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.)
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
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.
@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?
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. =
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.
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).
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.
Many years, many students...Box method easiest for them to use and remember. I teach all of these using a diamond method and the Box method can be used with least amount to remember. They do mot have the stamina to guess and check anymore. I had to change from guess and check after so many were getting frustrated.
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
Also, factorising and solving are essentially the same. (3x-1)(2x+3)=0 3x-1=0 or 2x+3=0 3x=1 or 2x=-3 x=1/3 or x=-3/2 x=1/3 or x=-3/2 3x=1 or 2x=-3 3x-1=0 or 2x+3=0 (3x-1)(2x+3)=0 If you know how to, factorising is normally the fastest with whole numbers or easy fractions. CTS and the formula tend to take longer due to the algebraic manipulation.
You don't. This method is full of nothing but trial and error. 12x²-7x-12, to look at another example, could be x & 12x, 2x & 6x, or 3x & 4x. Likewise, the constant term could be 1 & 12, 2 & 6, or 3 & 4 and each pair of constants has two combinations of plus/minus signs. By my figuring, this means there are 18 combinations of binomials you could form to try to find the right combination. To me, it seems like the most inefficient choice.
@@richskater: Although I wouldn't recommend it, you _can_ factor by solving. If the solutions are r and s, then it factors as a(x−r)(x−s). If r and s are integers, then you're done; otherwise, their denominators will be factors of a, which you can distribute. (Of course, if r and s are irrational, then it won't factor.)
@@Steve_Stowersyes, you are right, i thought you will get the same factors if we divide the whole equation by 6, but the factors you get by quadratic formula are 1/6th of the actual factors And as a maths student I am really embarrassed right now
How to use the quadratic formula to factor trinomials: 👉 ruclips.net/video/FkzNRzRNV0A/видео.html
I've been a math teacher for 29 years, and I've never seen the "slide and divide" method. Love it!!
Can you explain sir why does it works?
I prefer the first method because it doesn't require any "magical" steps, just basic algebra.
Here because I'm a first-year CS student and I forgot how to solve a quadratic
😂😂 probably me in the future
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.)
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
the pq quadratic formula that they don't teach you in school
ruclips.net/video/NlX_VR-e8qo/видео.html
They do teach it. It's just that the teachers don't want to teach.
I would rather always prefer method 6 (the quadratic formula). Because it works in all cases, such as the cases of irrational roots or complex roots.
If you show this video to twitter users they will literally explode. Also, you're a great teacher, thanks a lot!
It’s always nice to see different ways to solve common problems. Thanks for sharing.
All you need for know for factoring polynomials: 👉 ruclips.net/video/9Uaqso-4W64/видео.html
Is there any numerical method available to calculate gamma function of non analytical numbers like (1/3), (1/5) ?
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.
The best method is not any of these 5, or the quadratic formula. Completing the square is goat
I think first method makes most sense. It is also very easy.
@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?
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. =
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.
I love the 4th and I know why it works always
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).
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.
This sounds like an exhausting, tedious slog. 12x²-7x-12 for example has, by my calculation, 18 potential combinations of binomials to be tested.
@jimbobago Yeah, it can be, but I also enjoy working with numbers, so it's fun in it's own way.
@@AzureKyle Who am I to argue with someone who finds math fun in its own way?
I've always done the "weird AC" and I've certainly always considered it weird, but I'm super used to it haha
me seeing the given question- :DD
me then seeing bprp's shirt- 😵💫😵💫
i am in 8th and know nothing about calculus lol
My algebra teacher taught us the Amazon Prime way, but this was in 2005 long before Amazon Prime became what it is today 😂
😂
Last one is the best way🙏
I'm accustomed to the first method
Many years, many students...Box method easiest for them to use and remember. I teach all of these using a diamond method and the Box method can be used with least amount to remember. They do mot have the stamina to guess and check anymore. I had to change from guess and check after so many were getting frustrated.
ari gatou math sensei
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
Those are legit!!!!
