Special Products of Polynomials

 

Learning Objective(s)

·         Identify and multiply binomial products.

 

Introduction

 

Certain types of binomial multiplication sometimes produce results that are called special products. Special products have predictable terms. Although the distributive property can always be used to multiply any binomials, recognition of those that produce special products provides a problem-solving shortcut.

 

Square of a Sum

 

We call multiplying a number by itself “squaring” because we can represent this multiplication as a square. The number being squared is the length of the side of the square and the product is represented by the area of that square. Consider a square whose side length is described by the binomial x + 3:

 

x + 3

 

 

We can zoom into this square and show a little more detail by creating an image that shows the variable and constant terms as follows:

 

 

 

x

1

1

1

 

 

x

 

 

 

 

x2

 

 

 

 

x

 

 

x

 

 

x

1

x

1

1

1

1

x

1

1

1

1

x

1

1

1

 

From this area model, we can see that the area could be described by the sum of the red, green and yellow pieces. That is, the area is x2 + 6x + 9. So (x + 3)2 = x2 + 6x + 9.

 

We can also use what we know about multiplying binomials to find the square of a sum of two terms by using the distributive property. An example is shown below.

 

Example

Problem

Find (x + 5)2

 

 

 

 

(x + 5)(x + 5)

 

Multiply the binomial by itself – square it.

 

 

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

 

Use the distributive property to multiply each term of the first binomial by the second binomial.

 

x2 + 5x + 5x + 25

 

Group like terms together.

 

x2 + 10x + 25

 

Combine like terms.

Answer

x2 + 10x + 25

 

 

 

 

So far, we have two examples of squaring a sum of two terms, one from constructing an area model and one from algebraic calculations:

 

(x + 3)2 = x2 + 6x + 9

 

(x + 5)2 = x2 + 10x + 25

 

Notice a pattern? In both cases, the first term in the binomial is squared to get the first term in the product (x2 = x2). The last term in the binomial is also squared to get the last term in the product (32 = 9 and 52 = 25).

 

Now what about that middle term? Is there a pattern there, too? It isn't as obvious, but there is one. The middle term of the product is twice the product of the terms of the binomial (2 • 3x = 6x, and 2 • 5x = 10x).

 

Let’s look at another example to see if these patterns hold true.

 

Example

Problem

(2x +6)2

 

 

 

(2x + 6) (2x + 6)

 

Multiply the binomial by itself.

 

2x(2x + 6) + 6(2x + 6)

 

Use the distributive property to multiply each term of the first binomial by the second binomial.

 

4x2 + 12x + 12x + 36

 

Notice there are two 12x terms. One comes from 2x(x + 6) and the other from 6(2x + 6).

 

4x2 + 24x +36

 

Combine like terms.

Answer

4x2 + 24x +36

 

 

 

The same pattern showed up in this product. The term 2x is squared to get 4x2, and 6 is squared to get 36. 2x is multiplied by 6 in two instances, resulting in a term 24x.

 

This pattern will hold true for the square of the sum of any two terms and can be generalized as follows:

 

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

 

Square of a Difference

 

Is there also a pattern when you square the difference between two terms? Yes! Since subtraction can be expressed as adding the opposite, a similar pattern does occur.

 

Consider the square of the binomial (x – 7):

 

Example

Problem

(x – 7)2

 

 

 

(x – 7)(x – 7)

 

Rewrite as multiplication

 

[x + (-7)] [x + (-7)]

 

Rewrite subtraction of 7 as adding (-7)

 

x[x + (-7)] + (-7) [x + (-7)]

 

Multiply each term in one binomial by the other binomial

 

x2 + (-7)x + (-7)x + 49

 

Distribute

 

x2 + (-14)x + 49

 

Combine like terms

Answer

x2 – 14x + 49

 

 

 

If we compare the answer to the original problem, we can see the pattern: the first term, x, is squared to get x2. The middle term in the answer is twice the product of (-7) and x, or (-14x). The last term in the answer is 49, which is the second term of the binomial, (-7), squared. So, the square of a difference can be generalized as follows:

