Product of a Sum and a Difference
If we multiply the sum a + b and the difference a  b by using FOIL, we get
(a + b)(a  b) 
= a^{2}  ab + ab  b^{2} 

= a^{2}  b^{2} 
The inner and outer products add up to zero, canceling each other out. So the
product of a sum and a difference is the difference of two squares, as shown in the
following rule.
Rule for the Product of a Sum and a Difference
(a + b)(a  b) = a^{2}  b^{2}
Example 1
Finding the product of a sum and a difference
Find the products.
a) (x + 3)(x  3)
b) (a^{3} + 8)(a^{3}  8)
c) (3x^{2}  y^{3})(3x^{2} + y^{3})
Solution
a) (x + 3)(x  3) = x^{2}  9
b) (a^{3} + 8)(a^{3}  8) = a^{6}  64
c) (3x^{2}  y^{3})(3x^{2} + y^{3}) = 9x^{4}
 y^{6}
Helpful hint
You can use
(a + b)(a  b) = a^{2}  b^{2} to perform mental arithmetic
tricks such as
59 Â· 61 = 3600  1 = 3599. What is 49 Â· 51? 28 Â· 32?
The square of a sum, the square of a difference, and the product of a sum and a
difference are referred to as special products. Although the special products can be
found by using the distributive property or FOIL, they occur so frequently in algebra
that it is essential to learn the new rules. In the next example we use the special
product rules to multiply two trinomials and to square a trinomial.
Example 2
Using special product rules to multiply trinomials
Find the products.
a) [(x + y) + 3][(x + y)  3]
b) [(m  n) + 5]^{2}
Solution
a) Use the rule (a + b)(a  b) = a^{2}  b^{2} with a = x
+ y and b = 3:
[(x + y) + 3][(x + y)  3] 
= (x + y)^{2}  32 

= x^{2} +2xy + y^{2}  9 
b) Use the rule (a + b)^{2} = a^{2} + 2ab + b^{2} with a
= m  n and b = 5:
[(m  n) + 5]^{2} 
= (m  n)^{2} + 2(m  n)5 + 5^{2} 

= m^{2}  2mn + n^{2} + 10m  10n + 25 
