -4x = 2x\cdot x - 2\cdot 2\cdot x = 2x(x-2)
- \displaystyle \frac{y-x}{x-y} = \frac{-(x-y)}{x-y} = \frac{-1}{1} = -1
Squaring rules
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:
Example 4
- \displaystyle (x+1)(x-2) = x\cdot x + x \cdot (-2) + 1 \cdot x + 1 \cdot (-2)
= x^2 -2x+x-2
\displaystyle \phantom{(x+1)(x-2)}{}=x^2 -x-2 - \displaystyle 3(x-y)(2x+1) = 3(x\cdot 2x + x\cdot 1 - y \cdot 2x - y \cdot 1)
= 3(2x^2 +x-2xy-y)
\displaystyle \phantom{3(x-y)(2x+1)}{}=6x^2 +3x-6xy-3y - \displaystyle (1-x)(2-x) = 1\cdot 2 + 1 \cdot (-x) -x\cdot 2 - x\cdot (-x)
= 2-x-2x+x^2
\displaystyle \phantom{(1-x)(2-x)}{}=2-3x+x^2 where we have used \displaystyle -x\cdot (-x) = (-1)x \cdot (-1)x = (-1)^2 x^2 = 1\cdot 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
Squaring rules
\displaystyle (a+b)^2 = a^2 +2ab + b^2 |
\displaystyle (a-b)^2 = a^2 -2ab + b^2 |
These formulas are called the first and second squaring rules.
Example 5
- \displaystyle (x+2)^2 = x^2 + 2\cdot 2x+ 2^2 = x^2 +4x +4
- \displaystyle (-x+3)^2 = (-x)^2 + 2\cdot 3(-x) + 3^2 = x^2 -6x +9
- where \displaystyle (-x)^2 = ((-1)x)^2 = (-1)^2 x^2 = 1 \cdot x^2 = x^2\,\mbox{.}
- \displaystyle (x^2 -4)^2 = (x^2)^2 - 2 \cdot 4x^2 + 4^2 = x^4 -8x^2 +16
- \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 - \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 - \displaystyle (x-2)^3 = (x-2)(x-2)^2 = (x-2)(x^2-4x+4)
\displaystyle \phantom{(x-2)^3}{}=x \cdot x^2 + x\cdot (-4x) + x\cdot 4 - 2\cdot x^2 - 2 \cdot (-4x)-2 \cdot 4
\displaystyle \phantom{(x-2)^3}{}=x^3 -4x^2 + 4x-2x^2 +8x -8 = x^3-6x^2 + 12x -8
The squaring rules are also used in the reverse direction to factorise expressions.
Example 6
- \displaystyle x^2 + 2x+ 1 = (x+1)^2
- \displaystyle x^6-4x^3 +4 = (x^3)^2 - 2\cdot 2x^3 +2^2 = (x^3-2)^2
- \displaystyle 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
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
- \displaystyle (x-4y)(x+4y) = x^2 -(4y)^2 = x^2 -16y^2
- \displaystyle (x^2+2x)(x^2-2x)= (x^2)^2 - (2x)^2 = x^4 -4x^2
- \displaystyle (y+3)(3-y)= (3+y)(3-y) = 3^2 -y^2 = 9-y^2
- \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
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,
\displaystyle \frac{a}{b} \cdot \frac{c}{d}
= \frac{a\cdot c}{b\cdot d} \quad \mbox{and} \quad \frac{\displaystyle\ \frac{a}{b}\ }{\displaystyle\frac{c}{d}} = \frac{a\cdot d}{b\cdot c} \; \mbox{.} |
Example 8
- \displaystyle \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)}
- \displaystyle \frac{\displaystyle \frac{a}{x}}{\displaystyle \frac{x+1}{a}} = \frac{a^2}{x(x+1)}
- \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 |
The opposite of this is cancellation. Here we delete 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
- \displaystyle \frac{x}{x+1} = \frac{x}{x+1} \cdot \frac{x+2}{x+2} = \frac{x(x+2)}{(x+1)(x+2)}
- \displaystyle \frac{x^2 -1}{x(x^2-1)}= \frac{1}{x}
- \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 converted so that they have the same denominator. Only then can the numerators be combined together.
\displaystyle \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{.} |
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.
Example 10
- \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*}
- \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{.}
- \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*}
- \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 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.
Example 11
- \displaystyle \frac{1}{x-2} - \frac{4}{x^2-4}
= \frac{1}{x-2} - \frac{4}{(x+2)(x-2)}
= \left\{\,\mbox{LCD}
= (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} - \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}
- \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}
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