 

(a – b)2 = a2 – 2ab + b2

 

Let’s look at an example in which the coefficient is not 1. Consider the example of squaring the binomial (4s – 3):

 

Example

Problem

(4s – 3)2

 

 

 

(4s – 3) (4s – 3)

 

Rewrite as multiplication

 

 

16s2 – 12s – 12s + 9

 

Multiply each term in one binomial by the other binomial

 

16s2 – 24s + 9

 

Combine like terms

Answer

16s2 – 24s + 9

 

 

 

Notice that when a variable term has a coefficient, it is included in the multiplication. In the example above, the first term in the answer is the entire first term of the binomial squared (4s • 4s = 16s2).  The second term of the answer is twice the product of both terms of the binomial 2(4s • 3). Finally, the last term of the answer is the last term of the binomial squared (3 • 3 = 9).

 

Sound familiar? It's still the same generalized formula as squaring a difference:

 

(a – b)2 = a2 – 2ab + b2

 

Just remember that if the first term has a coefficient, it must be included in the multiplication.

 

Find the product:

(2a – 9)2

 

A) 4a2 – 81

 

B) 4a – 18

 

C) 4a2 – 36a + 81

 

D) 4a2 + 18a – 18

 

Show/Hide Answer

A) Incorrect. When you square binomials you are multiplying. You must multiply each term in the binomial by each term in the binomial. That includes multiplying the 2a by -9 and the -9 by 2a to get -36a as a middle term in the answer. The correct answer is C, 4a2 – 36a + 81.

 

B) Incorrect. The exponent 2 means the binomial is to be squared. You must multiply (2a – 9) by itself.  The correct answer is C, 4a2 – 36a + 81.

 

C) Correct.  (a – b)2 = a2 – 2ab + b2, so (2a – 9)2 = 4a2 – 36a + 81.

 

D) Incorrect. The terms 18a and -18 are incorrect.  The correct answer is C, 4a2 – 36a + 81.

 

 

 

Product of a Sum and a Difference

 

There is a third “special” product to consider among binomials. This one arises from multiplication of the sum of two terms by the difference of the same two terms. In this case, there is also a pattern that describes the product. Here is an example:

 

Example

Problem

(x + 8)(x – 8)

 

 

 

x(x – 8) + 8(x – 8)

 

Multiply each term in the first binomial by the second binomial

 

x2 – 8x + 8x – 64

 

Distribute.

 

x2  – 64

 

Add like terms.

Answer

X2 – 64

 

 

 

 

Notice that the answer to this binomial product is a binomial itself—the difference of two perfect squares. There is no middle term in this case. Why does this happen? Because one is a sum and the other is a difference, but otherwise they use the same numbers, the middle terms are opposites (one positive and one negative). They add to zero. The product of a sum and a difference can be generalized as follows:

 

(a + b)(a – b) = a2 – b2

 

 

Special Binomial Products

 

Product of a Sum: (a + b)2 = a2 + 2ab + b2

 

Product of a Difference: (a – b)2 = a2 – 2ab + b2

 

Product of a Sum & a Difference: (a + b)(a – b) = a2 – b2

 

 

 

Identify Special Products

 

Let’s examine the characteristics of polynomials that result from the “special” products we have been discussing. Could we look at a polynomial and tell whether it is the square of a sum or the square of a difference—either  (a + b)2  or  (a – b)2  —or the product of a sum and a difference, (a + b)(a – b)?

 

One characteristic of special products is that the first and last terms of these polynomials are always perfect squares (a2 and b2). If the first and last terms of a polynomial are perfect squares, the polynomial could be the result of a special product. (To determine if the terms are perfect squares, the polynomial needs to be written with the variable terms in order of decreasing exponents. For example, as x2 + 2x + 1, not 2x + x2 + 1.)

 

If the first and last terms are NOT perfect squares, then the polynomial is not the result of a special product and we can stop looking at other features of the polynomial.

