Solutions to Factoring of Special Polynomials

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

REVIEW - 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 + ab + b 2)

Sum of two cubes: a 3 + b 3 = (a + b)(a 2 - ab + 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 = 5x 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) = (-5x + 3)(5x + 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 writte 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 - ab + 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 Middle School Math (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers

More High School Math (Grades 10, 11 and 12) - Free Questions and Problems With Answers

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

Author - e-mail

Home Page


Updated: 20 August 2017 (A Dendane)