Factor Special Products

The strategy for factoring we developed in the last section will guide you as you factor most binomials, trinomials, and polynomials with more than three terms. We have seen that some binomials and trinomials result from special products — squaring binomials and multiplying conjugates. If you learn to recognize these kinds of polynomials, you can use the special products patterns to factor them much more quickly.

Factor perfect square trinomials

Some trinomials are perfect squares. They result from multiplying a binomial times itself. You can square a binomial by using FOIL, but using the Binomial Squares pattern you saw in a previous chapter saves you a step. Let’s review by squaring a binomial using FOIL:

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

The trinomial $9x^2 + 24x + 16$ is called a perfect square trinomial. It is the square of the binomial $3x + 4$.

When you square a binomial, the product is a perfect square trinomial. In this chapter you are learning to factor — now you will start with a perfect square trinomial and factor it into its prime factors. You could factor this trinomial using the methods of the last section, since it is of the form $ax^2 + bx + c$. But if you recognize that the first and last terms are squares and the trinomial fits the perfect square trinomials pattern, you will save yourself a lot of work. Here is the pattern — the reverse of the Binomial Squares pattern.

To make use of this pattern, you have to recognize that a given trinomial fits it. Check first to see if the leading coefficient is a perfect square, $a^2$. Next check that the last term is a perfect square, $b^2$. Then check the middle term — is it twice the product, $2ab$? If everything checks, you can easily write the factors.

Example. Factor: $9x^2 + 12x + 4$.

Does the trinomial fit the perfect square trinomials pattern, $a^2 + 2ab + b^2$? The first term $9x^2$ is a perfect square, $(3x)^2$, and the last term $4$ is a perfect square, $(2)^2$. The middle term $12x$ is twice the product of $3x$ and $2$, so it matches $2ab$. Write it as the square of a binomial:

Check by multiplying: $(3x+2)^2 = (3x)^2 + 2 \cdot 3x \cdot 2 + 2^2 = 9x^2 + 12x + 4$. ✓

The sign of the middle term determines which pattern we will use. When the middle term is negative, we use the pattern $a^2 - 2ab + b^2$, which factors to $(a-b)^2$. The steps are summarized here.

Example. Factor: $81y^2 - 72y + 16$.

The first and last terms are squares. The middle term is negative, so the binomial square would be $(a-b)^2$. Write the first term as $(9y)^2$ and the last term as $(4)^2$; the middle term $72y$ is $2 \cdot 9y \cdot 4$, so the pattern matches:

The next example is a perfect square trinomial with two variables.

Example. Factor: $36x^2 + 84xy + 49y^2$.

Test each term to verify the pattern. The first term is $(6x)^2$ and the last term is $(7y)^2$; the middle term $84xy$ is $2 \cdot 6x \cdot 7y$. It fits, so we write the square of the binomial:

Remember the very first step in our strategy for factoring polynomials — ask “is there a greatest common factor?” and, if there is, factor the GCF out before going any further. Perfect square trinomials may have a GCF in all three terms and it should be factored out first. Sometimes, once the GCF has been factored, you will recognize a perfect square trinomial.

Example. Factor: $36x^2y - 48xy + 16y$.

There is a GCF of $4y$, so factor it out first:

The trinomial in parentheses is a perfect square: $9x^2 = (3x)^2$, $4 = 2^2$, and $12x = 2 \cdot 3x \cdot 2$. Factor it, keeping the factor $4y$:

Factor differences of squares

The other special product you saw in the previous chapter was the Product of Conjugates pattern. You used this to multiply two binomials that were conjugates, for example:

When you multiply conjugate binomials, the middle terms of the product add to $0$. All you have left is a binomial, the difference of squares. Multiplying conjugates is the only way to get a binomial from the product of two binomials.

To factor, we use the product pattern “in reverse” to factor the difference of squares. Remember, “difference” refers to subtraction. So, to use this pattern you must make sure you have a binomial in which two squares are being subtracted.

Example. Factor: $x^2 - 4$.

Does the binomial fit the pattern? It is a difference, and both terms are perfect squares: $x^2 = (x)^2$ and $4 = 2^2$. Write them as squares, then write the product of conjugates:

Check by multiplying: $(x-2)(x+2) = x^2 - 4$. ✓

It is important to remember that sums of squares do not factor into a product of binomials. There are no binomial factors that multiply together to get a sum of squares. After removing any GCF, the expression $a^2 + b^2$ is prime!

Example. Factor: $64y^2 - 1$.

