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Section 9.1 Lots of Methods, Not All Good

Factoring quadratic polynomials is generally seen as an important marker for a student's algebra skill level. Because of this, various teachers have tried to design particular methods to help students factor correctly. Unfortunately, many of those methods are answer-oriented, which leads to students getting into the habit of making incorrect algebraic manipulations.

In this section, the focus is going to be on understanding the process of factoring and the ideas behind it. Although getting the right answer is also important, the right answer is the outcome of a proper understanding of the concepts, not the ultimate end in itself.

To emphasize this point, we're going to use a purely geometric framework and sidestep all of the algebra. A couple sections ago, we introduced algebra tiles. These are geometric images that represent different quantities to help us think about algebraic relationships.

Algebra tiles come with a set of "rules" for how they can be manipulated, which is especially important when you're working with negative values. However, for the purposes of illustration, we're just going to focus on the situation where all of the values are positive. This means that we're going to be thinking about three different tiles (both tiles should be viewed as the same piece, just rotated relative to each other):

A representation of a quadratic polynomial using the algebra tiles is simply a matter of having the right number of pieces:

The challenge is then to arrange all of these pieces into a rectangle where the small squares are at the bottom right, the large square (or squares) are on the top left, and the rectangles fill in the spaces on the top right and lower left. Once a rectangle is found, it can be used to determine the factorization.

Activity 9.1.1. Factoring a Quadratic by Diagram.

We're not going to focus on trying to give you a "method" for finding these rectangles. We're going to trust that your geometric intuition will lead you to the appropriate conclusion.

Try it!

Use a diagram of algebra tiles to factor \(x^2 + 7x + 12\text{.}\) Draw the diagram and write the final equation.

Solution.

Definition 9.1.1. Monic and Non-Monic Quadratic Polynomials.

A quadratic polynomial of the form \(ax^2 + bx + c\) is called monic quadratic polynomial if \(a = 1\text{.}\) If \(a \neq 1\text{,}\) then we call it a non-monic quadratic polynomial.

Activity 9.1.2. Factoring a Non-Monic Quadratic by Diagram.

Algebra tiles can be used to factor both monic and non-monic polynomials. It all comes down to finding the right rectangle.

Try it!

Use a diagram of algebra tiles to factor \(2x^2 + 9x + 4\text{.}\) Draw the diagram and write the final equation.

Solution.

While the use of diagrams (or physical manipulatives) may help us to get answers, it doesn't necessarily help us to think in an organized and logical manner. So we are going to look at a structured algebraic approach to the same challenge.

But before we do that, let's take a look at why factoring is difficult. Let's take a look at the steps of expanding out a product.

\begin{equation*} \begin{aligned} 2x^2 + 7x + 6 \amp = 2x^2 + x + 6x + 6 \\ \amp = 2x^2 + 2x + 5x + 6 \\ \amp = 2x^2 + 3x + 4x + 6 \end{aligned} \end{equation*}

At this point, you can still factor by grouping to get back the original product.

\begin{equation*} \begin{aligned} 2x^2 + 7x + 6 \amp = 2x^2 + x + 6x + 6 \\ \amp = 2x^2 + 2x + 5x + 6 \\ \amp = 2x^2 + 3x + 4x + 6 \end{aligned} \end{equation*}

The challenge arises once the like terms are combined. We no longer have access to the rationale behind uncombining the like terms. In fact, there are lots of ways to uncombine like terms!

\begin{equation*} \begin{aligned} 2x^2 + 7x + 6 \amp = 2x^2 + x + 6x + 6 \\ \amp = 2x^2 + 2x + 5x + 6 \\ \amp = 2x^2 + 3x + 4x + 6 \end{aligned} \end{equation*}

Some students learn how to do this with monic polynomials with a guessing method, but it very often falls apart on them when they are working on non-monic polynomials. At the core, most of the problems arise from the students learning to simply "write down" their answer without understanding how or why it worked. We want to avoid that type of method because it does not enhance mathematical thinking.

