Algebraic Expressions
 
      Expanding algebraic expression by removing parentheses (i.e. brackets)
         The binomial expansion algorithm
         The difference of two squares, multiplying
         The difference of two squares, factoring
         The difference of two cubes
         The sum of two cubes
Expanding algebraic expression by removing parentheses (i.e. brackets)
The operation of multiplying out algebraic expressions that involve parentheses using the distributive property  is often described as expanding the brackets.
The binomial expansion algorithm - the binomial theorem
The binomial expansion of any positive integral power of a binomial, which represents a polynomial with n + 1 terms, 
or written in the form of the sum formula
 
 
is called the binomial theorem.
The binomial coefficients can also be obtained by using Pascal's triangle.
The triangular array of integers, with 1 at the apex, in which each number is the sum of the two numbers above it in the preceding row, as is shown in the initial segment in the diagram, is called Pascal's triangle.
So, for example the last row of the triangle contains the sequence of the coefficients of a binomial of the 5th power.
n 1
1 1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
5 1 5 10 10 5 1
- 1 - - - - - 1
Example:
 
 
 
 
 
The difference of two squares, multiplying
  (a - b) (a + b) = a2 - ab + ab - b2 = a2 - b2  
Examples:   a)  (x - 2y) (x + 2y)  = x2 - 2xy + 2xy - 4y2  = x2 - 4y2
b)  (3a + 1) (3a - 1)  = 9a2 + 3a - 3a - 1  = 9a2 - 1
The difference of two squares, factoring
  a2 - b2 = (a - b) (a + b  
Examples:   a)  1 - 16y2  = 12 - (4y)2  = (1 - 4y) (1 + 4y)
b)  1/9a4 - 0.0001  = (1/3a2)2 - (0.01)2  = (1/3a2 - 0.01) (1/3a2 + 0.01)
The difference of two cubes
  a3 - b3 = (a - b) (a2 + ab + b2)  
Examples:   a)  8x3 - 125  = (2x)3 - 53  = (2x - 5) (4x2 + 10x + 25)
b)  1 - 27a3  = 13 - (3a)3  = (1 - 3a) (1 + 3a + 9a2)
The sum of two cubes
  a3 + b3 = (a + b) (a2 - ab + b2)  
Examples:   a)  8 + x3  = 23 + x3  = (2 + x) (4 - 2x + x2)
b)  64a3 + 0.001  = (4a)3 + 0.13  = (4a + 0.1) (16a2 - 0.4a + 0.01)
Pre-calculus contents A
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