Square of a Binomial and Difference of Two Squares
Special Products
In Algebra, there are some special products that reoccur. For these special products, we will develop patterns to use to find their products.

Square of a Binomial
(a + b)2 = a2 + 2ab + b2
(a  b)2 = a2  2ab + b2
Difference of Two Squares

(a + b)(a  b) = a2  b2
To develop the patterns, we will use FOIL.
Since (a + b)2 = (a + b)(a + b) and (a  b)2 = (a  b)(a  b), using FOIL:
F L F L
(a + b)(a + b)
O I I O
a2 + ab + ab + b2
a2 + 2ab + b2
Reasoning
Using FOIL:
F a(a) = a2
O a(b) = ab
Ib(a) = ab
L b(b) = b2
Combine like terms:
ab + ab = 2ab
F L F L
(a  b)(a  b)
O I I O
a2  ab  ab + b2
a2  2ab + b2
Reasoning
Using FOIL:
F a(a) = a2
O a(b) = ab
Ib(a) = ab
L b(b) = b2
Combine like terms:
ab + ab = 2ab
Remember: When combining like terms, you add the coefficients.
Therefore, ab + ab = 1ab + 1ab = 2ab and ab  ab = 1ab  1ab = 2ab
You add exponents only if you are multiplying.
Square of a Binomial
We can think of the square of a binomial:
(a + b)2 = a2 + 2ab + b2 and (a  b)2 = a2  2ab + b2
in terms of a and b where a is the 1st term and b is the 2nd term.
So, (a + b)2 becomes (1st term)2 + 2(1st term)(2nd term) + (2nd term)2
and (a  b)2 becomes (1st term)2  2(1st term)(2nd term) + (2nd term)2
Whichever way you remember how to square a binomial, first identify the values of a and b and then substitute them
into the special products.
Examples

(m + n)2
Reasoning
a = m
b = n
(m)2 + 2(m)(n) + (n)2
m2 + 2mn + n2
Identify the values of a and b.
Since a is the 1st term, a = m
Since b is the 2nd term, b = n
Substitute m for a and
n for b in the special product
(a + b)2 = a2 + 2ab + b2 to obtain
(m + n)2 = (m)2 + 2(m)(n) + (n2)
Simplify: m2 + 2mn + n2

(3a + 2)2
Reasoning
a = 3a
b = 2
(3a)2 + 2(3a)(2) + (2)2
9a2 + 12a + 4
Identify the values of a and b.
Since a is the 1st term, a = 3a
Since b is the 2nd term, b = 2
Substitute 3a for a and 2 for b in the special product
(a + b)2 = a2 + 2ab + b2 to obtain
(3a + 2)2 = (3a)2 + 2(3a)(2) + (2)2
Simplify: 9a2 + 12a + 4
When simplifying (3a)2, remember to square both
3: (3 3 = 9) and a: (a a = a2).

(4x2 + 5y)2
Reasoning
a = 4x2
b = 5y
(4x2)2 + 2(4x2)(5y) + (5y)2
16x4 + 40x2y + 25y2
Identify the values of a and b.
Since a is the 1st term, a = 4x2
Since b is the 2nd term, b = 5y
Substitute 4x2 for a and 5y for b in the special product
(a + b)2 = a2 + 2ab + b2 to obtain
(4x2 + 5y)2 = (4x2)2 + 2(4x2)(5y) + (5y)2
Simplify: 16x4 + 40x2y + 25y2
When simplifying (4x2)2, remember to square both the 4: (42 = 4 4 = 16) and x2. To square the x2,
remember that parentheses tell us to multiply, so we multiply exponents 2 and 2, that is 2(2) = 4; so (x2)2=x4

(4c  7)2
Reasoning
a = 4c
b = 7
(4c)2  2(4c)(7) + (7)2
16c2  56c + 49
Identify the values of a and b.
Since a is the 1st term, a = 4c
Since b is the 2nd term, b = 7
Substitute 4c for a and 7 for b in the special product
(a  b)2 = a2  2ab + b2 to obtain
(4c  7)2 = (4c)2  2(4c)(7) + (7)2
Simplify: 16c2  56c + 49
Notice when using the special product
(a  b)2 = a2  2ab + b2, the value of b is not negative. The minus sign is taken care of for us by the
special product.