 

If the first and last terms are perfect squares, we need to look further at other features of the polynomial. We need to see if there is a middle term.

 

No middle term: If there is no middle term and the binomial is a difference, then the binomial must be the product of a sum and a difference of two terms. Be careful, though, because the binomial MUST be a difference.

 

Example

Problem

Is 16y2 – t2 a square of a binomial or a product of a sum and a difference?

 

 

 

16y2 = (4y)(4y)

 

t2 = t • t.

 

Are the first and last terms perfect squares?

 

Yes – proceed

 

 

 

 

16y2 – t2

 

Is there a middle term in the polynomial?

 

No. This means it could be the product of a sum and a difference.

 

 

 

 

 

16y2 – t2

 

Is the binomial a difference? In other words, are the two terms combined by subtraction?

 

Yes. This means the binomial is the product of a sum and a difference of two terms.

 

Answer

Yes, 16y2 – t2 is the product of a sum and a difference

(4y + t)(4y – t).

 

 

 

 

Middle term is present: If there is a middle term in the polynomial, we first check to see if the last term is added or subtracted. If the last term is subtracted, then the polynomial is not the result of a special product and we do not need to examine the polynomial further.

 

If the last term is added, the polynomial may be a special product and we have to go yet another step further.

 

Next, we check to see if the middle term is double the product of the square root of the first and the last terms. If the first term is a2 and the last term is b2, is the middle term 2ab? If it is, then the polynomial is either the square of a sum or the square of a difference.

 

That's quite a list of requirements! Let's try an example and see how it works:

 

Example

Problem

Is 4x2 + 20xy + 25y2 the square of binomials or the product of a sum and a difference?

 

 

 

4x2 = (2x)(2x)

 

25y2 = (5y) (5y)

 

Are the first and last terms perfect squares?

 

Yes – proceed

 

 

 

 

4x2 + 20xy + 25y2

 

Is there a middle term in the polynomial?

 

Yes, 20xy.

 

 

 

 

4x2 + 20xy + 25y2

 

Is the last term added in the polynomial?

 

Yes.

 

 

 

 

 

Is the middle term equal to double the product of the square roots of the first and last terms?

 

Yes.  2 • = 20xy. This means the polynomial is either the square of a sum or the square of a difference.

 

Answer

4x2 + 20xy + 25y2 is the square of a binomial  (2x + 5y)

 

 

 

 

Which polynomials shown below are special products?

 

100x2 – 25

x2y2 + 4x + 16

y2 – 18y + 81

 

A) All three are special products.

 

B) Only 100x2 – 25 and  y2 – 18y + 81 are special products.

 

C) Only 100x2 – 25 and x2y2 + 4x + 16 are special products.

 

D) None of the polynomials shown are special products.

 

Show/Hide Answer

A) Incorrect. x2y2 + 4x + 16 is not a special product. The middle term would need to be 8xy. The correct answer is B) 100x2 – 25 and  y2 – 18y + 81 are special products.

 

B) Correct. 100x2 – 25 is the product of a sum and a difference.  y2 – 18y + 81

is a square of a binomial.

 

C) Incorrect. x2y2 + 4x + 16 is not a special product. The middle term would need to be 8xy.  y2 – 18y + 81 is a special product. The correct answer is B) 100x2 – 25 and  y2 – 18y + 81 are special products..

 

D) Incorrect. 100x2 – 25 and  y2 – 18y + 81 are special products. The correct answer is B) 100x2 – 25 and  y2 – 18y + 81 are special products..

 

 

Summary

 

Some products of multiplying binomials follow a predictable pattern that makes it easy to simplify them. These are known as special products.  There are three special products of binomials that each follow a specific formula:

 

Special Binomial Products

 

Product of a Sum: (a + b)2 = a2 + 2ab + b2

 

Product of a Difference: (a – b)2 = a2 – 2ab + b2

 

Product of a Sum & a Difference: (a + b)(a – b) = a2 – b2

 

Polynomials can be identified as special products by examining the characteristics of their terms.