Special Products — Elementary Algebra 2e

Square a Binomial Using the Binomial Squares Pattern

Mathematicians like to look for patterns that will make their work easier. A good example of this is squaring binomials. While you can always get the product by writing the binomial twice and using the methods of the last section, there is less work to do if you learn to use a pattern.

Step-by-step expansion of the binomial (x+9)^2 using the FOIL method, from definition to simplified quadratic expression.
Let's start by looking at ${(x + 9)}^{2}$ .
What does this mean?	${(x + 9)}^{2}$
It means to multiply $(x + 9)$ by itself.	$(x + 9) (x + 9)$
Then, using FOIL, we get:	$x^{2} + 9 x + 9 x + 81$
Combining like terms gives:	$x^{2} + 18 x + 81$

Step-by-step algebraic expansion of (y-7)^2, illustrating the FOIL method and simplification of terms.
Here's another one:	${(y - 7)}^{2}$
Multiply $(y - 7)$ by itself.	$(y - 7) (y - 7)$
Using FOIL, we get:	$y^{2} - 7 y - 7 y + 49$
And combining like terms:	$y^{2} - 14 y + 49$

This table demonstrates the step-by-step algebraic expansion of a binomial squared, illustrating the process including the FOIL method.
And one more:	${(2 x + 3)}^{2}$
Multiply.	$(2 x + 3) (2 x + 3)$
Use FOIL:	$4 x^{2} + 6 x + 6 x + 9$
Combine like terms.	$4 x^{2} + 12 x + 9$

Look at these results. Do you see any patterns?

What about the number of terms? In each example we squared a binomial and the result was a trinomial.

{(a + b)}^{2} =____+____+____

Now look at the first term in each result. Where did it come from?

This figure has three columns. The first column contains the expression x plus 9, in parentheses, squared. Below this is the product of x plus 9 and x plus 9. Below this is x squared plus 9x plus 9x plus 81. Below this is x squared plus 18x plus 81. The second column contains the expression y minus 7, in parentheses, squared. Below this is the product of y minus 7 and y minus 7. Below this is y squared minus 7y minus 7y plus 49. Below this is the expression y squared minus 14y plus 49. The third column contains the expression 2x plus 3, in parentheses, squared. Below this is the product of 2x plus 3 and 2x plus 3. Below this is 4x squared plus 6x plus 6x plus 9. Below this is 4x squared plus 12x plus 9.

The first term is the product of the first terms of each binomial. Since the binomials are identical, it is just the square of the first term!

{(a + b)}^{2} = a^{2} +____+____

To get the first term of the product, square the first term.

Where did the last term come from? Look at the examples and find the pattern.

The last term is the product of the last terms, which is the square of the last term.

{(a + b)}^{2} =____+____+ b^{2}

To get the last term of the product, square the last term.

Finally, look at the middle term. Notice it came from adding the “outer” and the “inner” terms—which are both the same! So the middle term is double the product of the two terms of the binomial.

\begin{matrix} {(a + b)}^{2} =____+ 2 a b +____ \\ {(a - b)}^{2} =____- 2 a b +____ \end{matrix}

To get the middle term of the product, multiply the terms and double their product.

Putting it all together:

To square a binomial:

square the first term
square the last term
double their product

A number example helps verify the pattern.

Steps for expanding and simplifying the binomial expression (10+4)^2 using algebraic identity.
	${(10 + 4)}^{2}$
Square the first term.	$10^{2} +___+$
Square the last term.	$10^{2} +___+ 4^{2}$
Double their product.	$10^{2} + 2 \cdot 10 \cdot 4 + 4^{2}$
Simplify.	$100 + 80 + 16$
Simplify.	$196$

To multiply ${(10 + 4)}^{2}$ usually you’d follow the Order of Operations.

\begin{matrix} {(10 + 4)}^{2} \\ {(14)}^{2} \\ 196 \end{matrix}

The pattern works!

Multiply: ${(x + 5)}^{2} .$

Solution


Square the first term.
Square the last term.
Double the product.
Simplify.

Multiply: ${(y - 3)}^{2} .$

Solution


Square the first term.
Square the last term.
Double the product.
Simplify.

Multiply: ${(4 x + 6)}^{2} .$

Solution


Use the pattern.
Simplify.

Multiply: ${(2 x - 3 y)}^{2} .$

Solution


Use the pattern.
Simplify.

Multiply: ${(4 u^{3} + 1)}^{2} .$

Solution


Use the pattern.
Simplify.

