# 4.7: Special Binomial Products

Three binomial products occur so frequently in algebra that we designate them as special binomial products. We have seen them before, but we will study them again because of their importance as time saving devices and in solving equations (which we will study in a later chapter).

These special products can be shown as the squares of a binomial

((a+b)^2) and (a-b)^2)

and as the sum and difference of two terms.

((a+b)(a-b))

There are two simple rules that allow us to easily expand (multiply out) these binomials. They are well worth memorizing, as they will save a lot of time in the future.

## Expanding ((a+b)^2) and ((a−b)^2)

Squaring a Binomial

To square a binomial:

1. Square the first term.

2. Take the product of the two terms and double it.

3. Square the last term.

4. Add the three results together

((a+b)^2 = a^2 + 2ab + b^2)

((a-b)^2 = a^2 - 2ab + b^2)

## Expanding (a+b)(a−b)

Sum and Difference of Two Terms.

To expand the sum and difference of two terms:†

1. Square the first term and square the second term.
2. Subtract the square of the second term from the square of the first term.

((a+b)(a-b) = a^2 - b^2)

## Sample Set A

Example (PageIndex{1})

(
(x+4)^{2}
)
Square the first term: (x^{2}).
The product of both terms is (4x). Double it: (8x).
Square the last term: 16.

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

Note that ((x+4)^{2} eq x^{2}+4^{2}). The (8x) term is missing!

Example (PageIndex{2})

(
(a-8)^{2}
)
Square the first term: (a^{2}).
The product of both terms is (-8a). Double it: (-16a).
Square the last term: 64.

Add them together: (a^2 + (-16a) + 64)

((a-8)^2 = a^2 - 16a + 64)

Notice that the sign of the last term in this expression is “(+).” This will always happen since the last term results from a number being squared. Any nonzero number times itself is always positive.

((+)(+) = +) and ((-)(-) = +)

The sign of the second term in the trinomial will always be the sign that occurs inside the parentheses.

Example (PageIndex{3})

(
(y-1)^{2}
)
Square the first term: (y^{2}).
The product of both terms is (-y). Double it: (-2y).
Square the last term: +1.

Add them together: (y^2 + (-2y) + 1)

Example (PageIndex{4})

(
(5x+3)^{2}
)
Square the first term: (25x^{2}).
The product of both terms is (15x). Double it: (30x).
Square the last term: 9.

Add them together: (25x^2 + 30x + 9)

Example (PageIndex{5})

(
(7b-2)^{2}
)
Square the first term: (49b^{2}).
The product of both terms is (-14b). Double it: (-28b).
Square the last term: 4.

Add them together: (49b^2 + (-28b) + 4)

Example (PageIndex{6})

(
(x+6)(x-6)
)
Square the first term: (x^2).
Subtract the square of the second term ((36)) from the square of the first term: (x^2 - 36)

((x+6)(x-6) = x^2 - 36)

Example (PageIndex{7})

(
(4a−12)(4a+12)
)
Square the first term: (16a^2).
Subtract the square of the second term ((144)) from the square of the first term: (16a^2-144)

((4a-12)(4a+12) = 16a^2 - 144)

Example (PageIndex{8})

(
(6x+8y)(6x−8y)
)
Square the first term: (36x^2).
Subtract the square of the second term ((64y^2)) from the square of the first term: (36x^2 - 64y^2)

((6x+8y)(6x-8y) = 36x^2 - 64y^2)

## Practice Set A

Find the following products.

Practice Problem (PageIndex{1})

((x+5)^2)

(x^2 + 10x + 25)

Practice Problem (PageIndex{2})

((x+7)^2)

(x^2 + 14x + 49)

Practice Problem (PageIndex{3})

((y-6)^2)

(y^2 - 12y + 36)

Practice Problem (PageIndex{4})

((3a+b)^2)

(9a^2 + 6ab + b^2)

Practice Problem (PageIndex{5})

((9m-n)^2)

(81m^2 - 18mn + n^2)

Practice Problem (PageIndex{6})

((10x - 2y)^2)

(100x^2 - 40xy + 4y^2)

Practice Problem (PageIndex{7})

((12a - 7b)^2)

(144a^2 - 168ab + 49b^2)

Practice Problem (PageIndex{8})

((5h - 15k)^2)

(25h^2 - 150hk + 225k^2)

## Exercises

For the following problems, find the products.

