The solutions to the questions on factoring of special polynomials in factoring of special polynomials are presented.

Review the Special Polynomials1) 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
Question a)Factor  25 x^{ 2} + 9Solution 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}  64Solution 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}  1Solution 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)
