1) Difference of two squares: a^{ 2} - b^{ 2} = (a - b)(a + b)
2)Trinomial Perfect Square
a) a^{ 2} + 2 a b + b^{ 2} = (a + b)^{ 2}
b) a^{ 2} - 2 a b + b^{ 2} = (a - b)^{ 2}
Difference of two cubes: a^{ 3} - b^{ 3} = (a - b)(a^{ 2} + a b + b^{ 2})
Sum of two cubes: a^{ 3} + b^{ 3} = (a + b)(a^{ 2} - a b + b^{ 2})

Factor the following special polynomials
a) - 25 x^{ 2} + 9
b) 16 y ^{ 4} - x^{ 4}
c) 36 y ^{ 2} - 60 x y + 25 x ^{ 2}
d) (1/2) x ^{ 2} + x + (1/2)
e) - y ^{ 3} - 64
f) x ^{ 6} - 1

Question a)

Factor - 25 x^{ 2} + 9
Solution
a) If we let a = 5 x and b = 3, the given polynomial may be written as:
- 25 x^{ 2} + 9 = - a^{ 2} + b^{ 2}
Use the special polynomial a^{ 2} - b^{ 2} = (a - b)(a + b) and factor the given polynomial as follows:
- 25 x^{ 2} + 9 = - a^{ 2} + b^{ 2} = (- a + b)(a + b) = (-5 x + 3)(5 x + 3)

Question b)

Factor 16 y ^{ 4} - x^{ 4}
Solution
b) The given polynomial has the form of the difference of two squares and may be writtesn as:
16 y ^{ 4} - x^{ 4} = (4 y^{ 2})^{ 2} - (x^{ 2})^{ 2}
Use the special polynomial a^{ 2} - b^{ 2} = (a - b)(a + b) and factor the given polynomial as follows:
16 y ^{ 4} - x^{ 4} = (4 y^{ 2})^{ 2} - (x^{ 2})^{ 2} = (4y^{ 2} - x^{ 2})(4y^{ 2} + x^{ 2})
The term (4y^{ 2} + x^{ 2}) in the above is the sum of two squares and cannot be factored using real numbers. However the term (4y^{ 2} - x^{ 2}) is the difference of two squares and can be further factored. Hence the given polynomial is factored as follows:
16 y ^{ 4} - x^{ 4} = (2 y - x)(2 y + x)(4y^{ 2} + x^{ 2})

Question c)

Factor 36 y ^{ 2} - 60 x y + 25 x ^{ 2}
Solution
c) The given polynomial may be written as:
36 y ^{ 2} - 60 x y + 25 x ^{ 2} = (6 y)^{ 2} - 2(6 y)(5 x) + (5 x)^{ 2}
Use the special trinomial a^{ 2} - 2 a b + b^{ 2} = (a - b)^{ 2} to factor the given polynomial as follows:
36 y ^{ 2} - 60 x y + 25 x ^{ 2} = (6 y)^{ 2} - 2(6 y)(5 x) + (5 x)^{ 2} = (6 y - 5 x)^{ 2}

Question d)

Factor (1/2) x ^{ 2} + x + (1/2)
Solution
d) Factor (1/2) out and rewrite the given polynomial as:
(1/2) x ^{ 2} + x + (1/2) = (1/2) x ^{ 2} + 2 (1/2) x + (1/2) = (1/2)( x ^{ 2} + 2 x + 1)
Use the special trinomial a^{ 2} + 2 a b + b^{ 2} = (a + b)^{ 2} to factor x ^{ 2} + 2 x + 1 = x ^{ 2} + 2(x)(1) + 1^{ 2} and the given polynomial as follows:
(1/2) x ^{ 2} + x + (1/2) = (1/2)( x ^{ 2} + 2 x + 1) = (1/2)(x + 1)^{ 2}

Question e)

Factor - y ^{ 3} - 64
Solution
d) Factor - 1 out and rewrite the given polynomial as:
- y ^{ 3} - 64 = - (y ^{ 3} + 64) = - ( y ^{ 3} + 4 ^{ 3})
Use a^{ 3} + b^{ 3} = (a + b)(a^{ 2} - a b + b^{ 2}) to factor the given polynomial as follows:
- y ^{ 3} - 64 = - (y ^{ 3} + 64) = - ( y ^{ 3} + 4 ^{ 3}) = -(y + 4)(y^{ 2} - (y)(4) + 4^{ 2}) = -(y + 4)(y^{ 2} - 4 y + 16)

Question f)

Factor x ^{ 6} - 1
Solution
f) Let us write the given polynomial as the difference of two squares as follows:
x ^{ 6} - 1 = (x^{ 3})^{ 2} - (1)^{ 2}
Use the special difference of squares polynomial a^{ 2} - b^{ 2} = (a - b)(a + b) and factor the given polynomial as follows:
x ^{ 6} - 1 = (x^{ 3})^{ 2} - (1)^{ 2} = (x^{ 3} - 1)(x^{ 3} + 1)
In the above we have the product of the sum and difference of two cubes. Hence
x ^{ 6} - 1 = (x^{ 3})^{ 2} - (1)^{ 2} = (x^{ 3} - 1)(x^{ 3} + 1) = (x - 1)(x^{ 2} + x + 1)(x + 1)(x^{ 2} - x + 1)