Factoring Special Products

Learning Objective

• Identify and factor special products of binomials.

One of the keys to factoring is finding relationships among the `a`, `b`, and `c` values of trinomialsA three-term polynomial. in `ax^2 + 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 binomialsA sum of two monomials, such as `3x^2 + 7`. , provides a shortcut to finding their factors.

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

Perfect squaresAny of the squares of the integers. Since `1^2 = 1`, `2^2 = 4`, `3^2 = 9`, etc., `1`, `4`, and `9` are 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 `x^2 + bx + c` is a perfect square, then it may be a perfect square trinomialA trinomial that is the product of a binomial times itself, such as `r^2 + 2rs + s^2` (from `(r + s)^2`), and `r^2 - 2rs + s^2` (from `(r - s)^2`). , a trinomial that is the product of a polynomial times itself. If it is, it'll be easy to factor.

Consider the trinomial, `x^2 + 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 `x^2 + bx + c`, find two integers, `r` and `s`, whose sum is `b` and whose product is `c`. Rewrite the trinomial as `x^2 + 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 `9x^2 - 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:

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 `r^2 + s^2` as `(r + s)^2`. Similarly, looking quickly at the binomial `(r - s)^2`, you may think that it evaluates to `r^2 - s^2`. 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 `r^2 + rs + rs + s^2`, or simply `r^2 + 2rs + s^2`. 

Polynomials in the form `r^2 - s^2` 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 `9x^2 - 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 `ax^2 + 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 `9x^2 - 4`, we will rewrite this polynomial as `9x^2 + 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.

`9x^2 + 0x - 4` fits our standard form of a trinomial, `ax^2 + 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`, and that is `6` and `-6`. Let’s expand our `0x` term into `6x - 6x` as we keep factoring.

So `9x^2 - 4` factors to `(3x + 2)(3x - 2)`! Notice that `3x` is the square root of `9x^2` and `2` is the square root of `4`. This pattern will continue for all binomials in the form `r^2 - s^2`. 

Some polynomials may seem more complicated to factor because they have more terms. For example, the polynomial `3x^3 + 9x^2 - 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.

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, `r^2 + 2rs + s^2` and `r^2 - 2rs + s^2`, which factor as `(r + s)^2` and `(r - s)^2`, respectively;
• the difference of squares binomial, `r^2 - s^2`, which factors as `(r + s)(r - s)`. 

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