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Study Guides > Intermediate Algebra

Factoring Special Cases

Learning Objectives

  • Special Cases - Squares
    • Factor a polynomial of the form: [latex]{a}^{2}+2ab+{b}^{2}[/latex]
    • Factor a polynomial of the form: [latex]{a}^{2}-{b}^{2}[/latex]
  • Special Cases - Cubes
    • Factor the sum of cubes.
    • Factor the difference of cubes

Why learn how to factor special cases?

Repeated pattern of interlocking plus signs, each row a different color following the rainbow spectrum.   Some people like to find patterns in the world around them, like a game.  There are some polynomials that, when factored, follow a specific pattern.    

These include:

Perfect square trinomials of the form: [latex]{a}^{2}+2ab+{b}^{2}[/latex]

A difference of squares: [latex]{a}^{2}-{b}^{2}[/latex]

A sum of cubes: [latex]{a}^{3}+{b}^{3}[/latex]

A difference of cubes: [latex]{a}^{3}-{b}^{3}[/latex]

  In this lesson you will see you can factor each of these types of polynomials following a specific pattern.  You will also learn how to factor polynomials that have negative or fractional exponents. Picture of a sidewalk leading to a parking lot. There is a path through the grass to teh right of the sidewalk through the trees that has been made by people walking on the grass. The shortcut to the parking lot is the preferred way. Some people find it helpful to know when they can take a shortcut to avoid doing extra work.  There are some polynomials that will always factor a certain way, and for those we offer a shortcut.  Most people find it helpful to memorize the factored form of a perfect square trinomial or a difference of squares.  The most important skill you will use in this section will be recognizing when you can use the shortcuts.

Factoring a Perfect Square Trinomial

A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.
[latex]\begin{array}{ccc}\hfill {a}^{2}+2ab+{b}^{2}& =& {\left(a+b\right)}^{2}\hfill \\ & \text{and}& \\ \hfill {a}^{2}-2ab+{b}^{2}& =& {\left(a-b\right)}^{2}\hfill \end{array}[/latex]
We can use this equation to factor any perfect square trinomial.

A General Note: Perfect Square Trinomials

A perfect square trinomial can be written as the square of a binomial:
[latex]{a}^{2}+2ab+{b}^{2}={\left(a+b\right)}^{2}[/latex]
In the following example we will show you how to define a, and b so you can use the shortcut.

Exercises

Factor [latex]25{x}^{2}+20x+4[/latex].

Answer: First, notice that [latex]25{x}^{2}[/latex] and [latex]4[/latex] are perfect squares because [latex]25{x}^{2}={\left(5x\right)}^{2}[/latex] and [latex]4={2}^{2}[/latex]. This means that [latex]a=5x\text{ and }b=2[/latex] Next, check to see if the middle term is equal to [latex]2ab[/latex], which it is:

[latex]2ab = 2\left(5x\right)\left(2\right)=20x[/latex].

  Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{\left(a+b\right)}^{2}={\left(5x+2\right)}^{2}[/latex].

Answer

[latex-display]25{x}^{2}+20x+4={\left(5x+2\right)}^{2}[/latex-display]

In the next example, we will show that we can use [latex]1 = 1^2[/latex] to factor a polynomial with a term equal to 1.

Example

Factor [latex]49{x}^{2}-14x+1[/latex].

Answer: First, notice that [latex]49{x}^{2}[/latex] and [latex]1[/latex] are perfect squares because [latex]49{x}^{2}={\left(7x\right)}^{2}[/latex] and [latex]1={1}^{2}[/latex]. This means that [latex]a=7x[/latex], we could say that [latex]b=1[/latex], but would that give a middle term of [latex]-14x[/latex]? We will need to choose [latex]b = -1[/latex] to get the results we want:

[latex]2ab = 2\left(7x\right)\left(-1\right)=-14x[/latex].

Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{\left(a-b\right)}^{2}={\left(7x-1\right)}^{2}[/latex].

