# Solutions to Factoring of Special Polynomials

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})
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} - 64f) 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 and factor the given polynomial as follows:^{ 2} - b^{ 2} = (a - b)(a + b)- 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 and factor the given polynomial as follows:^{ 2} - b^{ 2} = (a - b)(a + b)16 y ^{ 4} - x^{ 4} = (4 y^{ 2})^{ 2} - (x^{ 2})^{ 2} = (4y^{ 2} - x^{ 2})(4y^{ 2} + x^{ 2})The term (4y in the above is the sum of two squares and cannot be factored using real numbers. However the term (4y^{ 2} + x^{ 2})^{ 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)Factor36 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 to factor the given polynomial as follows:^{ 2} - 2 a b + b^{ 2} = (a - b)^{ 2}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 to factor ^{ 2} + 2 a b + b^{ 2} = (a + b)^{ 2}x and the given polynomial as follows:^{ 2} + 2 x + 1 = x ^{ 2} + 2(x)(1) + 1^{ 2}(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 to factor the given polynomial as follows:^{ 3} + b^{ 3} = (a + b)(a^{ 2} - a b + b^{ 2}) - 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)Factorx ^{ 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 and factor the given polynomial as follows:^{ 2} - b^{ 2} = (a - b)(a + b)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) |

### More References and links

Middle School Maths (Grades 6, 7, 8, 9) - Free Questions and Problems With AnswersHigh School Maths (Grades 10, 11 and 12) - Free Questions and Problems With Answers

Primary Maths (Grades 4 and 5) with Free Questions and Problems With Answers

Home Page