# Factoring of Special Polynomials

Questions With Detailed Solutions

How to use special polynomial forms to factor other polynomials? Grade 11 maths questions are presented along with detailed Solutions and explanations. We will use five special polynomial forms.

## 1 - Difference of two squaresa^{ 2} - b^{ 2} = (a - b)(a + b)Examples: Factor the polynomial. 16 x^{ 2} - 9 y^{ 2}Solution: Note that 16 x^{ 2} = (4 x)^{ 2} and 9 y^{ 2} = (3 y)^{ 2}We can write 16 x^{ 2} - 9 y^{ 2} = (4 x)^{ 2} - (3 y)^{ 2}Now that we have written the given polynomial as the the difference of two squares, we use formula above to factor the given polynomial as follows: 16 x^{ 2} - 9 y^{ 2} = (4 x)^{ 2} - (3 y)^{ 2} = (4 x - 3 y)(4 x + 3 y)2) ## 2 - Trinomial Perfect Squarea) a^{ 2} + 2 a b + b^{ 2} = (a + b)^{ 2}b) a^{ 2} - 2 a b + b^{ 2} = (a - b)^{ 2}Examples: Factor the polynomials. 4 x^{ 2} + 20 x y + 25 y^{ 2}Solution: Note that the monomials making the given polynomial may be written as follows: 4 x^{ 2} = (2 x)^{ 2}, 20 x y = 2(2 x)(5 y) and 25 y^{ 2} = (5 y)^{ 2}.
We now write the given polynomial as follows 4 x^{ 2} + 10 x y + 25 y^{ 2} = (2 x)^{ 2} + 2(2 x)(5 y) + (5 y)^{ 2}Use the formula a to write the given polynomial as a square as follows:
^{ 2} + 2 a b + b^{ 2} = (a + b)^{ 2}4 x^{ 2} + 20 x y + 25 y^{ 2} = (2 x)^{ 2} + 2(2 x)(5 y) + (5 y)^{ 2} = (2 x + 5 y)^{ 2}Example: Factor the polynomials. 1 - 6 x + 9 x^{ 2}Solution: Note that the monomials making the given polynomial may be written as follows: 1 = 1, ^{ 2} - 6 x = - 2(3)x and 9 x.
^{ 2} = (3 x)^{ 2}The given polynomial may be written as follows 1 - 6 x + 9 x^{ 2} = 1^{ 2} - 2(3) x + (3 x)^{ 2}Use the formula a to write the given polynomial as a square as follows:
^{ 2} - 2 a b + b^{ 2} = (a - b)^{ 2}1 - 6 x + 9 x^{ 2} = 1^{ 2} - 2(3) x + (3 x)^{ 2} = (1 - 3 x)^{ 2}## 3 - Difference of two cubesa^{ 3} - b^{ 3} = (a - b)(a^{ 2} + a b + b^{ 2})Example: Factor the polynomial. 8 - 27 x^{ 3}Solution: Note that the monomials making the given polynomial may be written as follows: 8 = (2)^{ 3} and 27 x^{ 3} = (3 x)^{ 3}The given polynomial may now be written as follows 8 - 27 x^{ 3} = (2)^{ 3} - (3 x)^{ 3}Use the formula a to write the given polynomial in factored as follows:
^{ 3} - b^{ 3} = (a - b)(a^{ 2} + ab + b^{ 2})8 - 27 x^{ 3} = (2)^{ 3} - (3 x)^{ 3} = (2 - 3 x)( (2)^{ 2} + (2)(3x) + (3 x)^{ 2}) = (2 - 3 x)(9 x^{ 2} + 6x + 4)## 4 - Sum of two cubesa^{ 3} + b^{ 3} = (a + b)(a^{ 2} - a b + b^{ 2})Example: Factor the polynomial. 8 y^{ 3} + 1Solution: The two monomials making the given polynomial may be written as follows: 8 y and ^{ 3} = (2 y)^{ 3}1 = (1)^{ 3}The polynomial to factor may now be written as follows 8 y^{ 3} + 1 = (2 y)^{ 3} + (1)^{ 3}Use the formula a to write the given polynomial in factored as follows:
^{ 3} + b^{ 3} = (a + b)(a^{ 2} - ab + b^{ 2})8 y^{ 3} + 1 = (2 y)^{ 3} + (1)^{ 3} = (2 y + 1)( (2 y)^{ 2} - (2 y)(1) + (1)^{ 2}) = (2 y + 1)(4 y^{ 2} - 2 y + 1)
## Factor the following special polynomialsDetailed Solutions and explanations to these questions. 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} - 64f) x ^{ 6} - 1Detailed Solutions and explanations to these questions. |

### 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

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