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)
Curious if you know the source of the slip and slide factoring method. Blew my mind away. Thanks.
No one seems to know. I actually have the students factor those trinomials on their own (guess and check basically) before I show them the short cut, so they’ll better understand the concept behind factoring, and so they’ll appreciate the simplicity of this method. The only flaw with this slip and slide method is students often forget to factor out the GCF, they just want to dive right in…anyone out there know anything about this?
THANK YOU!!! I have not known how to factor my entire high school career and now a freshmen in college and have a test tommrow!! And I just learned the “Slide and Divide,” method it works great! I’ve always been taught some guess and check method that always got confusing for me, and my current professor taught another method i didn’t understand…. So all in all thanks so much looking forward to tommorows test! =)
This works because you make the substitution y=ax. Where a is the leading coefficient. If you are going to teach this method you should mention this.
Of course. Thank you for clarifying.
Wow … o.o;;; I’ve been tutoring FOREVER and I had never heard about this method.
Thanks a bunch! This is way cool. Either that or I really, really need to get more sleep.
I tried this today!
LOVE it, but had a little trouble when the leading coefficient (a) is negative. It stills works but takes some extra explanation.
this method doesnt work. it will work only 30% of the time to i recommend using a diff method
This was fantastic! I am teaching this tomorrow and it is so difficult to teach the guess and check method!
Thats terrific!
The method is so simple and easy!
Thanks for uploading this.
Tanisha
Vedic Maths Forum India
In response to mike – I’d love to see a problem in which this doesn’t work. All the ones in a typical Algebra1 book have been factorable with this method.
I want to know how y=ax is the substitution… are we assuming y=ax^2+bx+c? or is y a new variable? I would really like to see this written out if possible!!
I got this directly off of Math Forum:
Here’s an example. We’ll factor
6x^2 – x – 12
We take the 6 from the first term and multiply the last by it:
x^2 – x – 72
Factoring this, we see that 72 = 8*9 and 8 – 9 = -1, so
x^2 – x – 72 = (x + 8)(x – 9)
Replacing x with 6x and pulling out common factors,
(6x + 8)(6x – 9)
becomes
(3x + 4)(2x – 3)
This is the desired factorization.
Now I’ll prove that it works:
Suppose you are factoring
ax^2 + bx + c
You factor
x^2 + bx + ac = (x-m)(x-n)
(In my example, m = -8, n = 9.)
Now you replace x with ax:
(ax-m)(ax-n) = (ax)^2 + b(ax) + ac
We’ll have to assume that when you remove common factors, the product
of the factors you divide out is equal to a. (In my example I took
out 2 in the first factor and 3 in the second, whose product is 6.)
I think this is a consequence of the assumption that a, b, and c have
no common factors, but I’m not going to bother to prove it.
With this assumption, the resulting factorization is the same as
(ax-m)(ax-n)/a = [a^2x^2 + abx + ac]/a = ax^2 + bx + c
So if you can follow your process, you have indeed factored the
original trinomial.
From Ask Dr. Math
I teach it by doing the same as when “a” is 1, substituting the factors for “bx” and then grouping.
ex. 6^2 – x – 12
6x^2 – 9x + 8x – 12
3x( 2x – 3) + 4(2x – 3)
(3x + 4)(2x – 3)
I can see my students forgetting to put “a” back and divide
Yes, I don’t teach factoring using the slide and divide anymore, although it seems the most popular method. I’ve used your method now for years, but I’m just wondering – how do you know to use the factors -9 and +8? I have my students multiply 6 and -12, so they know they need factors that when multiplied = -72, and added = -1. Is this the same method that you use?
Yes, but instead of “plugging” the -9 and +8 in parenthesis, they use the fact that they are a substitution for the -1 and use grouping to get their factors.