Exercise (PageIndex{1})

((x+3)^2)

(x^2 + 6x + 9)

Exercise (PageIndex{2})

((x+5)^2)

Exercise (PageIndex{3})

((x+8)^2)

(x^2 + 16x + 64)

Exercise (PageIndex{4})

((x+6)^2)

Exercise (PageIndex{5})

((y+9)^2)

(y^2 + 18y + 81)

Exercise (PageIndex{6})

((y+1)^2)

Exercise (PageIndex{7})

((a-4)^2)

(a^2 - 8a + 16)

Exercise (PageIndex{8})

((a-6)^2)

Exercise (PageIndex{9})

((a-7)^2)

(a^2 - 14a + 49)

Exercise (PageIndex{10})

((b+10)^2)

Exercise (PageIndex{11})

((b+15)^2)

(b^2 + 30b + 225)

Exercise (PageIndex{12})

((a-10)^2)

Exercise (PageIndex{13})

((x-12)^2)

(x^2 - 24x + 144)

Exercise (PageIndex{14})

((x+20)^2)

Exercise (PageIndex{15})

((y-20)^2)

(y^2 - 40y + 400)

Exercise (PageIndex{16})

((3x + 5)^2)

Exercise (PageIndex{17})

((4x + 2)^2)

(16x^2 + 16x + 4)

Exercise (PageIndex{18})

((6x - 2)^2)

Exercise (PageIndex{19})

((7x - 2)^2)

(49x^2 - 28x + 4)

Exercise (PageIndex{20})

((5a - 6)^2)

Exercise (PageIndex{21})

((3a - 9)^2)

(9a^2 - 54a + 81)

Exercise (PageIndex{22})

((3w - 2z)^2)

Exercise (PageIndex{23})

((5a - 3b)^2)

(25a^2 - 30ab + 9b^2)

Exercise (PageIndex{24})

((6t - 7s)^2)

Exercise (PageIndex{25})

((2h - 8k)^2)

(4h^2 - 32hk + 64k^2)

Exercise (PageIndex{26})

((a + dfrac{1}{2})^2)

Exercise (PageIndex{27})

((a + dfrac{1}{3})^2)

(a^2 + dfrac{2}{3}a + dfrac{1}{9})

Exercise (PageIndex{28})

((x + dfrac{3}{4})^2)

Exercise (PageIndex{29})

((x + dfrac{2}{5})^2)

(x^2 + dfrac{4}{5}x + dfrac{4}{25})

Exercise (PageIndex{30})

((x - dfrac{2}{3})^2)

Exercise (PageIndex{31})

((y-dfrac{5}{6})^2)

(y^2 - dfrac{5}{3}y + dfrac{25}{36})

Exercise (PageIndex{32})

((y + dfrac{2}{3})^2)

Exercise (PageIndex{33})

((x + 1.3)^2)

(x^2 + 2.6x + 1.69)

Exercise (PageIndex{34})

((x + 5.2)^2)

Exercise (PageIndex{35})

((a + 0.5)^2)

(a^2 + a + 0.25)

Exercise (PageIndex{36})

((a + 0.08)^2)

Exercise (PageIndex{37})

((x - 3.1)^2)

(x^2 - 6.2x + 9.61)

Exercise (PageIndex{38})

((y - 7.2)^2)

Exercise (PageIndex{39})

((b - 0.04)^2)

(b^2 - 0.08b + 0.0016)

Exercise (PageIndex{40})

((f - 1.006)^2)

Exercise (PageIndex{41})

((x + 5)(x - 5))

(x^2 - 25)

Exercise (PageIndex{42})

((x+6)(x-6))

Exercise (PageIndex{43})

((x+1)(x−1))

(x^2 - 1)

Exercise (PageIndex{44})

((t−1)(t+1))

Exercise (PageIndex{45})

((f+9)(f−9))