Multiply Conjugates Using the Product of Conjugates Pattern

We just saw a pattern for squaring binomials that we can use to make multiplying some binomials easier. Similarly, there is a pattern for another product of binomials. But before we get to it, we need to introduce some vocabulary.

What do you notice about these pairs of binomials?

\begin{matrix} (x - 9) (x + 9) & (y - 8) (y + 8) & (2 x - 5) (2 x + 5) \end{matrix}

Look at the first term of each binomial in each pair.

This figure has three products. The first is x minus 9, in parentheses, times x plus 9, in parentheses. The second is y minus 8, in parentheses, times y plus 8, in parentheses. The last is 2x minus 5, in parentheses, times 2x plus 5, in parentheses

Notice the first terms are the same in each pair.

Look at the last terms of each binomial in each pair.

Notice the last terms are the same in each pair.

Notice how each pair has one sum and one difference.

This figure has three products. The first is x minus 9, in parentheses, times x plus 9, in parentheses. Below the x minus 9 is the word “difference”. Below x plus 9 is the word “sum”. The second is y minus 8, in parentheses, times y plus 8, in parentheses. Below y minus 8 is the word “difference”. Below y plus 8 is the word “sum”. The last is 2x minus 5, in parentheses, times 2x plus 5, in parentheses. Below the 2x minus 5 is the word “difference” and below 2x plus 5 is the word “sum”.

A pair of binomials that each have the same first term and the same last term, but one is a sum and one is a difference has a special name. It is called a conjugate pair and is of the form $(a - b), (a + b)$ .

There is a nice pattern for finding the product of conjugates. You could, of course, simply FOIL to get the product, but using the pattern makes your work easier.

Let’s look for the pattern by using FOIL to multiply some conjugate pairs.

\begin{matrix} (x - 9) (x + 9) & (y - 8) (y + 8) & (2 x - 5) (2 x + 5) \\ x^{2} + 9 x - 9 x - 81 & y^{2} + 8 y - 8 y - 64 & 4 x^{2} + 10 x - 10 x - 25 \\ x^{2} - 81 & y^{2} - 64 & 4 x^{2} - 25 \end{matrix}

This figure has three columns. The first column contains the product of x plus 9 and x minus 9. Below this is the expression x squared minus 9x plus 9x minus 81. Below this is x squared minus 81. The second column contains the product of y minus 8 and y plus 8. Below this is the expression y squared plus 8y minus 8y minus 64. Below this is y squared minus 64. The third column contains the product of 2x minus 5 and 2x plus 5. Below this is the expression 4x squared plus 10x minus 10x minus 25. Below this is 4x squared minus 25.

Each first term is the product of the first terms of the binomials, and since they are identical it is the square of the first term.

\begin{matrix} (a + b) (a - b) = a^{2} -____ \\ To get the first term, square the first term . \end{matrix}

The last term came from multiplying the last terms, the square of the last term.

\begin{matrix} (a + b) (a - b) = a^{2} - b^{2} \\ To get the last term, square the last term . \end{matrix}

What do you observe about the products?

The product of the two binomials is also a binomial! Most of the products resulting from FOIL have been trinomials.

Why is there no middle term? Notice the two middle terms you get from FOIL combine to 0 in every case, the result of one addition and one subtraction.

The product of conjugates is always of the form $a^{2} - b^{2}$ . This is called a difference of squares.

This leads to the pattern:

Let’s test this pattern with a numerical example.

Comparison of two methods (difference of squares vs. order of operations) to simplify the expression (10-2)(10+2).
	$(10 - 2) (10 + 2)$
It is the product of conjudgates, so the result will be the difference of two squares.	$____-____$
Square the first term.	$10^{2} -____$
Square the last term.	$10^{2} - 2^{2}$
Simplify.	$100 - 4$
Simplify.	$96$
What do you get using the order of operations?
	$\begin{matrix} (10 - 2) (10 + 2) \\ (8) (12) \\ 96 \end{matrix}$

Notice, the result is the same!

Multiply: $(x - 8) (x + 8) .$

Solution

First, recognize this as a product of conjugates. The binomials have the same first terms, and the same last terms, and one binomial is a sum and the other is a difference.

It fits the pattern.
Square the first term, x.
Square the last term, 8.
The product is a difference of squares.

Multiply: $(2 x + 5) (2 x - 5) .$

Solution

Are the binomials conjugates?

It is the product of conjugates.
Square the first term, 2x.
Square the last term, 5.
Simplify. The product is a difference of squares.

The binomials in the next example may look backwards – the variable is in the second term. But the two binomials are still conjugates, so we use the same pattern to multiply them.

Find the product: $(3 + 5 x) (3 - 5 x) .$

Solution

It is the product of conjugates.
Use the pattern.
Simplify.

