Factoring Trinomials – Connect 4 Game

Let your students play Connect 4 while factoring trinomials. All trinomials have a leading coefficient of 1. Directions and game board included.

It calls for a set of dice for each pair of students, but I think the game can be played without dice. Students could simply pick their positions, blocking each other as needed. If it’s too easy to get 4 in a row, then maybe see who can get the most 4 in a row.

Teaching Factoring

I had written a post a while back on how to factor trinomials using the slide and divide method (or slip and slide, it has numerous names). However, I did not use that method this year when I taught this unit. The reason for that is because I don’t think it enables students to understand factoring, it was simply a memorized algorithm. That’s not always a bad thing, but they weren’t able to relate this skill to other types of factoring. First, there are the basic trinomials that have a leading coefficient of 1. Those are fairly easy to do quickly, once you understand the concept of finding factors of ac whose sum = b. Then, for trinomials with a leading coefficient other than 1, there was the slide and divide. Then they had to understand to factor out any GCF’s, which sounds simple enough, but most students want to jump right into the problem and don’t take the time. Then they get to factor by grouping, which doesn’t seem related to all the factoring of trinomials that they’ve been doing, so by the end of the unit, many students simply became confused.

This year, I went strictly with factoring by grouping. As strange as it might seem, I taught trinomials with coefficients other than 1 first. For the problem 2x^2 + 13x + 20, students multiplied 2(20) = 40. So they had to find factors of 40 that had a sum of 13. They used the numbers to rewrite the problem as 2x^2 + 8x + 5x + 20. Then they factored by grouping: 2x(x + 4) + 5(x + 4). So the two factors are (2x + 5)(x + 4).

Next I taught the basic trinomials (with leading coefficient of 1). These were a lot easier.
For the problem x^2 – 12x – 28, students still multiplied the leading coefficient with the constant, but since the leading coefficient = 1, then the number they arrive at is -28. So factors of -28 that had a sum of -12 are -14 and + 2. They rewrite the problem as x^2 – 14x + 2x – 28 and then factored by grouping. x(x – 14) + 2(x – 14). So the 2 factors are (x + 2)(x – 14). For some reason, when you add an extra step to a method, students complain about it being so hard. But when you go from “harder” to “easier,” the students love how it’s so much “easier.” By the way, it is at this point that I explain the relationship of multiplying 2 binomials and factoring, which is basically “undoing” multiplying 2 binomials, so that they understand what factoring is.

Now when you have trinomials with a GCF, it doesn’t matter. Take 4x^2 – 32x + 64. Students tend to forget pulling out the GCF, but with this method, it’s no big deal. The product of 4(64) = 256. (Okay, yes it makes for large numbers). So factors of 256 that have a sum of -32 are -16 and -16. They rewrite: 4x^2 – 16x – 16x + 64, then factor. 4x(x – 4) – 16(x – 4). So the 2 factors are (4x – 16)(x – 4). So there’s an added step. If a factor can be broken down further, then they must do that. So it ends up being 4(x – 4)(x – 4). All done, and you can see that it doesn’t matter if they forget to factor the GCF first. Of course, for many problems, they’ll want to since they’ll just end up with huge numbers. I tell them that if they end up with such a large number, that’s a clue that maybe they should factor out a GCF first.

Factoring special cases involves no extra steps and work as any other trinomials. Take x^2 – 100. Since the product of 1(-100) = -100, then they’re looking for factors of -100 that = 0. (since there is no x term, there are zero x’s and so b = 0) So it must be 10 and -10. Rewrite: x^2 – 10x + 10x – 100 then factor: x(x – 10) + 10(x – 10). (Many students will recognize by now the pattern and will understand that the 2 factors are (x + 10)(x – 10) without doing that middle step). What about those with a GCF? 3x^2 – 363 for example. Well, 3(-363) = -1089. Students might discover that the factors of -1089 that have a sum of zero are 33 and -33, so that 3x^2 + 33x – 33x – 363 = 3x(x + 11) – 33(x + 11). The factors would be (3x – 33)(x +11) but they must break the first factor down to 3(x – 11)(x + 11)) However, -1089 is such a huge number that most students will now be hoping that there is a GCF and there is: 3(x^2 – 121). So now they can continue factoring: 3(x^2 – 11x + 11x – 121) which factored leads to the same answer of 3(x – 11)(x + 11).

And finally, by the time they get to the factor by grouping method, they completely understand and find these easier than the rest. Take for example 4x^3 – 20x^2 + 3x – 15. They don’t have to multiply, search for factors or anything. It’s already set up for them: 4x^2(x – 5) + 3(x – 5). So the 2 factors are: (4x^2 + 3)(x – 5).

Also, keep in mind that 1 is still a factor. In the problem 2x^2 + 5x + 3, 2(3) = 6, so factors of 6 that have a sum of 5 are 3 and 2. Rewrite: 2x^2 + 3x + 2x + 3. Now the first half has a common factor of x. But the second half, which seemingly has no common factors, does have that common factor of 1 and since you MUST factor something, factor out the 1.
x(2x + 3) + 1(2x + 3). Which makes the factors (x + 1)(2x + 3)

So although slide and divide seems an easy enough algorithm, in the long run, the more connected each topic seems, the easier it will be on the students (and on you!). Hope this helps.

algebrafunsheets

Slide and Divide Method of Factoring Trinomials

Someone asked about how to teach factoring trinomials where the leading coefficient is not 1, so I thought I would post it here as well.

The best method I’ve found is often referred to as the slide and divide method, for trinomials of the type ax^2 + bx + c

Anyways, take the trinomial 3x^2 + x – 10.
(Always ensure to pull out any common factors before continuing with this method).
First you slide the leading coefficient to the end and multiply it by the constant:
You get x^2 + x – 30
Then factor like normal
(x + 6)(x – 5)
You have to “put back” the number you slid out, so to speak. You do this by dividing the constant in each factor by the leading coefficient you “slid” out of the way.
(x + 6/3)(x – 5/3)
Simplify the fractional terms you end up with.
(x + 2)(x – 5/3)
Once it’s simplified, if there’s a fraction left, the denominator becomes the coefficient of the variable term.
ANSWER: (x + 2)(3x – 5)

Here’s another example without the explanations:

2x^2 – 7x + 5
x^2 – 7x + 10
(x – 5)(x – 2)
(x – 5/2)(x – 2/2)
(x – 5/2)(x – 1)
ANSWER: (2x – 5)(x – 1)

Finally, an example where you have to pull out a common factor first:
12x^10 + 42x^9 + 18x^8
6x^8(2x^2 + 7x + 3)
6x^8(x^2 + 7x + 6)
6x^8(x + 6)(x + 1)
6x^8(x + 6/2)(x + 1/2)
6x^8(x + 3)(x + 1/2)
6x^8(x + 3)(2x + 1)