Which method would you use for 10x^2-57x+54?
Tic tac toe. (5x-6)(2x-9)
@@bprpmathbasics Does that mean this method is the most versatile?
I definitely like the tictactoe method the most.
Just a question, are we solving for X?
Not solving. Just factoring.
If this was part of an equation, you would need to factor prior to solving for X.
@@charleskrueger5523 Ah, okay.
Also, factorising and solving are essentially the same.
(3x-1)(2x+3)=0
3x-1=0 or 2x+3=0
3x=1 or 2x=-3
x=1/3 or x=-3/2
x=1/3 or x=-3/2
3x=1 or 2x=-3
3x-1=0 or 2x+3=0
(3x-1)(2x+3)=0
If you know how to, factorising is normally the fastest with whole numbers or easy fractions. CTS and the formula tend to take longer due to the algebraic manipulation.
@@darranrowe174 Oh, okay.
Mathmagical!!!
Regarding the tic tac toe....
what times what equals 6x^2....
6x * x
3x * 2x
How do you know if to check 6x and x or 3x and 2x ??
You don't. This method is full of nothing but trial and error. 12x²-7x-12, to look at another example, could be x & 12x, 2x & 6x, or 3x & 4x. Likewise, the constant term could be 1 & 12, 2 & 6, or 3 & 4 and each pair of constants has two combinations of plus/minus signs. By my figuring, this means there are 18 combinations of binomials you could form to try to find the right combination. To me, it seems like the most inefficient choice.
just complete the square yall x
I think , sticking one method and fully mastery it is the best .
I love maths❤
What does common factor mean in the screen-text?
as in 2x^2 + 4x + 6 = 0 would need to be reduced to x^2 + 2x + 3 = 0 first?
Hmm I kinda like the first (or first three) methods. I always just go to quadratic formula when a=/=1
I usually use method 6: trial and error until I finally stumble across the solution.
Have you taught the method taught by Po Shen Lo?
I LOVE YOU SOOO SOSOOSO SOOOO MUCH THANK YOU SO MUCH FOR THIS IM SO GOOD AT MATH NOW I WENT FROM 0% TO 88.9% THANK YUOU LORD AD SAVOURS
not even 89%?
@@Fatmuffintart I'm just better than you
Happy to hear and love your excitement! Keep up your good work!! 👍
My teacher calls AC 'Product Sum', where you find ax²+(m+an)x+mn
I would rather always prefer method 6 (the quadratic formula). Because it works in all cases, such as the cases of irrational roots or complex roots.
4th is the best method by far.
agreed
First one is the easiest method.
You can also use formula method which is equal to [-b+-sqrt(b²-4ac)]/2a
The video was on methods for factoring, not solving.
Factoring out is way quicker almost every time than formula. It is only troublesome when you get factors in roots and fractions
@@richskater: Although I wouldn't recommend it, you _can_ factor by solving. If the solutions are r and s, then it factors as a(x−r)(x−s). If r and s are integers, then you're done; otherwise, their denominators will be factors of a, which you can distribute. (Of course, if r and s are irrational, then it won't factor.)
Tell me you don't understand the point of the video without *saying* you don't understand the point of the video.
@@jimbobago yeah i know. It's about factorizing with different methods taught us in 6th grade.
I used the 4th way to factor which i found to be the easiest.
For C, that doesn't work when you can define the smallest non zero amount.
-(-2)= +2 but why
First
Not first.
I am 31
Why am I sitting here acting like a student????
Or you can just use quadratic formula
The quadratic formula isn't really a way of factoring a trinomial. Although you could work backwards from what it gives you to figure out the factors.
@@Steve_Stowersyes, you are right, i thought you will get the same factors if we divide the whole equation by 6, but the factors you get by quadratic formula are 1/6th of the actual factors
And as a maths student I am really embarrassed right now
@@Steve_Stowers it is one💀.
You only need one of the zeros and then apply ruffini bruh
Box method is for the weird kids
(2x+3)(3x-1)