Answer

[latex-display]49{x}^{2}-14x+1={\left(7x-1\right)}^{2}[/latex-display]  

In the following video we provide another short description of what a perfect square trinomial is, and show how to factor them using a the formula. https://youtu.be/UMCVGDTxxTI We can summarize our process in the following way:

Given a perfect square trinomial, factor it into the square of a binomial.

  1. Confirm that the first and last term are perfect squares.
  2. Confirm that the middle term is twice the product of [latex]ab[/latex].
  3. Write the factored form as [latex]{\left(a+b\right)}^{2}[/latex], or[latex]{\left(a-b\right)}^{2}[/latex].

Factoring a Difference of Squares

A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.
[latex]{a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)[/latex]
We can use this equation to factor any differences of squares.

A General Note: Differences of Squares

A difference of squares can be rewritten as two factors containing the same terms but opposite signs.
[latex]{a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)[/latex]

Example

Factor [latex]9{x}^{2}-25[/latex].

Answer: Notice that [latex]9{x}^{2}[/latex] and [latex]25[/latex] are perfect squares because [latex]9{x}^{2}={\left(3x\right)}^{2}[/latex] and [latex]25={5}^{2}[/latex]. This means that [latex]a=3x,\text{ and }b=5[/latex] The polynomial represents a difference of squares and can be rewritten as [latex]\left(3x+5\right)\left(3x - 5\right)[/latex]. Check that you are correct by multiplying. [latex-display]\left(3x+5\right)\left(3x - 5\right)=9x^2-15x+15x-25=9x^2-25[/latex-display]

Answer

[latex-display]9{x}^{2}-25=\left(3x+5\right)\left(3x - 5\right)[/latex-display]

The most helpful thing for recognizing a difference of squares that can be factored with the shortcut is knowing which numbers are perfect squares, as you will see in the next example.

Example

Factor [latex]81{y}^{2}-144[/latex].

Answer: Notice that [latex]81{y}^{2}[/latex] and [latex]144[/latex] are perfect squares because [latex]81{y}^{2}={\left(9x\right)}^{2}[/latex] and [latex]144={12}^{2}[/latex]. This means that [latex]a=9x,\text{ and }b=12[/latex] The polynomial represents a difference of squares and can be rewritten as [latex]\left(9x+12\right)\left(9x - 12\right)[/latex]. Check that you are correct by multiplying. [latex-display]\left(9x+12\right)\left(9x - 12\right)=81x^2-108x+108x-144=81x^2-144[/latex-display]

Answer

[latex-display]81{y}^{2}-144=\left(9x+12\right)\left(9x - 12\right)[/latex-display]

 
In the following video we show another example of how to use the formula for fact a difference of squares. https://youtu.be/Li9IBp5HrFA We can summarize the process for factoring a difference of squares with the shortcut this way:

How To: Given a difference of squares, factor it into binomials.

  1. Confirm that the first and last term are perfect squares.
  2. Write the factored form as [latex]\left(a+b\right)\left(a-b\right)[/latex].

Think About It

Is there a formula to factor the sum of squares, [latex]a^2+b^2[/latex], into a product of two binomials? Write down some ideas for how you would answer this in the box below before you look at the answer. [practice-area rows="1"][/practice-area]

Answer: There is no way to factor a sum of squares into a product of two binomials, this is because of addition - the middle term needs to "disappear" and the only way to do that is with opposite signs.  to get a positive result, you must multiply two numbers with the same signs. The only time a sum of squares can be factored is if they share any common factors, as in the following case: [latex-display]9x^2+36[/latex-display] [latex-display]9x^2={(3x)}^2, \text{ and }36 = 6^2[/latex-display] The only way to factor this expression is by pulling out the GCF which is 9. [latex-display]9x^2+36=9(x^2+4)[/latex-display]

Cubes

Some interesting patterns arise when you are working with cubed quantities within polynomials. Specifically, there are two more special cases to consider: [latex]a^{3}+b^{3}[/latex] and [latex]a^{3}-b^{3}[/latex]. Let’s take a look at how to factor sums and differences of cubes.