This is a difference, and both terms are perfect squares — don’t forget that $1$ is a perfect square. Write $64y^2$ as $(8y)^2$ and $1$ as $1^2$, then factor as the product of conjugates:

Example. Factor: $121x^2 - 49y^2$.

Is this a difference of squares? Yes — $121x^2 = (11x)^2$ and $49y^2 = (7y)^2$. Factor as the product of conjugates:

The binomial in the next example may look “backwards,” but it is still the difference of squares.

Example. Factor: $100 - h^2$.

Is this a difference of squares? Yes — $100 = 10^2$ and $h^2 = (h)^2$. Factor as the product of conjugates:

Be careful not to rewrite the original expression as $h^2 - 100$.

To completely factor the binomial in the next example, we factor a difference of squares twice!

Example. Factor: $x^4 - y^4$.

Is this a difference of squares? Yes — $x^4 = (x^2)^2$ and $y^4 = (y^2)^2$. Factor it as the product of conjugates. Notice the first binomial is also a difference of squares! The last factor, the sum of squares, cannot be factored:

As always, you should look for a common factor first whenever you have an expression to factor. Sometimes a common factor may “disguise” the difference of squares and you won’t recognize the perfect squares until you factor the GCF.

Example. Factor: $8x^2y - 98y$.

Is there a GCF? Yes, $2y$ — factor it out. The binomial that remains is a difference of squares. Factor it as a product of conjugates:

Remember, a sum of squares does not factor. After removing the GCF, if what remains is a sum of squares, it is prime.

Example. Factor: $6x^2 + 96$.

Is there a GCF? Yes, $6$ — factor it out:

Is the binomial in parentheses a difference of squares? No — it is a sum of squares. Sums of squares do not factor, so $x^2 + 16$ is prime and $6(x^2+16)$ is the complete factorization.

Factor sums and differences of cubes

There is another special pattern for factoring, one that we did not use when we multiplied polynomials. This is the pattern for the sum and difference of cubes. We can check these formulas by multiplication; for the sum of cubes, distributing $(a+b)$ over $\left(a^2 - ab + b^2\right)$ gives $a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3 = a^3 + b^3$.

The two patterns look very similar. But notice the signs in the factors. The sign of the binomial factor matches the sign in the original binomial. And the sign of the middle term of the trinomial factor is the opposite of the sign in the original binomial. The trinomial factor in the sum and difference of cubes pattern cannot be factored.

It can be very helpful if you learn to recognize the cubes of the integers from $1$ to $10$, just like you have learned to recognize squares:

Example. Factor: $x^3 + 64$.

Does the binomial fit the pattern? It is a sum, and both terms are perfect cubes: $x^3 = (x)^3$ and $64 = 4^3$. Write the terms as cubes, use the sum of cubes pattern, and simplify inside the parentheses:

Be careful to use the correct signs in the factors of the sum and difference of cubes.

Example. Factor: $x^3 - 1000$.

This binomial is a difference. The first and last terms are perfect cubes: $x^3 = (x)^3$ and $1000 = 10^3$. Use the difference of cubes pattern and simplify:

Example. Factor: $512 - 125p^3$.

This binomial is a difference. The first and last terms are perfect cubes: $512 = 8^3$ and $125p^3 = (5p)^3$. Use the difference of cubes pattern and simplify:

Example. Factor: $27u^3 - 125v^3$.

This binomial is a difference. Both terms are perfect cubes: $27u^3 = (3u)^3$ and $125v^3 = (5v)^3$. Use the difference of cubes pattern and simplify:

In the next example, we first factor out the GCF. Then we can recognize the sum of cubes.

Example. Factor: $5m^3 + 40n^3$.

Factor the common factor $5$ first. The binomial that remains is a sum, and its terms are perfect cubes: $m^3 = (m)^3$ and $8n^3 = (2n)^3$. Use the sum of cubes pattern and simplify:

Key terms

perfect square trinomial — a trinomial of the form $a^2 + 2ab + b^2$ or $a^2 - 2ab + b^2$; it factors to $(a+b)^2$ or $(a-b)^2$. difference of squares — a binomial of the form $a^2 - b^2$; it factors to the conjugate pair $(a-b)(a+b)$. sum of squares — a binomial of the form $a^2 + b^2$; it does not factor and is prime. sum of cubes — $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$. difference of cubes — $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$.

This section is adapted from Elementary Algebra 2e, Section 7.4: Factor 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, kept the Perfect Square Trinomials, Difference of Squares, and Sum and Difference of Cubes patterns and How To procedures as callouts, recreated the cubes reference table as a markdown table; 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.