(f^2 - 81)

Exercise (PageIndex{46})

((y−7)(y+7))

Exercise (PageIndex{47})

((2y+3)(2y−3))

(4y^2 - 9)

Exercise (PageIndex{48})

((5x+6)(5x−6))

Exercise (PageIndex{49})

((2a−7b)(2a+7b))

(4a^2 - 49b^2)

Exercise (PageIndex{50})

((7x+3t)(7x−3t))

Exercise (PageIndex{51})

((5h−2k)(5h+2k))

(25h^2 - 4k^2)

Exercise (PageIndex{52})

((x + dfrac{1}{3})(x - dfrac{1}{3}))

Exercise (PageIndex{53})

((a + dfrac{2}{9})(a - dfrac{2}{9}))

(a^2 - dfrac{4}{81})

Exercise (PageIndex{54})

((x + dfrac{7}{3})(x - dfrac{7}{3}))

Exercise (PageIndex{55})

((2b + dfrac{6}{7})(2b - dfrac{6}{7}))

(4b^2 - dfrac{36}{49})

Exercise (PageIndex{56})

Expand ((a+b)^2) to prove it is equal to (a^2 + 2ab + b^2).

Exercise (PageIndex{57})

Expand ((a-b)^2) to prove it is equal to (a^2 - 2ab + b^2).

((a-b)(a-b) = a^2 - ab - ab + b^2 = a^2 - 2ab + b^2)

Exercise (PageIndex{58})

Expand ((a+b)(a-b)) to prove it is equal to (a^2-b^2).

Exercise (PageIndex{59})

Fill in the missing label in the equation below.

First term squared

Exercise (PageIndex{60})

Label the parts of the equation below.

Exercise (PageIndex{61})

Label the parts of the equation below.

a) Square the first term.

b) Square the second term and subtract it from the first term.

## Exercises for Review

Exercise (PageIndex{62})

Simplify ((x^3y^0z^4)^5).

Exercise (PageIndex{63})

Find the value of (10^{-1} cdot 2^{-3})

(dfrac{1}{80})

Exercise (PageIndex{64})

Find the product.

((x+6)(x-7)).

Exercise (PageIndex{65})

Find the product.

((5m - 3)(2m + 3))

(10m^2 + 9m - 9)

Exercise (PageIndex{66})

Find the product.

((a+4)(a^2 - 2a + 3))

Let's take a look at a special rule that will allow us to find the product without using the FOIL method.

The square of a binomial is the sum of: the square of the first terms, twice the product of the two terms, and the square of the last term.

I know this sounds confusing, so take a look..

If you can remember this formula, it you will be able to evaluate polynomial squares without having to use the FOIL method. It will take practice.

Now let's take a look at Example 1 and find the product using our special rule.

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## FOIL Calculator- Multiplying Binomials

This binomial calculator calculates the product of a binomial raised either to the 2nd power or the 3rd power using the FOIL method. The product of the binomial expression is obtained, as with all products, by multiplying two binomial expressions together.

To use this calculator above, we follow the format (ax + b) n . A user just enters the a and b values. S/he can also change the sign and the exponent to which the binomial is raised. By default, the sign and exponent are "+" and "2". The user, however, can change the sign to "-" and the exponent to "3". Thus, the calculator allows for dynamic input.

Once the user clicks "Calculate", the answer will automatically be computed.

### Multiplying Different Binomials

This binomial calculator calculates the product of two binomials which may be either the same or different. If they are the same, you could use the first calculator, but if they are different, then you must use this one.

Again, this calculates the product of the binomials by the FOIL method, the steps which are explained below.

### FOIL Explained

Below is a reference visual of how FOIL works:

The FOIL is a method of binomial product calculation which utilizes the following steps shown below:

Let's say we have the following binomial which is shown above.

If we expand it without using the exponent, it will look like the following below:

Using FOIL, we will calculate this binomial product by the following steps:

First- We take the first term from each binomial and multiply them together. With the binomial expression (ax + b)(ax +b), the first terms are ax and ax. This yields the product a 2 x 2 .