Sum of Cubes

The term “cubed” is used to describe a number raised to the third power. In geometry, a cube is a six-sided shape with equal width, length, and height; since all these measures are equal, the volume of a cube with width x can be represented by [latex]x^{3}[/latex]. (Notice the exponent!) Cubed numbers get large very quickly. [latex]1^{3}=1[/latex], [latex]2^{3}=8[/latex], [latex]3^{3}=27[/latex], [latex]4^{3}=64[/latex], and [latex]5^{3}=125[/latex]. Before looking at factoring a sum of two cubes, let’s look at the possible factors. It turns out that [latex]a^{3}+b^{3}[/latex] can actually be factored as [latex]\left(a+b\right)\left(a^{2}–ab+b^{2}\right)[/latex]. Let’s check these factors by multiplying.

Example

Does [latex](a+b)(a^{2}–ab+b^{2})=a^{3}+b^{3}[/latex]?

Answer: Apply the distributive property.

[latex]\left(a\right)\left(a^{2}–ab+b^{2}\right)+\left(b\right)\left(a^{2}–ab+b^{2}\right)[/latex]

Multiply by a.

[latex]\left(a^{3}–a^{2}b+ab^{2}\right)+\left(b\right)\left(a^{2}-ab+b^{2}\right)[/latex]

Multiply by b.

[latex]\left(a^{3}–a^{2}b+ab^{2}\right)+\left(a^{2}b–ab^{2}+b^{3}\right)[/latex]

Rearrange terms in order to combine the like terms.

[latex]a^{3}-a^{2}b+a^{2}b+ab^{2}-ab^{2}+b^{3}[/latex]

Simplify.

Answer

[latex]a^{3}+b^{3}[/latex]

Did you see that? Four of the terms cancelled out, leaving us with the (seemingly) simple binomial [latex]a^{3}+b^{3}[/latex]. So, the factors are correct. You can use this pattern to factor binomials in the form [latex]a^{3}+b^{3}[/latex], otherwise known as “the sum of cubes.”

The Sum of Cubes

A binomial in the form [latex]a^{3}+b^{3}[/latex] can be factored as [latex]\left(a+b\right)\left(a^{2}–ab+b^{2}\right)[/latex].

Examples:

The factored form of [latex]x^{3}+64[/latex] is [latex]\left(x+4\right)\left(x^{2}–4x+16\right)[/latex]. The factored form of [latex]8x^{3}+y^{3}[/latex] is [latex]\left(2x+y\right)\left(4x^{2}–2xy+y^{2}\right)[/latex].

Example

Factor [latex]x^{3}+8y^{3}[/latex].

Answer: Identify that this binomial fits the sum of cubes pattern: [latex]a^{3}+b^{3}[/latex]. [latex]a=x[/latex], and [latex]b=2y[/latex] (since [latex]2y\cdot2y\cdot2y=8y^{3}[/latex]).

[latex]x^{3}+8y^{3}[/latex]

Factor the binomial as [latex]\left(a+b\right)\left(a^{2}–ab+b^{2}\right)[/latex], substituting [latex]a=x[/latex] and [latex]b=2y[/latex] into the expression.

[latex]\left(x+2y\right)\left(x^{2}-x\left(2y\right)+\left(2y\right)^{2}\right)[/latex]

Square [latex](2y)^{2}=4y^{2}[/latex].

[latex]\left(x+2y\right)\left(x^{2}-x\left(2y\right)+4y^{2}\right)[/latex]

Multiply [latex]−x\left(2y\right)=−2xy[/latex] (writing the coefficient first).

Answer

[latex]\left(x+2y\right)\left(x^{2}-2xy+4y^{2}\right)[/latex]

And that’s it. The binomial [latex]x^{3}+8y^{3}[/latex] can be factored as [latex]\left(x+2y\right)\left(x^{2}–2xy+4y^{2}\right)[/latex]! Let’s try another one. You should always look for a common factor before you follow any of the patterns for factoring.

Example

Factor [latex]16m^{3}+54n^{3}[/latex].

Answer: Factor out the common factor 2.

