Copyright 2013, 2010, 2007, 2005, Pearson, Education, Inc.
5.4
Special Products
The FOIL Method
When multiplying 2 binomials, the distributive
property can be easily remembered as the FOIL
method.
F – product of First terms
O – product of Outside terms
I – product of Inside terms
L – product of Last terms
Example
Multiply (y – 12)(y + 4).
(y – 12)(y + 4)
Product of First terms is y2
(y – 12)(y + 4)
Product of Outside terms is 4y
(y – 12)(y + 4)
Product of Inside terms is –12y
(y – 12)(y + 4)
Product of Last terms is – 48
F O
I
L
(y – 12)(y + 4) = y2 + 4y – 12y – 48
= y2 – 8y – 48
Example
Multiply (2x – 4)(7x + 5).
L
F
F
O
I
L
(2x – 4)(7x + 5) 2x(7x) + 2x(5) – 4(7x) – 4(5)
=
I
= 14x2 + 10x – 28x – 20
O
= 14x2 – 18x – 20
We multiplied these same two binomials together in the
previous section, using a different technique, but arrived
at the same product.
Special Products
In the process of using the FOIL method on
products of certain types of binomials, we see
specific patterns that lead to special products.
Special Products
Squaring a Binomial
A binomial squared is equal to the square of the first term
plus or minus twice the product of both terms plus the
square of the second term.
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
Example
Multiply. (x + 6)2
(x + 6)2 = (x + 6)(x + 6)
F
=
O
I
L
x2 + 6x + 6x + 36
The inner and outer products are the same.
=
x2 + 12x + 36
Example
Multiply.
a. (12a – 3)2 = (12a)2 – 2(12a)(3) + (3)2
= 144a2 – 72a + 9
b. (x + y)2 = x2 + 2xy + y2
Special Products
Multiplying the Sum and Difference of Two
Terms
The product of the sum and difference of two terms
is the square of the first term minus the square of
the second term.
(a + b)(a – b) = a2 – b2
Example
Multiply.
(2x + 4)(2x – 4)
=
(2x)(2x) + (2x)(– 4) + (4)(2x) + (4)(– 4)
F
=
O
4x2 + (– 8x) + 8x + (–16)
The inner and outer products cancel.
=
4x2 – 16
I
L
Example
Multiply.
a. (5a + 3)(5a – 3) = (5a)2 – 32
= 25a2 – 9
b. (8c + 2d)(8c – 2d) = (8c)2 – (2d)2
= 64c2 – 4d2
Example
Use a special product to multiply, if possible.
a. (7a + 4)2
= 49a2 + 56a + 16
b. (c + 0.2d)(c – 0.2d) = c2 – 0.04d2
c.  y 3  2   2 y 2  1   2 y 5  1 y 3  4 y 2  2

5 
5
5
5
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