Outer- Next, we take the outer or outside terms of the two binomials. With the binomial expression (ax + b)(ax +b), the outer terms are ax and b. This yields the product abx.

Inner- Next, we take the inner or insider terms of the two binomials. With the binomial expression (ax + b)(ax +b), the inner terms are b and ax. This yields the product abx. We now can combine the outer and inner terms, since they are like terms. Therefore, they simply add up to give the term 2abx.

Last- Next, we take the last terms of the two binomials. With the binomial expression (ax + b)(ax +b), the last terms are b and b. This yields the product b 2 .

So now in total, once we add all the terms up, from this binomial expression, we get the end product of a 2 x 2 + 2abx + b 2 .

FOIL
First- (3x)(3x)=9x 2
Outer- (3x)(4)= 12x
Inner- (4)(3x)= 12x
Last- (4)(4)=16

Total terms added together: 9x 2 + 12x + 12x +16 = 9x 2 + 24x + 16

FOIL
First- (2x)(5x)=10x 2
Outer- (2x)(-7)= -14x
Inner- (3)(5x)= 15x
Last- (3)(-7)= -21

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## Lane ORCCA (2020–2021): Open Resources for Community College Algebra

Since we are now able to multiply polynomials together, we will look at a few special cases of polynomial multiplication.

### Subsection 6.6.1 Squaring a Binomial

###### Example 6.6.1 .

To “square a binomial” is to take a binomial and multiply it by itself. We know that exponent notation means that (4^2=4cdot 4 ext<.>) Applying this to a binomial, we'll see that ((x+4)^2=(x+4)(x+4) ext<.>) To expand this expression, we'll simply distribute ((x+4)) across ((x+4) ext<:>)

Similarly, to expand ((y-7)^2 ext<,>) we'll have:

These two examples might look like any other example of multiplying binomials, but looking closely we can see that something very specific (or special) happened. Focusing on the original expression and the simplified one, we can see that a specific pattern occurred in each:

egin left( y-7 ight)^2 amp= y^2 -highlight<7>y - highlight<7>y + highlight<7cdot 7> left( y-highlight <7> ight)^2 amp= y^2 -2(highlight<7>y) + highlight<7>^2 end

Notice that the two middle terms are not only the same, they are also exactly the product of the two terms in the binomial. Furthermore, the last term is the square of the second term in each original binomial.

What we're seeing is a pattern that relates to two important phrases: The process is called , and the result is called a . The first phrase is a description of what we're doing, we are literally squaring a binomial. The second phrase is a description of what you end up with. This second name will become important in a future chapter.

###### Example 6.6.2 .

The general way this pattern is presented is by squaring the two most general binomials possible, ((a+b)) and ((a-b) ext<.>) We will establish the pattern for ((a+b)^2) and ((a-b)^2 ext<.>) Once we have done so, we will be able to substitute anything in place of (a) and (b) and rely upon the general pattern to simplify squared binomials.

We first must expand ((a+b)^2) as ((a+b)(a+b)) and then we can multiply those binomials:

Notice the final simplification step was to add (ab+ba ext<.>) Since these are like terms, we can combine them into (2ab ext<.>)

Similarly, we can find a general formula for ((a-b)^2 ext<:>)

###### Fact 6.6.3 . Squaring a Binomial Formulas.

If (a) and (b) are real numbers or variable expressions, then we have the following formulas:

These formulas will allow us to multiply this type of special product more quickly.

###### Remark 6.6.4 .

Notice that when both ((a+b)^2) and ((a-b)^2) are expanded in Example 6.6.2, the last term was a positive (b^2) in both. This is because any number or expression, regardless of its sign, is positive after it is squared.

### Subsection 6.6.2 Further Examples of Squaring Binomials

###### Example 6.6.5 .

Expand ((2x-3)^2) using the squaring a binomial formula.

