Previous to this lesson, students have learned how to recognize and factor perfect square trinomials such as x² – 6x + 9. They have also learned to factor non-perfect square trinomials in the form Ax² + Bx + C, where A = 1. This latter lesson may be spread over two days to look at all the possible cases:
1. All signs positive. Example: x² + 5x + 6. Factored form: (x + 3)(x + 2)
2. Middle term negative, last term positive. Example: x² – 5x + 6
Factored form: (x – 3)(x – 2)
3. Last term negative. Examples: a) x² – x – 6, Factored form: (x – 3)(x + 2)
b) x² + x – 6 Factored form: (x +3)(x – 2)
Students find that factoring trinomials of this form is a “reverse FOIL” and find the FOIL mechanism helpful. For example for the trinomial x² + x – 12, students know that since the last term is negative, that the product must be a negative and a positive. This leads to writing a partial solution: (x + __)(x – __). Since the last term is –12, there are several possible factors. The factors must sum to –1. After eliminating factors such as –6 and 2, and –12 and 1, we come down to –3 and 4 and –4 and 3. Of these, –4 and 3 sums to –1 so the factored form is (x + 3)(x – 4).
Today’s lesson covers the general form of factoring Ax² + Bx + C in which A > 1, where A is an integer. These are more difficult for students since they have to consider factors for A as well as C.
Warm-Ups.
Problem 1 emphasizes the squaring of a power, and also that the middle term is found by multiplying 2(2x³)(-4y), which is -16x³y. Problem 2 requires raising the term in parentheses to the third power first, and then multiplying by 2x². Some students may view the negative sign in the parentheses as a minus sign and try to distribute; this problem should therefore be used as an example that it is a multiplication, not a distribution. Problem 3 uses squares of 18 and 14. There should be no chart of squares on the board, so that students must rely on memorization. Problem 4 eludes some students because of the obvious factors of 10 being 10 and 1—or in this case -10 and 1. Problem 5 revisits equations with fractions and how to eliminate the denominators by multiplying all terms on both sides by a common denominator—in this case, 20.
Factoring Trinomials When A and/or C are Prime Numbers. “So far we’ve been factoring trinomials where the first term has a coefficient of 1. Can you guess what we’re going to do today?”
Most students figure out immediately where I’m going with this. “I figured you would know this. So let’s look at the easy ones first. You’ve no doubt discovered that prime numbers like 3, 5, 11 and so forth only have one set of factors.”
Saying it this way opens me up to someone pointing out that if the number is negative there are two sets of factors; e.g., for -5 it is 5 and -1, and 1 and -5. I take note of such students and will try to give them more challenging problems. In the case where no one points this out, then I do so myself.
“Let’s say we have these two problems to solve.”
“The first problem, the leading coefficient is 3. There are only two factors of 3. What are they?” As the class shouts out 3 and 1 I write:
“You’ll notice I didn’t put in a plus or minus sign in the parentheses because that’s my next step. What are the factors of - 1?”
I hear -1 and 1.
“So where do I put the -1 and where does the 1 go?” This is where “reverse FOIL” comes in handy, since they are essentially trying out where the numbers go and finding the middle value by “inner product” + “outer product”.
I have them work it out in their notebooks and they get the answer fairly quickly:
Students can see they need 3x to be positive in order to have 3x – x, and the only way to get a positive 2x is for the second binomial to be (x + 1).
“Let’s try the second one. But in this case, the last number is not prime. Oh no! What’re you going to do? Well we know that 2 is prime so let’s do the first step.”
“Now what are the possibilities for the factors of 9, given that the middle term must be negative?”
I hear 9 and -1, -9 and 1, 3 and -3.
“So let’s try some combinations. Does 9 and -1 or -9 and 1 seem like a possibility?”
They quickly see that the middle value will be too high no matter where they put the 9 or -9. That leaves 3 and -3, which I have them work on by themselves. They quickly come up with the answer:
“Now on a test or quiz if you ask me ‘Is this right?’ I won’t answer you, but how would you find out if you’re right?” They know that they simply multiply the binomials.
I give more examples, with increasing complexity. We do the first together and the rest they do independently.
Examples:
I throw in a non-factorable problem as well. Something like x² + x + 1, to keep them on their toes.
More Difficult Factoring. I generally devote a separate lesson for this but include it in this section.
Trinomials for which A and/or C are not prime numbers are generally more confusing to students because of all the possible combinations of factors.
I start by writing a problem on the board:
“This problem is different from the ones we were doing yesterday. The last number is a prime, but the leading coefficient is not. But before we get into that, notice that the last number is positive but the middle number is negative. What does that tell us?”
Since they have learned this strategy already, students are ready to tell me that both numbers in the binomials will be negative.
“OK, so we will have something like this:”
“Now all we have to do is find the factors that go in the blanks. So what are the factors of 8?”
I hear 4 and 2, and 8 and 1. I try 8 and 1 first:
“What happens if I change the 1 and the 3 around?” They try this and quickly conclude that won’t work because there will be a -24x and –x.
“OK, let’s try 4 and 2. Try it in your notebooks.” I walk around and monitor how they’re doing. I’ll pick a student who gets it right to write it on the board.
I try another, a bit more involved, and work with them on it.
I establish with the students that here will be two negative factors. Also, the factors for 14 are -14 and -1; and -7 and -2. Factors for 6 are 6 and 1 and 3 and 2.
“Can you tell if we should switch the 14 and 1 around?” I want them to be able to see that if they do so they’ll have -84x and -1x which will not sum to -25x.
Similarly they should be able to see that -7 and -2 will result in too big a middle term.
“So let’s try 3x and 2x as our first terms.”
“Can we see that -14 and -1 are going to result in too big a middle term? So what should we do next?”
They will suggest -7 and -2. I’ll have them do it and walk around to check. Correct answer:
Examples. I will work with them on the first and then have them work independently, but will provide guidance as necessary.
This last one can be done easily be recognizing that the trinomial is a perfect trinomial square which they have had practice identifying. Many students do not recognize these at first, so it is beneficial to have them practice this.
Homework. For the first day, problems should be a mix of trinomials with leading coefficient equal to 1, and trinomials with prime numbers for the leading coefficient and final number. For the second day, the mix should include trinomials in which the numbers are non-prime.