2.1 Algebraic expressions
From Förberedande kurs i matematik 1
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* Squaring rules | * Squaring rules | ||
*Difference of two squares | *Difference of two squares | ||
| - | * Rational | + | * Rational expressions |
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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 | + | *Simplify complex algebraic expressions. |
| - | *Factorise expressions using squaring rules and | + | *Factorise expressions using squaring rules and the difference of two squares rule. |
| - | *Expand expressions using squaring rules and | + | *Expand expressions using squaring rules and the difference of two squares rule. |
}} | }} | ||
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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 | + | The rule says that a bracket with a minus sign in front 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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== Squaring rules == | == Squaring rules == | ||
| - | The distributive law occasionally has to be used repeatedly to deal with larger | + | The distributive law occasionally has to be used repeatedly to deal with larger expressions. |
If we consider | If we consider | ||
{{Fristående formel||<math>(a+b)(c+d)</math>}} | {{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 | + | and regard <math>a+b</math> as a factor that multiplies the bracketed expression <math>(c+d)</math> we get |
{{Fristående formel||<math>\eqalign{ | {{Fristående formel||<math>\eqalign{ | ||
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{{Fristående formel||<math> \frac{a}{b} \cdot \frac{c}{d} | {{Fristående formel||<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>}} | ||
Revision as of 17:08, 11 August 2008
| Theory | Exercises |
Content:
- 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 using squaring rules and the difference of two squares rule.
- Expand expressions using squaring rules and the difference of two squares rule.
Distributive Law
right 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 bracket with a minus sign in front 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 “factor” the expression. One often would like to to factorise out as large a numerical factor as possible.
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
The distributive law occasionally 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, where 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 before the numerators can 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{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} - \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:
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
Learn more about algebra in the English Wikipedia
Understanding Algebra - English text on the Web
Useful web sites
