5 easy ways of factoring a trinomial ax^2+bx+c

Learn how to factor a trinomial in ax^2+bx+c when a is not 1, including the AC method, factoring by grouping, the box method, slide and divide, and the tic-tac-toe method. Factoring a trinomial is an important part of your algebra class especially when you need to solve a polynomial equation with a degree 2 or higher. This algebra lesson will help you master your factoring skills for your algebra class. Subscribe to @bprpmathbasics for more math tutorials.

How to use the quadratic formula to factor trinomials: 👉
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#math #algebra #mathbasics

0:00 Helping you master factoring trinomials
0:09 1st, AC method with grouping
2:08 2nd, AC method with the box
3:35 3rd, AC method, but weird version
4:52 4th, slide & divide
6:33 5th, tic-tac-toe factoring method

[jimhaskell]

I've been a math teacher for 29 years, and I've never seen the "slide and divide" method. Love it!!

[ndrslvn]

I prefer the first method because it doesn't require any "magical" steps, just basic algebra.

[DD]

Here because I'm a first-year CS student and I forgot how to solve a quadratic

[tobybartels]

I always do the a-c method, because it's systematic; there's no trial and error, just going through a list of divisors. Plus, if it can't be factored, then you'll know it; you won't worry that you missed an option. (Generally I do a-c by grouping, to not introduce new elements, although the box is slightly faster.)

[ivannovalery]

I usually teach the final method like this

( a + c) | bc
( b + d) | ad
————— —— +
6x^2 - 3 |+7x
a and b are factors of 6x^2.
c and d are factors of -3
Cross multiply these factor to get bc and ad. add them together and use +7x to check if your selection of factors are correct

[elliekittymeow]

If you show this video to twitter users they will literally explode. Also, you're a great teacher, thanks a lot!

[ASChambers]

It’s always nice to see different ways to solve common problems. Thanks for sharing.

[bprpmathbasics]

the pq quadratic formula that they don't teach you in school

[johnroome]

I normally use the Factor Theorem, where if f(y) = 0, then x-y is a factor. Can be difficult because you sometimes need fractions, ie f(y/3) = 0 means 3x-y is a factor. You also have to use substitution and poly long division, but it’s a satisfying method and works for polynomials of all orders, not just quadratics.

[schlingel]

I think first method makes most sense. It is also very easy.

[nathanbarnes]

The best method is not any of these 5, or the quadratic formula. Completing the square is goat

[tanelkagan]

@bprpmathbasics

Great video as always, but a few observations:-

1. In Method 1 - AC with grouping - having found two numbers that sum to 7 (being -2 and 9) you say "if one of the numbers is negative, write it down first". I can't quite figure out why that's important.

Clearly in step 2, if you were to write instead 6x² + 9x - 2x - 3 this becomes 3x (2x + 3) - (2x + 3), which again factors in exactly the same way to (3x - 1)(2x + 3). I haven't tested many cases, but it seems to me that if the quadratic *does* factor in a nice way like this, then it won't matter how you group, you'll always get the same result. It links to Method 2 - all that's happening is that we're either grouping "across rows" or "across columns", but either way should work equally well, right? Do you have a particular reason for wanting to write the negative number first?

2. Purely my opinion, but Method 3 (the "weird" AC method) and 4 ("Slide and divide") would seem to involve more things to remember, and more things that could go therefore wrong. They're probably not methods I would recommend to anyone that isn't already very familiar with factoring - are there cases where these methods are easier to use?

3. In Method 5 ("Tic Tac Toe") - you say "We ask ourselves 'what times what will give us 6x²?' and the correct combination is 2x and 3x". Well, that is *one* combination, sure, but the reason why it's the *correct* combination is rather much glossed over. As a beginner, it's not obvious, and I think it would be useful to establish why you've rejected the combination 6x and x. In other cases, there might be three or more combinations (eg. 12x² might be factored to x and 12x, 2x and 6x, or 3x and 4x.

Of course the wrong combination is not going to work as you progress through the method, but in your example if you were merely selecting 2x and 3x because you already knew the answer, that obviously wouldn't be a valid approach for someone looking at the problem for the first time.

Thoughts?

[WilliamMarkRock]

I forgot the one I took at school but I also didn't like any other method because they didn't seem familiar. The one in the school textbook was different from the one that was explained by the teacher, and I couldn't understand it. This one has both, plus other methods I found on the internet, and it explains them all very well. =

[lumina_]

I've always done the "weird AC" and I've certainly always considered it weird, but I'm super used to it haha

[jeffw]

I would just use the quadratic method, if the factors are not immediately apparent. This one can be done mentally if you know that the square root of 121 is 11.

[ZipplyZane]

Yeah, I don't like the last method because you now have more things to check. You have to check all the factors of a *and* all the factors of c, mixing and matching.

Discovering the first method was a revelation. It works. If I can't do that, then I use the sixth method: the quadratic formula. If I actually need the factors, I can reverse it from the answer, eg. if I get x=3/4 as one answer, then one factor is (4x-3).

[teengrogu]

I love the 4th and I know why it works always

[ezequielgerstelbodoha]

I remember cracking my head with the weird method that was the only one my teacher used in school. If I just knew there were much easier ways to do the same

[donq]

My algebra teacher taught us the Amazon Prime way, but this was in 2005 long before Amazon Prime became what it is today 😂

[AzureKyle]

These are all interesting methods, however the last method is the one that's closest to what I usually do. I don't use a tic tac toe grid, I just look at the x^2 and ask what times what equals my x^2 (in this case 6x^2) For this problem, I'd get x and 6x, as well as 2x and 3x. Then I'd look at the -3 and get either 1 and -3, or -1 and 3 Then I'd make every combination of the factors ie. (6x+1)(x-3) etc. and then using FOIL, solve each one until I got the original problem (in this case, (3x-1)(2x+3)) However, I might utilize the first method from now on, as that seems the easiest one to do.