 # 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 Polynomials

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)