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
Thanks for the tip about starting hard then go easier. It is so true! I will use the slip and slide method after they learn it the “right” way, or as “survival math” (when a kid just can’t get it.
Please keep blogging. It is a treasure for us math teachers.
I was searching through google for methods to teach factoring because although I have a cutsie method that always works (Tic Tac toe Method) I agree with you the kids last year didn’t really understand why it worked. As a result I found both of your blogs on factoring and thus your worksheets of which I am now a subsciber. Still tossing and turning about which direction to go on factoring, but I’m with Sue W — keep blogging. It helped me tremendously today.
To be honest, even though I finished the factoring on a considerably more successful note than usual, I’m still not satisfied with the results. Most of my students can’t factor x^2 + 3x + 2 without going through the process of factors of 2 that have a sum of 3 are +2 and +1,
so x^2 + 2x + 1x + 2 = x(x + 2) + 1(x + 2) = (x + 1)(x + 2). I guess that’s not a bad thing, since all it does is take them a little longer, but I don’t know why they just don’t see the pattern that if the two factors are 2 and 1, then those are what the constant of their two binomials are going to be when they’re dealing with leading coefficients of 1. Students are so methodical when it comes to math that they rarely step back to think about it and see the bigger picture. I guess I just traded the ability for students to factor quickly for the ability to factor “correctly,” especially so they can factor those harder quadratics. Is that a bad trade-off?
A question: Why is it with the slide and divide method that you MUST factor out a GCF, otherwise the answer will be incorrect?
I’m answering without really sitting down and analyzing this, but I believe that it has to do with the fraction portion of the slide and divide (the last step where you “put back what you slid out of the way). For instance, even though 5/10 and 2/4 are both equivalent to 1/2, when you simplify in that fraction step of this method, you lose track which fraction you needed in the first place. For example, in the following problem (done without factoring GCF and therefore the wrong answer):
4x^2 – 14x + 10
x^2 – 14x + 40
(x – 10)(x – 4)
(x – 10/4)(x – 4/4)
(x – 5/2)(x – 1)
(Wrong) Answer: (2x – 5)(x – 1)
If you multiply it back out, you would get 2x^2 – 7x + 5, which even though it is obviously wrong, it is only wrong by a factor of 2 (which was the GCF that needed to be factored out first).
Now here it is done correctly:
4x^2 – 14x + 10
2(2x^2 – 7x + 5)
2(x^2 – 7x + 10)
2(x – 5)(x – 2)
2(x – 5/2)(x – 2/2)
2(2x – 5)(x – 1)
I suppose you could simply not simplify that final fraction and leave it as 10/4, then input the rule that if one of your binomial factors is not simplified, then take out the common factor at the end, but you see the problem it might create for student, because how do you distinguish which fractions to simplify and which to leave alone. In this problem, I could possibly leave it as 10/4 but I had to simplify 4/4 as 1, (otherwise the second factor would seem to be 4x – 4 instead of x – 1). I suppose the rule of thumb is maybe you only simplify those fractions that simplify to a whole number?? I haven’t checked this out yet, that might not be the case)
Trying the problem again, without simplifying that first fraction.
4x^2 – 14x + 10
x^2 – 14x + 40
(x – 10)(x – 4)
(x – 10/4)(x – 4/4)
(4x – 10)(x – 1) (Now you need to factor 4x – 10 by pulling out a 2
2(2x – 5)(x – 1)
Thanks, will try this method!
You must always factor out any GCF first or the resulting factors will not be prime. That’s true in any method of factoring you use.
4x^2 – 14x + 10
= 2(2x^2-7x+5) GCF factoring
= 2(x^2 -7x+10) Slide
= 2(x-5)(x-2) Factor
= 2(x-5/2)(x-2/2) Divide
= 2(2x-5)(x-1) Completely factored.
So after you get the two middle terms how do you know where to put them. So for your first example
“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″
How do you know its the 8x than the 5x? I can see why after you factor by how do you tell kids that? Do you make them do it both ways or is there a trick I’m missing? Thanks!
Great question – the order of the two numbers doesn’t matter:
2x^2 + 8x + 5x + 20
= 2x(x + 4) + 5(x + 4)
= (2x + 5)(x + 4)
OR
2x^2 + 5x + 8x +20
= x(2x + 5) + 4(2x + 5)
= (x + 4)(2x + 5)
I hope I answered the question you were asking.
Awesome, thank you very much! I just got done teaching it this semester so next semester I will try this out.
Thank you so much. I am tutoring someone and his teacher is not doing a good job at showing him this so I was looking for a different method. I think he will understand this much better.