Multiplying Special Binomials
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 have a sum of 0. So the product of a sum and a
difference of the same two terms is equal to the difference of two squares.
The Product of a Sum and a Difference
(a + b)(a  b) = a^{2 } b^{2}
Example 1
Product of a sum and a difference
Find each product
a) (x + 2)(x  2)
b) (b + 7)(b  7)
c) (3x  5)(3x + 5)
Solution
a) (x + 2)(x  2) = x^{2}  4
b) (b + 7)(b  7) = b^{2}  49
c) (3x  5)(3x + 5) = 9x^{2}  25
Higher Powers of Binomials
To find a power of a binomial that is higher thatn 2, we can use the rule for
squaring a binomial along with the method of multiplying binomials using the
distributive property. Finding the second or higher power of a binomial is
called expandin the binomial because the result has more terms than the
original.
Example 2
Higher powers of a binomial
Expand each binomial.
a) (x + 4)^{3}
b) (y  2)^{4}
Solution
a) (x + 4)^{3} 
= (x + 4)^{2}(x + 4) 

= (x^{2} + 8x + 16)(x + 4) 

= (x^{2} + 8x + 16)x + (x^{2} + 8x + 16)4 

= x^{3} + 8x^{2} + 16x + 4x^{2} +
32x + 64 

= x^{3} + 12x^{2} + 48x + 64 
b) (y  2)^{4} 
= (y  2)^{2}(y  2)^{2} 

= (y^{2}  4y + 4)(y^{2}  4y + 4) 

= (y^{2}  4y + 4)(y^{2}) + (y^{2}
 4y + 4)(4y) + (y^{2}  4y + 4)(4) 

= y^{4}  4y^{3} + 4y^{2 } 4y^{3}
+ 16y^{2 } 16y + 4y^{2}  16y + 16 

= y^{4}  8y^{3} + 24y^{2} + 16 
