## 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)

1. Andrea burnes says:

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?

3. Nick M says:

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! =)

4. James says:

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.

6. Amanda S says:

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.

7. Ally H says:

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.

8. mike says:

this method doesnt work. it will work only 30% of the time to i recommend using a diff method

9. lynnie says:

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!

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.

12. Kassidy says:

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!!

13. 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.

14. Dana says:

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

15. 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?

16. Dana says:

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.

17. Sue says:

I can’t justify this explanation to students. In your explanation, how do you justify that (ax-m)(ax-n) = (ax)^2 + b(ax) + ac? Why not split the middle tern and factor by grouping? I feel that this method is logical and easily justifiable.

18. I agree with you in that the explanation for this method is hard for students to understand. I prefer to teach the factor by grouping method as well – but this slide and divide is used by a lot of teachers who need different methods to reach different students, even if it just allows them to find the answers versus truly understanding the math behind it.

19. June Nicholas says:

Hi everyone,

I’m not crazy about this method because it “breaks rules.” Two methods that work better are given here. I use the first one in my college algebra classes, and it is a modification of a similar method from some of Sullivan’s books. The second one I just learned from my son’s algebra teacher and requires greater manipulation of the products for A and C. It is from some textbook she had and was used for enrichment. Both are better than any other methods I’ve ever seen, and NEITHER method requires dividing out common factors first.

For a trinomial Ax^2 + Bx + C where A MAY or MAY NOT be equal to 1:

“AC” Product Method
• Step 1 – find the value of AC
• Step 2 – find integers a and b whose PRODUCT is AC and whose SUM adds up to B (order of arrangement of a and b WILL NOT matter) – can do in a vertical column for easier visibility. Don’t forget to account for signs!
• Step 3 – rewrite as Ax^2 + ax + bx + C
• Step 4 – factor by grouping where (Ax^2 + ax) + (bx + C) and factor out GCF from each group. You will be left with a common factor (kx + m) for both groups. NOTE: Sign in the middle may be left as positive and b can be negative, but factor out -1 accordingly from (bx + C).
• Step 5 – factor out the common factor (kx + m) from each group and you are left with the other factor (lx + n)

“KLMN” Method – best to do steps 1-3 in a horizontal table for easier visibility
• Step 1 – find the factors of A and call them (k, l)
• Step 2 – find the factors of C and call them (m, n)
• Step 3 – multiply and add kn + ml = B, or “outside” + “inside”
• Step 4 – write as (kx + m) (lx + n)
NOTE: you might have to reverse the order of k and l and/or m and n a few times in order to get the correct kn + ml = B combination. Also a little harder to account for signs.

Happy teaching!

20. Thanks June, you’re absolutely right. I taught the slide and divide method the year I wrote this, but was unhappy with the results. I changed to the factor by grouping method (your first example, which I wrote about in another post) but that post is not as popular as this one. I’ve left this post up, though, because it’s good for teachers to have several methods up their sleeve just in case, maybe as a last resort for students who just don’t get it.

21. Kevin says:

this is good, and I like this method, but I was also wondering if you have taught the students the “completing the square” method as well. That one is also amazing

22. I would love it if you would share that method with us here, I don’t think I’ve heard of it.

23. James says:

4x^2 +12x+ 8

1. x^2 + 12x +32
2. (x +4) (x+8)
3. (x + 4/4) (x+ 8/4)
4. (x+1) (x+2)

But (x+1) (x+2) doesn’t equal 4x^2 +12x+ 8

24. Okay – so the reason it doesn’t work is because there is a GCF that needs to be factored out and that should always be the first step. I think I gave an example where it did work without factoring out the GCF first but that was probably a fluke.

25. anonymous says:

will this work in trinomial?

26. Jose Harris says:

Even if you don’t remove the GCF as in the example you cited, you get the roots of the polynomial. If proper factoring is the point, then rigor must be followed; if only the roots are needed, then it’s not required. I prefer rigor.

