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^{ 2} + 2 a b + b^{ 2} = (a + b)^{ 2} to write the given polynomial as a square as follows: 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^{ 2}  2 a b + b^{ 2} = (a  b)^{ 2} to write the given polynomial as a square as follows: 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^{ 3}  b^{ 3} = (a  b)(a^{ 2} + ab + b^{ 2}) to write the given polynomial in factored as follows: 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} + 1 Solution The two monomials making the given polynomial may be written as follows: 8 y^{ 3} = (2 y)^{ 3} and 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^{ 3} + b^{ 3} = (a + b)(a^{ 2}  ab + b^{ 2}) to write the given polynomial in factored as follows: 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}  64 f) 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^{ 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) 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^{ 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}) 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} 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} e) 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) 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)