We will be using the "\(ac\) method" of factoring. The goal is to uncover the right way to uncombine like terms so that we can factor by grouping. Here is a representation of the method:

\begin{equation*} ax^2 + bx + c \longrightarrow \left\{ \begin{array}{l} \text{Multiply to $ac$} \\ \text{Add to $b$} \end{array} \right. \end{equation*}

As the text states, the goal is to find two numbers that multiply to and add to The values you obtain from this will be the way to uncombine the like terms, which will allow you to factor by grouping.

Activity 9.1.3. The \(ac\) Method with a Monic Quadratic.

We will start by applying this method to factor \(x^2 - 3x - 10\text{.}\) We will first translate our quadratic into the two target properties.

\begin{equation*} x^2 - 3x - 10 \longrightarrow \left\{ \begin{array}{l} \text{Multiply to -10} \\ \text{Add to -3} \end{array} \right. \end{equation*}

And here is where you simply have to use your experience with numbers to try to find the right combination. What two numbers multiply to -10 and add to -3? You should be able to determine that the values are -2 and 5. These two values are used to uncombine the like terms (\(-3x = -5x + 2x\)), which then allows us to complete the factorization using grouping. Here is the complete presentation.

\begin{equation*} \begin{aligned} x^2 - 3x - 10 \amp = x^2 - 5x + 2x - 10 \amp \eqnspacer \amp \text{The $ac$ method} \\ \amp = x(x - 5) + 2(x - 5) \amp \amp \text{Factor by grouping} \\ \amp = (x + 2)(x - 5) \amp \amp \text{Factor out the common factor} \end{aligned} \end{equation*}

Try it!

Use the \(ac\) method to factor \(x^2 - 3x - 10\) using a complete presentation, but instead of un-combining the middle term using \(-3x = -5x + 2x\text{,}\) swap the order and un-combine it using \(-3x = 2x - 5x\text{.}\)

Solution.
\begin{equation*} \begin{aligned} x^2 - 3x - 10 \amp = x^2 + 2x - 5x - 10 \amp \amp \text{The $ac$ method} \\ \amp = x(x + 2) - 5(x + 2) \amp \amp \text{Factor by grouping} \\ \amp = (x - 5)(x + 2) \amp \amp \text{Factor out the common factor} \end{aligned} \end{equation*}

One of the main values of the method is that it works for both monic and non-monic polynomials. This benefit allows us to simplify the process of factoring quadratic polynomials to a single approach that works for all situations. So by adopting the method, we don't have to create two different sets of methods to accomplish the same goal.

Activity 9.1.4. The \(ac\) Method with a Non-Monic Quadratic.

We will use this method to factor \(4x^2 + 5x - 6\text{.}\)

\begin{equation*} 4x^2 + 5x - 6 \longrightarrow \left\{ \begin{array}{l} \text{Multiply to -24} \\ \text{Add to 5} \end{array} \right. \end{equation*}

With a little bit of thought, you will find the combination of and will satisfy the goal.

\begin{equation*} \begin{aligned} 4x^2 + 5x - 6 \amp = 4x^2 + 8x - 3x - 6 \amp \amp \text{The $ac$ method} \\ \amp = 4x(x + 2) - 3(x + 2) \amp \amp \text{Factor by grouping} \\ \amp = (4x - 3)(x + 2) \amp \amp \text{Factor out the common factor} \end{aligned} \end{equation*}

Try it!

Use the \(ac\) method to factor \(2x^2 + 5x - 12\) using a complete presentation.

Solution.
\begin{equation*} \begin{aligned} 2x^2 + 5x - 12 \amp = 2x^2 - 3x + 8x - 12 \amp \amp \text{The $ac$ method} \\ \amp = x(2x - 3) + 4(2x - 3) \amp \amp \text{Factor by grouping} \\ \amp = (x + 4)(2x - 3) \amp \amp \text{Factor out the common factor} \end{aligned} \end{equation*}