[latex]16m^{3}+54n^{3}[/latex]

[latex]8m^{3}[/latex] and [latex]27n^{3}[/latex] are cubes, so you can factor [latex]8m^{3}+27n^{3}[/latex] as the sum of two cubes: [latex]a=2m[/latex], and [latex]b=3n[/latex].

[latex]2\left(8m^{3}+27^n{3}\right)[/latex]

Factor the binomial [latex]8m^{3}+27n^{3}[/latex] substituting [latex]a=2m[/latex] and [latex]b=3n[/latex] into the expression [latex]\left(a+b\right)\left(a^{2}-ab+b^{2}\right)[/latex].

[latex]2\left(2m+3n\right)\left[\left(2m\right)^{2}-\left(2m\right)\left(3n\right)+\left(3n\right)^{2}\right][/latex]

Square: [latex](2m)^{2}=4m^{2}[/latex] and [latex](3n)^{2}=9n^{2}[/latex].

[latex]2\left(2m+3n\right)\left[4m^{2}-\left(2m\right)\left(3n\right)+9n^{2}\right][/latex]

Multiply [latex]-\left(2m\right)\left(3n\right)=-6mn[/latex].

Answer

[latex]2\left(2m+3n\right)\left(4m^{2}-6mn+9n^{2}\right)[/latex]

Difference of Cubes

Having seen how binomials in the form [latex]a^{3}+b^{3}[/latex] can be factored, it should not come as a surprise that binomials in the form [latex]a^{3}-b^{3}[/latex] can be factored in a similar way.

The Difference of Cubes

A binomial in the form [latex]a^{3}–b^{3}[/latex] can be factored as [latex]\left(a-b\right)\left(a^{2}+ab+b^{2}\right)[/latex].

Examples

The factored form of [latex]x^{3}–64[/latex] is [latex]\left(x–4\right)\left(x^{2}+4x+16\right)[/latex]. The factored form of [latex]27x^{3}–8y^{3}[/latex] is [latex]\left(3x–2y\left)\right(9x^{2}+6xy+4y^{2}\right)[/latex].
Notice that the basic construction of the factorization is the same as it is for the sum of cubes; the difference is in the [latex]+[/latex] and [latex]–[/latex] signs. Take a moment to compare the factored form of [latex]a^{3}+b^{3}[/latex] with the factored form of [latex]a^{3}-b^{3}[/latex]. Factored form of [latex]a^{3}+b^{3}[/latex]: [latex]\left(a+b\right)\left(a^{2}-ab+b^{2}\right)[/latex] Factored form of [latex]a^{3}-b^{3}[/latex]: [latex]\left(a-b\right)\left(a^{2}+ab+b^{2}\right)[/latex] This can be tricky to remember because of the different signs—the factored form of [latex]a^{3}+b^{3}[/latex] contains a negative, and the factored form of [latex]a^{3}-b^{3}[/latex] contains a positive! Some people remember the different forms like this: “Remember one sequence of variables: [latex]a^{3}b^{3}=\left(a\,b\right)\left(a^{2}ab\,b^{2}\right)[/latex]. There are 4 missing signs. Whatever the first sign is, it is also the second sign. The third sign is the opposite, and the fourth sign is always [latex]+[/latex].” Try this for yourself. If the first sign is [latex]+[/latex], as in [latex]a^{3}+b^{3}[/latex], according to this strategy how do you fill in the rest: [latex]\left(a\,b\right)\left(a^{2}ab\,b^{2}\right)[/latex]? Does this method help you remember the factored form of [latex]a^{3}+b^{3}[/latex] and [latex]a^{3}–b^{3}[/latex]? Let’s go ahead and look at a couple of examples. Remember to factor out all common factors first.

Example

Factor [latex]8x^{3}–1,000[/latex].

Answer: Factor out 8.

[latex]8(x^{3}–125)[/latex]

Identify that the binomial fits the pattern [latex]a^{3}-b^{3}:a=x[/latex], and [latex]b=5[/latex] (since [latex]5^{3}=125[/latex]).