27. Lucas Buck says:

One of my fellow teacher solved the issue of students forgetting to remove the GCF first. Once they have finished factoring using slide and divide check for a GCF by multiply the two x terms of the factor. If it makes the x^2 term then there is no GCF. If it does not match the x^2 term then the GCF is the multiple that would give you the x^2 term. For instance the problem 18x^2 – 33x + 12. When you factor using slide and divide, you get (2x – 1)(3x – 4) it is missing the GCF of 3 that was not pulled out at the beginning. So you take 2x time 3x which gives 6x^2 what time this equals 18x^2 that would be 3 which is your GCF. Check it out pretty cool I think and a great way for students who always forget to take out the GCF.

28. Walter says:

Thank you for this. It will make factoring so much easier!

29. Coach J says:

Slip and slide has proven to be the easiest that I’ve seen. Takes 3 basic steps if done correctly.

For quadratics with a=1(meaning no common factors-I have them factor out GCF first-always), I teach students FACTOR/COMBO-a simple sentence that they can use and just have them fill in the blanks. “FIND FACTORS OF ________ THAT COMBINE BY __________ TO GIVE _______.” Fill in the first blank with factors of the constant term, second blank with ‘addition’ or ‘subtraction’(depending on the sign of the constant term), and the third blank with the positive value of the middle (linear) terms coefficient. I don’t worry about positive or negative values for the factors or the middle term. In your two binomials, put the sign of the middle term in the first binomial and the largest factor you use of the constant term. Now you know the sign in one of the binomials. If you combine by addition the other sign will be the same. If you combine by subtraction, the other sign will be different. This will take care of the negative or positive signs. I have students write pairs of factors that give the constant term until they find the pair that combine to give the middle terms positive value. Using a systematic approach helps keep students from missing any pairs. I also use this as a chance to review ways to test if a number is a factor(ends in 0 or 5, adding digits up to test for 3 or 9, etc.)

Example:
24
1 x 24
2 x 12
3 x 8
4 x 6 (when you get to the point where there are no more factors between a pair, then you’re done! Between 4 and 6 is 5 and it’s not a factor of 24, so I’m done!)
Slip and Slide Method
(Use this method when the quadratic term’s coefficient is NOT = 1)
Example: Factor 3×2 – 5x – 2
Step 1. Multiply the constant term by the quadratic term coefficient (a • c). Drop the quadratic coefficient.
__x^2 – 5x – 2(3)

Step 2. Factor: x^2 – 5x – 6 by factor/combo:
(x –__)(x __ __)
Bring the sign of the middle term down to the first binomial. FACTOR/COMBO: Find factors of 6 that combine by subtraction to give you 5 .
(6 & 1 since 6 – 1 = 5. Disregard the signs! )
Put the larger factor in the 1st binomial & the smaller factor in the 2nd binomial. Since you combined by subtraction the signs of the factors will be different. If you combined by addition, the signs would be the same!
So, x^2 – 5x – 6 = (x – 6) (x + 1)
Step 3. Divide each constant of the 2 binomials by the original leading coefficient (3)-> (x-6/3)(x+1/3), then reduce and/or slide the coefficient back.
Answer: 3x^2 – 5x – 2 = (x-6/3)(x+1/3)
= (x – 2)(3x + 1)
WORKS EVERY TIME!!

30. yes, this is for factoring trinomials, particularly ones with leading coefficients that students have the most difficulty with.

31. Brock England says:

I found one that does not work with this method

3x^2+13x-10

32. This set of numbers (factors of 30 that give 13) has always thrown students off because the pairs 2, 15 and 3, 10 are both feasible. If a student thinks of the 3, 10 pair first, they will frustrate themselves trying to make those factors fit and won’t realize there is a second pair that would work.

3x^2 + 13x – 10
x^2 + 13x – 30
(x + 15/3)(x – 2/3)
(x + 5)(3x – 2) = 3x^2 – 2x + 15x – 10 = 3x^2 + 13x – 10