For this example we need to recognize that to apply the formula ((a-b)^2 = a^2-2ab+b^2) in this situation, (a=2x) and (b=3 ext<.>) Expanding this, we have:

###### Remark 6.6.6 .

While we rely on the formula for squaring a binomial in Example 6.6.5, we will often omit the step of formally writing the formula and jump to the simplification, in this way:

###### Example 6.6.7 .

Multiply the following using the squaring a binomial formula:

(displaystyle egin[t] (5xy+1)^2 amp= (5xy)^2+2(5xy)(1)+1^2 amp= 25x^2y^2+10xy+1end )

With this expression, we will first note that the factor of (4) is outside the portion of the expression that is squared. Using the order of operations, we will first expand ((3x-7)^2) and then multiply that expression by (4 ext<:>)

###### Example 6.6.8 .

A circle's area can be calculated by the formula

where (A) stands for area, and (r) stands for radius. If a certain circle's radius can be modeled by (x-5) feet, use an expanded polynomial to model the circle's area.

The circle's area would be:

The circle's area can be modeled by (pi x^2-10pi x+25pi) square feet.

### Subsection 6.6.3 The Product of the Sum and Difference of Two Terms

To identify the next “special case” for multiplying polynomials, we'll look at a couple of examples.

###### Example 6.6.10 .

Multiply the following binomials:

(displaystyle egin[t] (x+5)(x-5) amp= x^2-5x+5x-25 amp= x^2 -25end )

(displaystyle egin[t] (y+8)(y-8) amp= y^2-8y+8y-4 amp= y^2 - 64end )

Notice that for each of these products, we multiplied the sum of two terms by the difference of the same two terms. Notice also in these three examples that once these expressions were multiplied, the two middle terms were opposites and thus canceled to zero.

These pairs, generally written as ((a+b)) and ((a-b) ext<,>) are known as . If we multiply ((a+b)(a-b) ext<,>) we can see this general pattern more clearly:

As with the previous special case, this one also has two names. This can be called the , because this pattern is built on multiplying two binomials that have the same two terms, except one binomial is a sum and the other binomial is a difference. The second name is a , because the end result of the multiplication is a binomial that is the difference of two perfect squares. As before, the second name will become useful in a future chapter when using exactly the technique described in this section will be pertinent.

###### Fact 6.6.11 . The Product of the Sum and Difference of Two Terms Formula.

If (a) and (b) are real numbers or variable expressions, then we have the following formula:

## Sunday, October 22, 2006

### Polynomials: Operations 4.7

4.7 OPERATIONS WITH POLYNOMIALS IN SEVERAL VARIABLES
a. Evaluate a polynomial in several variables for given values of the variables.
b. Identify the coefficients and the degrees of the terms of a polynomial and the degree of a polynomial.
c. Collect terms of a polynomial.
e. Subtract polynomials.
f. Multiply polynomials.

Objective a
Evaluate a polynomial in several variables for given values of the variables.

Objective b
Identify the coefficients and the degrees of the terms of a polynomial and the degree of a polynomial.

Objective c
Collect terms of a polynomial.

Example C Combine like terms.

Objective d

Objective e
Subtract polynomials.

Objective f
Multiply polynomials.

### Polynomials: Operations 4.6

4.6 SPECIAL PRODUCTS
a. Multiply two binomials mentally using the FOIL method.
b. Multiply the sum and the difference of two terms mentally.
c. Square a binomial mentally.
d. Find special products when polynomial products are mixed together.

Objective a
Multiply two binomials mentally using the FOIL method.

Objective b
Multiply the sum and the difference of two terms mentally.

Objective c
Square a binomial mentally.

Objective d
Find special products when polynomial products are mixed together.

### Polynomials: Operations 4.5

4.5 MULTIPLICATION OF POLYNOMIALS
a. Multiply monomials.
b. Multiply a monomial and any polynomial.
c. Multiply two binomials.
d. Multiply any two polynomials.

## Basics of the Binomial Option Pricing Model

With binomial option price models, the assumptions are that there are two possible outcomes—hence, the binomial part of the model. With a pricing model, the two outcomes are a move up, or a move down. The major advantage of a binomial option pricing model is that they’re mathematically simple. Yet these models can become complex in a multi-period model.

In contrast to the Black-Scholes model, which provides a numerical result based on inputs, the binomial model allows for the calculation of the asset and the option for multiple periods along with the range of possible results for each period (see below).