[latex]8\left(x^{3}–125\right)[/latex]

Factor [latex]x^{3}–125[/latex] as [latex]\left(a–b\right)\left(a^{2}+ab+b^{2}\right)[/latex], substituting [latex]a=x[/latex] and [latex]b=5[/latex] into the expression.

[latex]8\left(x-5\right)\left[x^{2}+\left(x\right)\left(5\right)+5^{2}\right][/latex]

Square the first and last terms, and rewrite [latex]\left(x\right)\left(5\right)[/latex] as [latex]5x[/latex].

[latex]8\left(x–5\right)\left(x^{2}+5x+25\right)[/latex]

Answer

[latex]8\left(x–5\right)\left(x^{2}+5x+25\right)[/latex]

Let’s see what happens if you don’t factor out the common factor first. In this example, it can still be factored as the difference of two cubes. However, the factored form still has common factors, which need to be factored out. As you can see, this last example still worked, but required a couple of extra steps. It is always a good idea to factor out all common factors first. In some cases, the only efficient way to factor the binomial is to factor out the common factors first. Here is one more example. Note that [latex]r^{9}=\left(r^{3}\right)^{3}[/latex] and that [latex]8s^{6}=\left(2s^{2}\right)^{3}[/latex].

Example

Factor [latex]r^{9}-8s^{6}[/latex].

Answer: Identify this binomial as the difference of two cubes. As shown above, it is. Using the laws of exponents, rewrite [latex]r^{9}[/latex] as [latex]\left(r^{3}\right)^{3}[/latex].

[latex]r^{9}-8s^{6}[/latex]

Rewrite [latex]r^{9}[/latex] as [latex]\left(r^{3}\right)^{3}[/latex] and rewrite [latex]8s^{6}[/latex] as [latex]\left(2s^{2}\right)^{3}[/latex].

[latex]\left(r^{3}\right)^{3}-\left(2s^{2}\right)^{3}[/latex]

Now the binomial is written in terms of cubed quantities. Thinking of [latex]a^{3}-b^{3}[/latex], [latex]a=r^{3}[/latex] and [latex]b=2s^{2}[/latex]. Factor the binomial as [latex]\left(a-b\right)\left(a^{2}+ab+b^{2}\right)[/latex], substituting [latex]a=r^{3}[/latex] and [latex]b=2s^{2}[/latex] into the expression.

[latex]\left(r^{3}-2s^{2}\right)\left[\left(r^{3}\right)^{2}+\left(r^{3}\right)\left(2s^{2}\right)+\left(2s^{2}\right)^{2}\right][/latex]

Multiply and square the terms.

[latex]\left(r^{3}-2s^{2}\right)\left(r^{6}+2r^{3}s^{2}+4s^{4}\right)[/latex]

Answer

[latex]\left(r^{3}-2s^{2}\right)\left(r^{6}+2r^{3}s^{2}+4s^{4}\right)[/latex]

In the following two video examples we show more binomials that can be factored as a sum or difference of cubes. https://youtu.be/tFSEpOB262M https://youtu.be/J_0ctMrl5_0 You encounter some interesting patterns when factoring. Two special cases—the sum of cubes and the difference of cubes—can help you factor some binomials that have a degree of three (or higher, in some cases). The special cases are:
  • A binomial in the form [latex]a^{3}+b^{3}[/latex] can be factored as [latex]\left(a+b\right)\left(a^{2}–ab+b^{2}\right)[/latex]
  • A binomial in the form [latex]a^{3}-b^{3}[/latex] can be factored as [latex]\left(a-b\right)\left(a^{2}+ab+b^{2}\right)[/latex]
Always remember to factor out any common factors first.  

Licenses & Attributions

CC licensed content, Original

CC licensed content, Shared previously

  • Factor Perfect Square Trinomials Using a Formula. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Ex: Factor a Difference of Squares. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.
  • Unit 12: Factoring, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education Located at: https://www.nroc.org/. License: CC BY: Attribution.
  • Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Ex 3: Factor a Sum or Difference of Cubes. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Factor Expressions with Negative Exponents. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Factor Expressions with Fractional Exponents. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Factor Expressions Using Substitution. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Ex: Factoring Polynomials with Common Factors. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.