Previous to this lesson, students will have learned to completely factor an expression. Thus, the expression 3x² - 27 requires two factoring steps: 3(x² - 9), which then can be further factored as 3(x – 3)(x + 3). A common mistake students make is to omit the 3 at the front of the factored expression.
Complete factoring also addresses expressions like – x² +4x – 4, in which – 1 is factored:
Complete factoring is especially important in making factoring easier as shown below:
This lesson addresses the zero property of multiplication; that is, if ab=0, and a does not equal zero, then b must equal zero—and vice versa. This property is used to solve quadratic equations that lend themselves to factoring and the zero property rule. Later, students will learn to solve quadratic equations which are not factorable. It is important that students continue to practice solving these types of factorable quadratics as preparation for the future chapter.
Warm-Ups.
Problem 1 requires factoring of each successive difference of squares. Problem 2 has two variables; students must identify the factors of -35y² as -7y and 5y, or 7y and -5y. Problem 3 can be solved by multiplying each term by a common denominator. Problem 4 requires students to see that x equals zero divided by the product of the numbers that are multiplied by x and serves as a segue to the day’s lesson. Problem 5 reviews negative exponents and the zero exponent. In the case of the zero exponent, it is important for students to realize that 2g⁰ is 2(1) and is not the same as (2g)⁰.
The Zero Property Rule. I point out that Problem 4 of the Warm-Ups illustrates the principle that anything times zero is zero. So if we have something like 32∙5x = 0, we can see that x must equal zero. This is called the Zero Property Rule.” I then write it on the board:
A product is zero if, and only if, at least one of the factors is zero.
“If we have an equation like xy = 0, the equation will be true if x = 0, or y = 0, or both. Suppose we have an equation like 5(x – 5) = 0. What does the Zero Property tell us?”
If there are no immediate responses, I supply a hint. “We have two factors, one of which is 5 and the other is (x – 5). Look at the Zero Property Rule.”
Generally a few students will see the connection and tell me that x – 5 is zero.
“Right; we know that 5 is not zero, so that only leaves one other factor. And since we now know that x – 5 = 0, can we solve it?”
This is a simple equation the likes of which they’ve solved before; yet for some students it suddenly seems unfamiliar. In the context of new information, even the familiar can seem strange, and students think they must do something different with the equation than they’ve done before. So I usually offer a hint. “How do we isolate x? What do we do with –5?”
This usually does it and they solve it. “You add 5 to both sides, but it’s even obvious without that step that x must equal 5.”
“How would you solve (x – 1)(x – 4) = 0 using the zero property?”
I have them look at the rule again if there are no immediate responses, but there usually are a few among the students who are catching on. Someone will say that either one of the factors must equal zero.
“Then that means that either (x – 1) is zero, or (x – 4) is zero. Knowing this, what are our possible solutions for x that make this equation true?”
I hear x = 1 and 4.
“Yes. We can see that’s true. If x equals 1, than plugging that in to the equation we get 0∙(x-4) which equals 0. Same deal with x = 4. We have two solutions for this equation. We call this the “solution set”; the solution set for this equation is 1, -4.”
Examples: Students are to solve the equations using the Zero Property Rule.
“This last equation is what is called a quadratic equation. These are equations that are in the following form.” I write on the board.
“These have a squared term, which makes this an equation of degree 2, since 2 is the highest exponent of all the monomials. Let’s try one.”
“What if we have something like this?”
“Can we do something to make this easier to factor? What did we do yesterday?”
This should ring a bell and they will tell me to factor out the 2. Doing so results in:
“We have two factors. We can divide by 2 if you want, or you can just look at it and using the Zero Product Property we know at least one of the factors—that is, 2 and x² + 5x - 14 must be zero. So which one is not zero?” They quickly tell me it is 2.
“Now factor the trinomial and tell me the solutions.”
Examples:
This is similar to Problem 3 above, except that it isn’t in standard form. “If you don’t put it in standard form, you might decide to divide both sides by x and get x = 3, and think you’re done. But you’re not. You’ve forgotten a solution. In standard form you have two solutions.”
Homework. I set aside time to work some of the homework problems with them so they get the hang of the various types of problems.