Now we’ll multiply conjugates that have two variables.

Find the product: $(5 m - 9 n) (5 m + 9 n) .$

Solution

This fits the pattern.
Use the pattern.
Simplify.

Find the product: $(c d - 8) (c d + 8) .$

Solution

This fits the pattern.
Use the pattern.
Simplify.

Find the product: $(6 u^{2} - 11 v^{5}) (6 u^{2} + 11 v^{5}) .$

Solution

This fits the pattern.
Use the pattern.
Simplify.

Recognize and Use the Appropriate Special Product Pattern

We just developed special product patterns for Binomial Squares and for the Product of Conjugates. The products look similar, so it is important to recognize when it is appropriate to use each of these patterns and to notice how they differ. Look at the two patterns together and note their similarities and differences.

Comparing the Special Product Patterns

This table summarizes formulas and key characteristics for binomial squares and products of conjugates, highlighting their algebraic properties and differences.
Binomial Squares	Product of Conjugates
${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$	$(a - b) (a + b) = a^{2} - b^{2}$
${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$
- Squaring a binomial	- Multiplying conjugates
- Product is a trinomial	- Product is a binomial
- Inner and outer terms with FOIL are the same.	- Inner and outer terms with FOIL are opposites.
- Middle term is double the product of the terms.	- There is no middle term.

Choose the appropriate pattern and use it to find the product:

ⓐ $(2 x - 3) (2 x + 3)$ ⓑ ${(8 x - 5)}^{2}$ ⓒ ${(6 m + 7)}^{2}$ ⓓ $(5 x - 6) (6 x + 5)$

Solution

ⓐ $(2 x - 3) (2 x + 3)$ These are conjugates. They have the same first numbers, and the same last numbers, and one binomial is a sum and the other is a difference. It fits the Product of Conjugates pattern.

This fits the pattern.

Use the pattern.

Simplify.
ⓑ ${(8 x - 5)}^{2}$ We are asked to square a binomial. It fits the binomial squares pattern.

Use the pattern.

Simplify.
ⓒ ${(6 m + 7)}^{2}$ Again, we will square a binomial so we use the binomial squares pattern.

Use the pattern.

Simplify.

ⓓ

(5 x - 6) (6 x + 5)

This product does not fit the patterns, so we will use FOIL.

This table illustrates the step-by-step multiplication of two binomials using the FOIL method, resulting in a simplified quadratic expression.
	$(5 x - 6) (6 x + 5)$
Use FOIL.	$30 x^{2} + 25 x - 36 x - 30$
Simplify.	$30 x^{2} - 11 x - 30$

Key Concepts

Binomial Squares Pattern
- If $a, b$ are real numbers,
- ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$
- ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$
- To square a binomial: square the first term, square the last term, double their product.
Product of Conjugates Pattern
- If $a, b$ are real numbers,
- $(a - b) (a + b) = a^{2} - b^{2}$
- The product is called a difference of squares.
To multiply conjugates:
- square the first term square the last term write it as a difference of squares

Practice Makes Perfect

Square a Binomial Using the Binomial Squares Pattern

In the following exercises, square each binomial using the Binomial Squares Pattern.

${(w + 4)}^{2}$

${(q + 12)}^{2}$

Solution

$q^{2} + 24 q + 144$

${(y + \frac{1}{4})}^{2}$

${(x + \frac{2}{3})}^{2}$

Solution

$x^{2} + \frac{4}{3} x + \frac{4}{9}$

${(b - 7)}^{2}$

${(y - 6)}^{2}$

Solution

$y^{2} - 12 y + 36$

${(m - 15)}^{2}$

${(p - 13)}^{2}$

Solution

$p^{2} - 26 p + 169$

${(3 d + 1)}^{2}$

${(4 a + 10)}^{2}$

Solution

$16 a^{2} + 80 a + 100$

${(2 q + \frac{1}{3})}^{2}$

${(3 z + \frac{1}{5})}^{2}$

Solution

$9 z^{2} + \frac{6}{5} z + \frac{1}{25}$

${(3 x - y)}^{2}$

${(2 y - 3 z)}^{2}$

Solution

$4 y^{2} - 12 y z + 9 z^{2}$

${(\frac{1}{5} x - \frac{1}{7} y)}^{2}$

${(\frac{1}{8} x - \frac{1}{9} y)}^{2}$

Solution

$\frac{1}{64} x^{2} - \frac{1}{36} x y + \frac{1}{81} y^{2}$

${(3 x^{2} + 2)}^{2}$

${(5 u^{2} + 9)}^{2}$

Solution

$25 u^{4} + 90 u^{2} + 81$

${(4 y^{3} - 2)}^{2}$

${(8 p^{3} - 3)}^{2}$

Solution

$64 p^{6} - 48 p^{3} + 9$

Multiply Conjugates Using the Product of Conjugates Pattern

In the following exercises, multiply each pair of conjugates using the Product of Conjugates Pattern.

