# 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 study five special polynomial forms.

## 1 - Difference of two squaresa^{ 2} - b^{ 2} = (a - b)(a + b)Question 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 - Trinomial Perfect Squarea)a^{ 2} + 2 a b + b^{ 2} = (a + b)^{ 2}b) a^{ 2} - 2 a b + b^{ 2} = (a - b)^{ 2}Question 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}Question 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})Question 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})Question 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)
## More Question of Factoring Special PolynomialsFactor the following special polynomialsa) - 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} - 1
## Solutions to Above Questionsa) 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 and factor the given polynomial as follows:^{ 2} - b^{ 2} = (a - b)(a + b)- 25 x^{ 2} + 9 = - a^{ 2} + b^{ 2} = (- a + b)(a + b) = (-5 x + 3)(5 x + 3)
b) The given polynomial has the form of the difference of two squares and may be written as: 16 y ^{ 4} - x^{ 4} = (4 y^{ 2})^{ 2} - (x^{ 2})^{ 2}Use the special polynomial a and factor the given polynomial as follows:^{ 2} - b^{ 2} = (a - b)(a + b)16 y ^{ 4} - x^{ 4} = (4 y^{ 2})^{ 2} - (x^{ 2})^{ 2} = (4y^{ 2} - x^{ 2})(4y^{ 2} + x^{ 2})The term (4y in the above is the sum of two squares and cannot be factored using real numbers. However the term (4y^{ 2} + x^{ 2})^{ 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})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 to factor the given polynomial as follows:^{ 2} - 2 a b + b^{ 2} = (a - b)^{ 2}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}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 to factor ^{ 2} + 2 a b + b^{ 2} = (a + b)^{ 2}x and the given polynomial as follows:^{ 2} + 2 x + 1 = x ^{ 2} + 2(x)(1) + 1^{ 2}(1/2) x ^{ 2} + x + (1/2) = (1/2)( x ^{ 2} + 2 x + 1) = (1/2)(x + 1)^{ 2}e) Factor - 1 out and rewrite the given polynomial as: - y ^{ 3} - 64 = - (y ^{ 3} + 64) = - ( y ^{ 3} + 4 ^{ 3})Use a to factor the given polynomial as follows:^{ 3} + b^{ 3} = (a + b)(a^{ 2} - a b + b^{ 2}) - 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)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 and factor the given polynomial as follows:^{ 2} - b^{ 2} = (a - b)(a + b)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) |

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