Factoring Trinomials Using Patterns
If we recognize that a given polynomial follows a certain pattern, we can
use that pattern to factor the polynomial.
Pattern â€”
To Factor a Perfect Square Trinomial
Perfect Square Trinomial: 
a^{2} + 2ab + b^{2} = (a + b)^{2}
a^{2}  2ab + b^{2} = (a  b)^{2} 
Example 1
Factor: 25y^{2}  10x^{2}y + x^{4}
Solution
Check the form of 25y^{2}  10x^{2}y + x^{4}:
â€¢ there are three terms;
â€¢ the first term is a perfect square: 25y^{2} = (5y)^{2};
â€¢ the third term is a perfect square: x^{4} = (x^{2})^{2}; and
â€¢ the second term is twice the product of the factors being squared:
10x^{2}y = 2(5y)(x^{2}).
So, the given polynomial is
a perfect square trinomial:
Substitute 5y for a and x2 for b: 
a^{2}  2ab + b^{2}
(5y)^{2}  2(5y)(x^{2}) + (x^{2})^{2} 
= (a + b)^{2} = (5y  x^{2})^{2} 
Thus, the factorization is (5y  x^{2})^{2}.
You can multiply to check the factorization.
Note:
We can use FOIL to check the
factorization:
(5y  x^{2})^{2} 
= (5y  x^{2})(5y  x^{2})
= 25y^{2}  5x^{2}y  5x^{2}y + x^{4}
= 25y^{2}  10x^{2}y + x^{4} 
The factorization checks.
Pattern â€”
To Factor the Difference of Two Perfect Squares
Difference of Two Perfect Squares: a^{2}  b^{2} = (a + b)(a
 b)
Example 2
Factor: 16w^{2}  81y^{6}.
Solution
Check the form of 16w^{2}  81y^{6}.
â€¢ there are two terms;
â€¢ the first term is a perfect square: 16w^{2} = (4w)^{2};
â€¢ the second term is a perfect square: 81y^{6} = (9y^{3})^{2}.
So, the given polynomial is a
difference of two squares:
Substitute 4w for a and 9y^{3} for b: 
a^{2}  b^{2}
(4w)^{2}  (9y^{3})^{2} 
= (a + b)(a  b) = (4w + 9y^{3})(4w  9y^{3}) 
Thus, the factorization is (4w + 9y^{3})(4w  9y^{3}). You can multiply to check the factorization.
Note:
The polynomial 16w^{2} + 81y^{6} is not a
difference of two squares because the
terms are added rather than subtracted.
The polynomial 16w^{2} + 81y^{6} cannot be
factored using integers.
We can also use patterns to factor the sum or difference of two cubes.
Patterns â€”
To Factor the Sum or Difference of Two Cubes
Sum of Two Cubes: a^{3} + b^{3} = (a + b)(a^{2}  ab
+ b^{2})
Difference of Two Cubes: a^{3}  b^{3} = (a  b)(a^{2} + ab
+ b^{2})
Example 3
Factor: 8y^{3}  27x^{6}
Solution
Check the form of 8y^{3}  27x^{6}:
â€¢ there are two terms;
â€¢ the first term is a perfect cube: 8y^{3} = (2y)^{3};
â€¢ the second term is a perfect cube: 27x^{6} = (3x^{2})^{3}.
So, the polynomial is a difference of two cubes.
Substitute 2y for a and 3x^{2} for b:
Simplify. 
a^{3}  b^{3}
(2y)^{3}  (3x^{2})^{3} 
= (a  b) (a^{2} + ab + b^{2})
= (2y  3x^{2})[(2y)^{2} + (2y)(3x^{2}) + (3x^{2})^{2}]
= (2y  3x^{2})(4y^{2} + 6x^{2}y + 9x^{4})

Thus, the factorization is (2y  3x^{2})(4y^{2} + 6x^{2}y
+ 9x^{4}).
You can multiply to check the factorization.
