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: a² - b² = (a - b)(a + b)
2)Trinomial Perfect Square
a) a² + 2 a b + b² = (a + b)²
b) a² - 2 a b + b² = (a - b)²
Difference of two cubes: a³ - b³ = (a - b)(a² + a b + b²)
Sum of two cubes: a³ + b³ = (a + b)(a² - a b + b²)

Factor the following special polynomials
a) - 25 x² + 9
b) 16 y ⁴ - x⁴
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 x² + 9
Solution
a) If we let a = 5 x and b = 3, the given polynomial may be written as:
- 25 x² + 9 = - a² + b²
Use the special polynomial a² - b² = (a - b)(a + b) and factor the given polynomial as follows:
- 25 x² + 9 = - a² + b² = (- a + b)(a + b) = (-5 x + 3)(5 x + 3)

Question b)

Factor 16 y ⁴ - x⁴
Solution
b) The given polynomial has the form of the difference of two squares and may be writtesn as:
16 y ⁴ - x⁴ = (4 y²)² - (x²)²
Use the special polynomial a² - b² = (a - b)(a + b) and factor the given polynomial as follows:
16 y ⁴ - x⁴ = (4 y²)² - (x²)² = (4y² - x²)(4y² + x²)
The term (4y² + x²) in the above is the sum of two squares and cannot be factored using real numbers. However the term (4y² - x²) is the difference of two squares and can be further factored. Hence the given polynomial is factored as follows:
16 y ⁴ - x⁴ = (2 y - x)(2 y + x)(4y² + x²)

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 a² - 2 a b + b² = (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 a² + 2 a b + b² = (a + b)² to factor x ² + 2 x + 1 = x ² + 2(x)(1) + 1² 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 a³ + b³ = (a + b)(a² - a b + b²) to factor the given polynomial as follows:
- y ³ - 64 = - (y ³ + 64) = - ( y ³ + 4 ³)
= -(y + 4)(y² - (y)(4) + 4²) = -(y + 4)(y² - 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 = (x³)² - (1)²
Use the special difference of squares polynomial a² - b² = (a - b)(a + b) and factor the given polynomial as follows:
x ⁶ - 1 = (x³)² - (1)² = (x³ - 1)(x³ + 1)
In the above we have the product of the sum and difference of two cubes. Hence
x ⁶ - 1 = (x³)² - (1)² = (x³ - 1)(x³ + 1)
= (x - 1)(x² + x + 1)(x + 1)(x² - x + 1)