2.6 Factoring

Introduction

In the previous section, we spent time using the distributive property to expand expressions into multiple terms that could then be simplified. In this section we'll use those insights to do the reverse: convert an expanded expression back into its original factors.

Identifying Factors.

We should start by saying exactly what a factor is. When two quantities are combined through multiplication, each of those quantities is called a factor. For example, consider the expression \(3 x y\). This expression has three factors: \(3\), \(x\) and \(y\) since they are quantites being mutliplied together to create the expression.

Sometimes the factors within an expression may not be immediately apparent. To identify factors it is helpful to think about what can divide evenly into the expression.

An example with numbers will illustrate this idea. We might ask if \(3\) is a factor of \(15\). To answer this question, we should ask ourselves if \(15\) can be divided by \(3\) without leaving a remainder. Alternatively, we can try to write \(15\) as the product of \(3\) and another number. Since \(\frac{15}{3}=5\) and \(3 \cdot 5=15\), it becomes clear that \(3\) is a factor of \(15\).

The same type of reasoning can be applied to expressions with variables and powers. For instance, is \(7x\) a factor of \(-21x^2\)? In other words, can we divide \(-21x^2\) by \(7x\) without any remainder?

The answer is yes, we can do that division and \(\frac{-21x^2}{7x}=-3x\) with no remainder. Using this result we can see that \(-21x^2=(7x)(-3x)\).

By expressing \(-21x^2\) as \((7x)(-3x)\) we are putting it in a factored form and we can say that we have factored the original expression.

There can be several different, yet equally valid ways to factor something. For example, here are several ways to factor \(6x^2\).

\begin{align} 6x^2 &= 6 \cdot x^2 \\ 6x^2 &= 6 \cdot x \cdot x \\ 6x^2 &= 3x \cdot 2x \\ 6x^2 &= 2 \cdot 3 \cdot x^2 \end{align}

You can probably come up with additional ways to factor \(6x\) beyond the four written here. While all of these representations differ in appearance, they all represent the same quantity and can be used interchangeably.

It is critical to point out that factoring involves expressing a quantity as the product of other quantities multiplied together. Although we could write \(6x^2=4x^2+2x^2\), that would not be considered a way to factor \(6x^2\) because it invoves adding terms rather than multiplying them.

Factoring the Greatest Common Factor

The examples we've seen so far contained a single term. Oftentimes, factoring comes into play when we are dealing with expression that have multiple terms. In such cases, it can be helpful to identify factors that are common to all of the terms. In particular, we will be looking for the largest common factor shared by the terms, called the greatest common factor or GCF.

Think about the expressions\(7x^2\) and \(14x\), what is their greatest common factor? The largest thing we can divide into both terms is \(7x\). How do we know that is the largest and that there isn't something bigger? If we divide both terms by \(7x\) we get:

\[\frac{7x^2}{7x} = x\] \[\frac{14x}{7x} = 2\]

Since \(x\) and \(2\) don't have any common factors, we can be confident that we've found the GCF.

Now that we can identify the greatest common factor between terms, let's move on to factoring the GCF out of an expression. Factoring out the greatest common factor involves dividing each term by the GCF and rewriting the expression in factored form.

If we take the expression \(6x^4+9x^3\) then we can use the results from our previous example since we already know the GCF is \(3x^2\).

\begin{align} 6x^4+9x^3 &= 3x^2\left(\frac{6x^4}{3x^2}\right) + 3x^2\left(\frac{9x^2}{3x^2}\right) \\ &= 3x^2\left(2x^2 + 3 \right) \end{align}

Notice that if we were to take our result and distribute the \(3x^2\) we would exactly reverse our steps and end up back with the original expression.

\begin{align} 3x^2(2x^2-3x) &= 3x^2 (2x^2) + 3x^2(-3) \\ &=6x^4+9x^2 \end{align}

We can always check our factoring by multiplying.

Factoring Quadratic Trinomials of the form \(x^2+bx+c\)

Although we should always begin by looking for a GCF, expressions may not have one. This is often the case with polynomials of the form \(x^2+bx+c\), which are sometimes called quadratic (due to the exponent of \(2\)) trinomials (because there are 3 terms). Quadratics play a pivotal role in various mathematical contexts and you will continue to make use of them in every math class from here on.

However, lack of a GCF does not mean that an expression cannot be factored. Take, for example, \(x^2+5x+6\) which has a GCF of \(1\) but can be factored as (\(x + 2)(x + 3)\). We can check this using the FOIL method or distribution.

\begin{align} (x+2)(x+3) &= x^2+2x+3x+6 \\ &= x^2+5x+6 \end{align}

This check is more than a trival thing. If we look at it closely we can unlock the process for factoring quadratic trinomials of the form \(x^2+bx+c\). Notice that the two numbers in our factors add up to \(5\) and multiply to \(6\), which are the coefficients \(b\) and \(c\) in \(x^2+5x+6\). That's the trick.

To factor trinomials of the form \(x^2+bx+c\) we need to find two numbers that add up to \(b\) and multiply to \(c\). Once we have those, the factored form will be \((x + \text{Number1})(x + \text{Number2})\).

In practice, we usually examine pairs of factors of \(c\) and see if any of them add up to \(b\). That is often quicker and easier than a guess-and-check method. Let's see this in work with a few examples.

Our final two examples showcase situations where it can be a bit harder, or even impossible, to factor a trinomial.

When an expression cannot be factored, we say that it is prime. In real life, most quadratic trinomials are prime. In class, however, the excercises are designed to allow you to practice the skills taught and most can be factored. So don't give up too early. Just because you can't find factors at first doesn't mean they don't exist. Don't delcare an expression "prime" until you have tried all possible values.

As the last example illustrates, it's only after we've considered both positive and negative factors that we can be sure we've tested every possible combination.

Conclusion

You might be wondering why we devote time to factoring when the result is equivalent to the original expression. The truth is that factored expressions offer distinct advantages in terms of simplicity. This will become especially evident when we tackle quadratic equations in Chaper 4, but it's also apparent when evaluating quadratic expressions here.

Suppose you were asked to evaluate \((x-1)(x+3)\) and \(x^2+2x-3\) for \(x=2\) by hand. Which would you find easier to work with?

Factored expressions are often more manageable for calculations because they frequently involve fewer terms or operations. Let's go through both evaluations side-by-side and see for ourselves.

\begin{align} (x-1)(x+3) &= (2-1)(2+3) & & & x^2+2x-3 &= 2^2+2(2)-3 \\ &= (1)(5) &&&&= 4 + 4 - 3 \\ &=5 &&&&= 5 \end{align}

While neither calculation was difficult, the factored expression was substantially simpler to evaluate. Lastly, if you hadn't noticed, the results are the same, because one expression is the factored version of the other!