Special Products

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 multiplying with the methods of the last section, there is less work to do if you learn to use a pattern.

Let’s start by looking at $(x+9)^2$. This means to multiply $(x+9)$ by itself, so $(x+9)^2 = (x+9)(x+9)$. Using FOIL and combining like terms:

Here’s another one. Squaring $(y-7)$ means multiplying $(y-7)$ by itself:

And one more, $(2x+3)^2$:

Look at these results. Do you see any patterns? In each example we squared a binomial and the result was a trinomial:

Now look at the first term in each result. Where did it come from? The first term is the product of the first terms of each binomial, and since the binomials are identical, it is just the square of the first term. To get the first term of the product, square the first term.

The last term is the product of the last terms, which is the square of the last term. 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. To get the middle term of the product, multiply the terms and double their product.

Putting it all together gives the pattern.

A number example helps verify the pattern. Squaring $(10+4)$ with the pattern gives $10^2 + 2 \cdot 10 \cdot 4 + 4^2 = 100 + 80 + 16 = 196$. Using the order of operations instead, $(10+4)^2 = (14)^2 = 196$. The pattern works!

Example. Multiply: $(x+5)^2$.

Here $a$ is $x$ and $b$ is $5$. Square the first term, square the last term, and double their product:

Example. Multiply: $(y-3)^2$.

Because the binomial is a difference, we use $(a-b)^2 = a^2 - 2ab + b^2$. Here $a$ is $y$ and $b$ is $3$:

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

Use the pattern with $a = 4x$ and $b = 6$:

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

Now both terms carry variables. Use the pattern with $a = 2x$ and $b = 3y$:

Example. Multiply: $\left(4u^3 + 1\right)^2$.

The pattern still applies when the first term has an exponent. With $a = 4u^3$ and $b = 1$:

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.

Look at these pairs of binomials:

Notice that the first terms are the same in each pair, and the last terms are the same in each pair. Notice too that each pair has one sum and one difference. 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. Let’s look for it by using FOIL to multiply some conjugate pairs:

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

What do you observe about the products? The product of the two binomials is also a binomial! 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.

Let’s test this pattern with a numerical example. The product $(10-2)(10+2)$ is a product of conjugates, so the result is a difference of squares: $10^2 - 2^2 = 100 - 4 = 96$. Using the order of operations, $(10-2)(10+2) = (8)(12) = 96$. The result is the same!

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

First, recognize this as a product of conjugates. The binomials have the same first term and the same last term, and one is a sum and the other is a difference. Square the first term, square the last term, and write the product as a difference of squares:

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

Are the binomials conjugates? Yes — so it is the product of conjugates. Square the first term $2x$ and the last term $5$:

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.

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

It is the product of conjugates. Square the first term $3$ and the last term $5x$:

Now we’ll multiply conjugates that have two variables.

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

This fits the pattern. Square the first term $5m$ and the last term $9n$:

Example. Find the product: $(cd-8)(cd+8)$.

This fits the pattern with first term $cd$ and last term $8$:

Example. Find the product: $\left(6u^2 - 11v^5\right)\left(6u^2 + 11v^5\right)$.

This fits the pattern with first term $6u^2$ and last term $11v^5$:

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.

Example. Choose the appropriate pattern and use it to find each product.

(a) $(2x-3)(2x+3)$ — these are conjugates. They 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 Product of Conjugates pattern:

(b) $(8x-5)^2$ — we are asked to square a binomial. It fits the Binomial Squares pattern:

(c) $(6m+7)^2$ — again we square a binomial, so we use the Binomial Squares pattern:

(d) $(5x-6)(6x+5)$ — this product does not fit either pattern (the binomials are not identical and are not conjugates), so we use FOIL:

Key terms

conjugate pair — two binomials of the form $(a-b)$, $(a+b)$ that have the same first term and the same last term, but one is a sum and the other a difference. difference of squares — a binomial of the form $a^2 - b^2$; it is the product of a conjugate pair. Binomial Squares Pattern — $(a+b)^2 = a^2 + 2ab + b^2$ and $(a-b)^2 = a^2 - 2ab + b^2$. Product of Conjugates Pattern — $(a-b)(a+b) = a^2 - b^2$.

This section is adapted from Elementary Algebra 2e, Section 6.4: Special Products by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recast the pattern-derivation walkthroughs and worked-example step tables as typeset display equations and a comparison table; kept the Binomial Squares, Conjugate Pair, and Product of Conjugates patterns as callouts; omitted the Be Prepared quiz, Self Check checklist, media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.