2.1 Algebraische Ausdrücke

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K (Robot: Automated text replacement (-[[2.1 Algebraic expressions +[[2.1 Algebraische Ausdrücke))
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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If we consider
If we consider
-
{{Displayed math||<math>(a+b)(c+d)</math>}}
+
{{Abgesetzte Formel||<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
-
{{Displayed math||<math>\eqalign{
+
{{Abgesetzte Formel||<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
Zeile 100: Zeile 100:
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,
-
{{Displayed math||<math>(a+b)c + (a+b)d = ac + bc + ad + bd \, \mbox{.}</math>}}
+
{{Abgesetzte Formel||<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:
Zeile 127: Zeile 127:
<div class="regel">
<div class="regel">
'''Squaring rules '''
'''Squaring rules '''
-
{{Displayed math||<math>(a+b)^2 = a^2 +2ab + b^2</math>}}
+
{{Abgesetzte Formel||<math>(a+b)^2 = a^2 +2ab + b^2</math>}}
-
{{Displayed math||<math>(a-b)^2 = a^2 -2ab + b^2</math>}}
+
{{Abgesetzte Formel||<math>(a-b)^2 = a^2 -2ab + b^2</math>}}
</div>
</div>
Zeile 175: Zeile 175:
<div class="regel">
<div class="regel">
'''Difference of two squares:'''
'''Difference of two squares:'''
-
{{Displayed math||<math>(a+b)(a-b) = a^2 -b^2</math>}}
+
{{Abgesetzte Formel||<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
-
{{Displayed math||<math>(a+b)(a-b)
+
{{Abgesetzte Formel||<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
Zeile 206: Zeile 206:
<div class="regel">
<div class="regel">
-
{{Displayed math||<math> \frac{a}{b} \cdot \frac{c}{d}
+
{{Abgesetzte 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{and} \quad
\quad \mbox{and} \quad
Zeile 229: Zeile 229:
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}
+
{{Abgesetzte Formel||<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)}
Zeile 235: Zeile 235:
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
-
{{Displayed math||<math>\frac{(x+2)(x+3)(x+4)}{(x+1)(x+3)(x+4) }
+
{{Abgesetzte 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+4)}{(x+1)(x+4)}
= \frac{x+2}{x+1} \mbox{.}</math>}}
= \frac{x+2}{x+1} \mbox{.}</math>}}
Zeile 256: Zeile 256:
-
{{Displayed math||<math>\frac{1}{x} - \frac{1}{x-1}
+
{{Abgesetzte 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{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)}
Zeile 271: Zeile 271:
= (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>
-
{{Displayed math||<math>\begin{align*}
+
{{Abgesetzte Formel||<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]
Zeile 280: Zeile 280:
= 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
-
{{Displayed math||<math>\frac{1}{x} + \frac{1}{x^2}
+
{{Abgesetzte Formel||<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>
Zeile 286: Zeile 286:
\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>
-
{{Displayed math||<math>\begin{align*}
+
{{Abgesetzte Formel||<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)}
Zeile 298: Zeile 298:
\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>
-
{{Displayed math||<math>\begin{align*}
+
{{Abgesetzte Formel||<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)}

Version vom 08:11, 22. Okt. 2008

 

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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 using squaring rules and the difference of two squares rule.
  • Expand expressions using squaring rules and the difference of two squares rule.

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\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}
  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 bracket with a minus sign in front can be eliminated if all the terms inside the brackets, switch signs.

Example 2

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

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

  1. \displaystyle 3x +9y = 3x + 3\cdot 3y = 3(x+3y)
  2. \displaystyle xy + y^2 = xy + y\cdot y = y(x+y)
  3. \displaystyle 2x^2 -4x = 2x\cdot x - 2\cdot 2\cdot x = 2x(x-2)
  4. \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:

[Image]

Example 4

  1. \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
  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
  3. \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

  1. \displaystyle (x+2)^2 = x^2 + 2\cdot 2x+ 2^2 = x^2 +4x +4
  2. \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{.}
  3. \displaystyle (x^2 -4)^2 = (x^2)^2 - 2 \cdot 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 \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

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

  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

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

  1. \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)}
  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

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

  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 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

  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 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

  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:

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

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