Factoring Special Products

 

Learning Objective(s)

·         Identify and factor special products of binomials.

 

 

One of the keys to factoring is finding relationships among the a, b, and c values of trinomials in ax² + bx + c form. Being able to think of possible number combinations rapidly, and then test those combinations, is an important skill that takes patience and practice.

 

Learning to recognize a few common polynomial types will cut down the amount of time it takes to factor them. Knowing the characteristic patterns of special products, for example trinomials that come from squaring binomials, provides a shortcut to finding their factors.

 

Curious about what these patterns are? Let’s take a look.

 

 

Perfect squares are numbers that are a whole number times itself. For example 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 are all perfect squares: they come from squaring each of the numbers from 1 to 10.

 

If the c term of a trinomial in x² + bx + c is a perfect square, then it may be a perfect square trinomial, a trinomial that is the product of a polynomial times itself. If it is, it'll be easy to factor.

 

Consider the trinomial, x² + 6x + 9. The c term, 9, is a perfect square. Let's factor it and see what happens. (Remember, to factor trinomials in the form x² + bx + c, find two integers, r and s, whose sum is b and whose product is c. Rewrite the trinomial as x² + rx + sx + c and then use grouping and the Distributive Property to factor the polynomial.)

 

 

Look at that—there's a lot of squares going on. The a and c terms are each a perfect square, the b term is twice the product of the square roots of those terms, and the trinomial factors as a perfect square. That sounds like it could be significant.

 

Let’s try one more example and see if it is.

 

We'll factor 9x² – 24x + 16.  For this trinomial, we need to find two numbers whose sum is -24 and whose product is 9 • 16, or 144.

 

Let’s try the number 12, since 12 is the square root of 144. While 12 • 12 = 144, 12 + 12 = 24. This is no good—we are looking for a sum of -24, not +24.

 

But wait—what about using -12 instead of +12? We know that -12 • -12 = 144, and that (-12) + (-12) = -24. So that works! We can split -24x into -12x and -12x and factor from there.

 

 

In this example, both the a term and the c term are perfect squares, the b term is twice the square root of the a term times the square root of the c term, and the trinomial factors to a perfect square. Yes, we do have a pattern!

 

This leads us to the general rule for factoring perfect square trinomials:

 

Perfect Square Trinomials   A trinomial in the form r2 + 2rs + s2 can be factored as (r + s)2. A trinomial in the form r2 – 2rs + s2 can be factored as (r – s)2. Examples: The factored form of 4x2 + 20x + 25 is (2x + 5)2. The factored form of x2 – 10x + 25 is (x – 5)2.

 

 

 

A quick word of warning: be careful of two very common errors that people make as they learn to work with perfect square trinomials.

 

Some people may be tempted to factor a binomial like r² + s² as (r + s)². Similarly, looking quickly at the binomial (r – s)², you may think that it evaluates to r² – s². These are incorrect! Let’s look at (r + s)² and (r – s)² in expanded form.

 

 

The expanded form of a squared binomial has a middle term—it's a trinomial! This is the product (literally) of applying the Distributive Property when we multiply the binomials together: we multiply r • r, then r • s, then s • r, then s • s, resulting in r² + rs + rs + s², or simply r² + 2rs + s².

 

Factor: a2 – 12a + 36   A) (a – 4)(a – 9)   B) (a + 6)2   C) (a – 6)2   D) (a + 6)(a – 6)   Show/Hide Answer A) Incorrect. While 4 • 9 gives the constant term 36, the middle term would be –13a rather than –12a. The correct answer is (a – 6)2.   B) Incorrect. While 62 is 36, the middle term in the original polynomial is negative, so you need a negative in the binomial you square. The correct answer is (a – 6)2.   C) Correct. This is a perfect square trinomial r2 – 2rs + s2 where r = a and s = 6. The factored form is (r – s)2, or (a – 6)2.   D) Incorrect. This is a perfect square trinomial r2 – 2rs + s2 where r = a and s = 6, so the factored form is (r – s)2, or (a – 6)2. Notice that if you expand (a + 6)(a – 6), you get a2 + 6a – 6a – 36. The 36 is subtracted rather than added, and the 6a – 6a gives a middle term of 0 (that results in no middle term at all).

 

 

 

Polynomials in the form r² – s² are also considered a special case. While it may be tempting to think that these polynomials can be factored as (r – s)², we just saw how and why that is not the case.

 

Let’s take a moment and see if we can factor this type of expression using what we already know about grouping and the Distributive Property.

 

We’ll factor 9x² – 4.

 

This polynomial looks a little different than some of the other problems we've done. For one thing, it does not look like a trinomial—there are only two terms! We are used to thinking about, and factoring, trinomials in the form ax² + bx + c. This polynomial seems to be missing the bx term.

 

Instead of thinking that the bx term is missing, however, what if we assign a value of 0 to b? If we think about the expression this way, we can put it in trinomial form, and then hopefully use what we know about trinomials to factor it. So instead of writing 9x² – 4, we will rewrite this polynomial as 9x² + 0x – 4. Representing this middle term as 0x does not change the value of the trinomial; we are not adding or subtracting anything, we are merely taking steps to represent the bx term in the polynomial.

 

9x² + 0x – 4 fits our standard form of a trinomial, ax² + bx + c. Let’s factor this trinomial the same way we would any other: find two integers, r and s, whose sum is b (0) and whose product is ac (9 • -4 = -36).

 

 

Looking down our list of possible factors, only one pair has a product of -36 and a sum of 0: 6 and -6. Let’s expand our 0x term into 6x – 6x as we keep factoring. 

 

 

So 9x² – 4 factors to (3x + 2)(3x – 2)! Notice that 3x is the square root of 9x² and 2 is the square root of 4. This pattern will continue for all binomials in the form r² – s².

 

Difference of Squares Binomial   A binomial in the form r2 – s2 has the factors (r + s)(r – s).   Examples: The factored form of x2 – 100 is (x + 10)(x – 10). The factored form of 49y2 – 25 is  (7y + 5)(7y – 5).

 

 

Some polynomials may seem more complicated to factor because they have more terms. For example, the polynomial 3x³ + 9x² – 12x + 36 has four terms. However, we have the tools to completely factor expressions like these already. Let’s look at this polynomial and see what we can do with it.

 

First, let’s review the tools we have so far.

 

Factoring Tools   Common factors A good first step when trying to factor a polynomial is to look for common factors among all the terms. This includes integers (such as 6, -3, or -1) and variables.   Special products Look for special products, that have shortcuts to factoring: r2 + 2rs + s2 = (r + s)2 r2 – 2rs + s2 = (r – s)2 r2 – s2 = (r + s)(r – s)   Grouping Use grouping to factor by finding common factors in pairs of terms rather than all of the terms. In the case of trinomials of the form ax2 + bx + c, split one of the terms (bx) into two terms and then group.

 

Now, let’s see if we can use those tools to factor the polynomial.

 

 

 

Learning to identify certain patterns in polynomials helps us factor some “special cases” of polynomials quickly. The special cases are:

 

• perfect square trinomials, r² + 2rs + s² and r² – 2rs + s², which factor as (r + s)² and (r – s)², respectively;
• the difference of squares binomial, r² – s², which factors as (r + s)(r – s).

 

For some polynomials, we may need to combine techniques (look for common factors, grouping, and use special products) to factor the polynomial completely.