$(m - 7) (m + 7)$

$(c - 5) (c + 5)$

Solution

$c^{2} - 25$

$(x + \frac{3}{4}) (x - \frac{3}{4})$

$(b + \frac{6}{7}) (b - \frac{6}{7})$

Solution

$b^{2} - \frac{36}{49}$

$(5 k + 6) (5 k - 6)$

$(8 j + 4) (8 j - 4)$

Solution

$64 j^{2} - 16$

$(11 k + 4) (11 k - 4)$

$(9 c + 5) (9 c - 5)$

Solution

$81 c^{2} - 25$

$(11 - b) (11 + b)$

$(13 - q) (13 + q)$

Solution

$169 - q^{2}$

$(5 - 3 x) (5 + 3 x)$

$(4 - 6 y) (4 + 6 y)$

Solution

$16 - 36 y^{2}$

$(9 c - 2 d) (9 c + 2 d)$

$(7 w + 10 x) (7 w - 10 x)$

Solution

$49 w^{2} - 100 x^{2}$

$(m + \frac{2}{3} n) (m - \frac{2}{3} n)$

$(p + \frac{4}{5} q) (p - \frac{4}{5} q)$

Solution

$p^{2} - \frac{16}{25} q^{2}$

$(a b - 4) (a b + 4)$

$(x y - 9) (x y + 9)$

Solution

$x^{2} y^{2} - 81$

$(u v - \frac{3}{5}) (u v + \frac{3}{5})$

$(r s - \frac{2}{7}) (r s + \frac{2}{7})$

Solution

$r^{2} s^{2} - \frac{4}{49}$

$(2 x^{2} - 3 y^{4}) (2 x^{2} + 3 y^{4})$

$(6 m^{3} - 4 n^{5}) (6 m^{3} + 4 n^{5})$

Solution

$36 m^{6} - 16 n^{10}$

$(12 p^{3} - 11 q^{2}) (12 p^{3} + 11 q^{2})$

$(15 m^{2} - 8 n^{4}) (15 m^{2} + 8 n^{4})$

Solution

$225 m^{4} - 64 n^{8}$

Recognize and Use the Appropriate Special Product Pattern

In the following exercises, find each product.

Solution

ⓐ $x^{10} - y^{10}$ ⓑ $m^{6} - 16 m^{3} n + 64 n^{2}$ ⓒ $81 p^{2} + 144 p q + 64 q^{2}$ ⓓ $r^{5} + r^{2} s^{2} - r^{3} s^{3} - s^{5}$

Everyday Math

Mental math You can use the product of conjugates pattern to multiply numbers without a calculator. Say you need to multiply 47 times 53. Think of 47 as $50 - 3$ and 53 as $50 + 3$ .

ⓐ Multiply $(50 - 3) (50 + 3)$ by using the product of conjugates pattern, $(a - b) (a + b) = a^{2} - b^{2}$ .
ⓑ Multiply $47 \cdot 53$ without using a calculator.

Mental math You can use the binomial squares pattern to multiply numbers without a calculator. Say you need to square 65. Think of 65 as $60 + 5$ .

ⓐ Multiply ${(60 + 5)}^{2}$ by using the binomial squares pattern, ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ .
ⓑ Square 65 without using a calculator.

Solution

Writing Exercises

How do you decide which pattern to use?

Why does ${(a + b)}^{2}$ result in a trinomial, but $(a - b) (a + b)$ result in a binomial?

Solution

Answers will vary.

Marta did the following work on her homework paper:

\begin{matrix} {(3 - y)}^{2} \\ 3^{2} - y^{2} \\ 9 - y^{2} \end{matrix}

Explain what is wrong with Marta’s work.

Use the order of operations to show that ${(3 + 5)}^{2}$ is 64, and then use that numerical example to explain why ${(a + b)}^{2} \neq a^{2} + b^{2}$ .

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This is a table that has four rows and four columns. In the first row, which is a header row, the cells read from left to right “I can…,” “Confidently,” “With some help,” and “No-I don’t get it!” The first column below “I can…” reads “square a binomial using the binomial squares pattern,” “multiply conjugates using the product of conjugates pattern,” and “recognize and use the appropriate special product pattern.” The rest of the cells are blank.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?