## Square of a Binomial SquareRootFunctions

#### 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.

1. Square of a Binomial

1. (a + b)2 = a2 + 2ab + b2

2. (a - b)2 = a2 - 2ab + b2

2. Difference of Two Squares

1. (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

I-b(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

1. (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

2. (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).

3. (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

4. (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.

5. (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

1. (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

2. (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

#### Find each product

1. (c + d)2

2. (6w + 7)2

3. (4ab - 5)2

4. (y - z)2

5. (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

1. (c + d)2

a = c and b = d

(c)2 + 2(c)(d) + (d)2

c2 + 2cd + d2

2. (6w + 7)2

a = 6w and b = 7

(6w)2 + 2(6w)(7) + (7)2

36w2 + 84w + 49

3. (4ab - 5)2

a = 4ab and b = 5

(4ab)2 - 2(4ab)(5) + (5)2

16a2b2 - 40ab + 25

4. (y - z)2

a = y and b = z

(y)2 - 2(y)(z) + (z)2

y2 - 2yz + z2

5. (3m2 - 8n2)2

a = 3m2 and b = 8n2

(3m2)2 - 2(3m2)(8n2) + (8n2)2

9m4 - 48m2n2 + 64n4

6. (5 + 4d)(5 - 4d)

a = 5 and b = 4d

(5)2 - (4d)2

25 - 16d2

7. #### Reasoning

8. 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