(5p3  3q2)2
Reasoning
a = 5p3
b = 3q2
(5p3)2  2(5p3)(3q2) + (3q2)2
25p6  30p3q2 + 9q4
Identify the values of a and b.
Since a is the 1st term, a = 5p3
Since b is the 2nd term, b = 3q2
Substitute 5p3 for a and 3q2 for b in the special product
(a  b)2 = a2  2ab + b2 to obtain
(5p3  3q2)2 = (5p3)2  2(5p3)(3q2) + (3q2)2
Simplify: 25p6  30p3q2 + 9q4
Again, when simplifying powers, remember to multiply exponents.
Difference of Two Squares
As in the square of a binomial, we can think of the difference of two squares (a + b)(a  b) = a2  b2 in terms of a
and b. So, (a + b)(a  b) becomes(1st term)2  (2nd term)2.
Whichever way you choose to simplify the difference of two squares, first identify the values of a and b and then substitute them into the special product.
Examples

(6a  5)(6a + 5)
Reasoning
a = 6a
b = 5
(6a)2  (5)2
36a2  25
Identify the values of a and b.
Since a is the 1st term, a = 6a
Since b is the 2nd term, b = 5
Substitute 6a for a and 5 for b in the special product
(a + b)(a  b) = a2  b2 to obtain
(6a  5)(6a + 5) = (6a)2  (5)2
Simplify: 36a2 – 25

(8a3b2 + 5)(8a3b2  5)
Reasoning
a = 8a3b2
b = 5
(8a3b2)2  (5)2
64a6b4  25
Identify the values of a and b.
Since a is the 1st term, a = 8a3b2
Since b is the 2nd term, b = 5
Substitute 8a3b2 for a and 5 for b in the special product
(a + b)(a  b) = a2  b2 to obtain
(8a3b2 + 5)(8a3b2  5) = (8a3b2)2  (5)2
Simplify: 64a6b4  25
Remember, when raising a product to a power (8a3b2)2 to multiply the exponent outside the parentheses by
each exponent inside.
(81a3b2)2
For 8: 1(2) = 2 or 82 = 64
For a: 3(2) = 6 or a6
For b: 2(2) = 4 or b4
We can use the square of a binomial to find the area of a square. A = s2.
Example: Find the area of a square when the length of a side is seven more than five times a number.

Reasoning

A = s2
s = 5n + 7
A = (5n + 7)2
a = 5n
b = 7
A = (5n)2 + 2(5n)(7) + (7)2
A = 25n2 + 70n + 49 
We must find the length of a side of the square. Seven more than tells us to add 7 to five times a number (5n), or simply 5n + 7 
Identify the values of a and b.
Since a is the 1st term, a = 5n
Since b is the 2nd term, b = 7
Substitute 5n for a and 7 for b in the special product
(a + b)2 = a2 + 2ab + b2 to obtain
(5n + 7)2 = (5n)2 + 2(5n)(7) + (7)2
Simplify: 25n2 + 70n + 49
Try these examples:
Find each product
(c + d)2
(6w + 7)2
(4ab  5)2
(y  z)2
(3m2  8n2)2
Problem Solving
Find the area of a square if the length of a side is 7 less than four times a number.
A = s2
Explain your reasoning.
Answers to Practice Problems
(c + d)2
a = c and b = d
(c)2 + 2(c)(d) + (d)2
c2 + 2cd + d2
(6w + 7)2
a = 6w and b = 7
(6w)2 + 2(6w)(7) + (7)2
36w2 + 84w + 49
(4ab  5)2
a = 4ab and b = 5
(4ab)2  2(4ab)(5) + (5)2
16a2b2  40ab + 25
(y  z)2
a = y and b = z
(y)2  2(y)(z) + (z)2
y2  2yz + z2
(3m2  8n2)2
a = 3m2 and b = 8n2
(3m2)2  2(3m2)(8n2) + (8n2)2
9m4  48m2n2 + 64n4
(5 + 4d)(5  4d)
a = 5 and b = 4d
(5)2  (4d)2
25  16d2
Reasoning
A = s2
A = (4n  7)2
A = 4n and b = 7
A = (4n)2  2(4n)(7) + (7)2
A = 16n2  56n + 49
The area of a square is A = s2 because the lengths of all the sides of a square are equal.
Find the length of a side from 7 less than which tells us to subtract 7 from 4 times a number 4n for 4n  7
Identify that a = 4n and b = 7
Substitute into the square of a binomial
(4n)2  2(4n)(7) + (7)2
Simplify: 16n2  56n + 49