The advantage of this multi-period view is that the user can visualize the change in asset price from period to period and evaluate the option based on decisions made at different points in time. For a U.S-based option, which can be exercised at any time before the expiration date, the binomial model can provide insight as to when exercising the option may be advisable and when it should be held for longer periods.

By looking at the binomial tree of values, a trader can determine in advance when a decision on an exercise may occur. If the option has a positive value, there is the possibility of exercise whereas, if the option has a value less than zero, it should be held for longer periods.

(x + 2) 2  is in the form of (a + b) 2

Comparing  (a + b) 2  and (x + 2) 2 , we get

Write the formula / expansion for  (a + b) 2 .

Substitute x for a and 2 for b.

So, the expansion of  (x + 2) 2  is

(x - 5) 2  is in the form of (a - b) 2

Comparing  (a - b) 2  and (x - 5) 2 , we get

Write the formula / expansion for  (a - b) 2 .

Substitute x for a and 5 for b.

So, the expansion of  (x - 5) 2  is

(5x + 3) 2  is in the form of (a + b) 2

Comparing  (a + b) 2  and (5x + 3) 2 , we get

Write the expansion for  (a + b) 2 .

Substitute 5x for a and 3 for b.

(5x + 3) 2   =  (5x) 2  + 2(5x)(3) + 3 2

So, the expansion of  (5x + 3) 2  is

(5x - 3) 2  is in the form of (a - b) 2

Comparing  (a - b) 2  and (5x - 3) 2 , we get

Write the expansion for  (a - b) 2 .

Substitute 5x for a and 3 for b.

(5x - 3) 2   =  (5x) 2  - 2(5x)(3) + 3 2

So, the expansion of  (5x - 3) 2  is

If a + b  =  7 and a 2 + b 2   =  29, then find the value of ab.

To get the value of ab, we can use the formula or expansion of  (a + b) 2 .

Write the formula / expansion for  (a + b) 2 .

Substitute 7 for (a + b)  and 29 for  (a 2 + b 2 ).

Subtract 29 from each side.

If a - b  =  3 and a 2  + b 2   =  29, then find the value of ab.

To get the value of ab, we can use the formula or expansion of  (a - b) 2 .

Write the formula / expansion for  (a - b) 2 .

Substitute 3 for (a - b)  and 29 for  (a 2  + b 2 ).

Subtract 29 from each side.

(√2 + 1/√2) 2   is in the form of (a + b) 2

Comparing  (a + b) 2  and  (√2 + (1/√2) 2 , we get

Write the expansion for  (a + b) 2 .

Substitute  √2 ਏor a and 1/ √2 ਏor b.

So, the value of  (√2 + 1/√2) 2 is

(√2 - 1/√2) 2   is in the form of (a - b) 2

Comparing  (a - b) 2  and  (√2 - 1/√2) 2 , we get

Write the formula / expansion for  (a - b) 2 .

Substitute  √2 ਏor a and 1/ √2 ਏor b.

So, the value of  (√2 - 1/√2) 2  is

Instead of multiplying 105 by 105 to get the value of ( 105) 2 , we can use the algebraic formula for (a + b) 2 and find the value of (105) 2  easily.

Write  (105) 2  in the form of (a + b) 2 .

Write the expansion for  (a + b) 2 .

Substitute 100 ਏor a and 5 ਏor b.

(100  + 5 ) 2   =  (100 ) 2  + 2(100) (5 ) + (5 ) 2

So, the value of  (10 5 ) 2  is

Instead of multiplying 95 by 95 to get the value of (9 5) 2 , we can use the algebraic formula for (a - b) 2  and find the value of (95) 2  easily.

Write  (95) 2  in the form of (a - b) 2 .

Write the formula / expansion for  (a - b) 2 .

Substitute 100 ਏor a and 5 ਏor b.

(100  - 5 ) 2   =  (100 ) 2  - 2(100) (5 ) + (5 ) 2

So, the value of  (9 5 ) 2  is

After having gone through the stuff given above, we hope that the students would have understood the formula or expansion of (a + b) 2 and the example problems on expansion of (a + b) 2 .

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