# 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)

### More References and links

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

High 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

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