I’m 78 and never learned this in high school nor engineering school. Don’t have a strong need to know this method, but it brought a big smile to my face. Thank you!
Yeah, I know another youtuber math person who blabs for 20 minutes for a single problem. None of it pertains to the math problem itself, just random stuff that doesn't help with the problem.
I've been teaching algebra for 38 years and I have never seen this method before!! I am using it in my lesson on Friday. Thank you, Professor H. I am now a subscriber.
@@marklyles3335that is not a real method. It is voodoo math and should be avoided. What he is doing is just a u-substitution in disguise. You are better off teaching the u-substitution and get the benefits later.
Nice one. But I prefer the good old sum and product method, followed by the factoring by grouping method, which is faster: 12x^2 + 17x + 6 The coefficients A and B and the constant term C are as follows: A = 12 B = 17 C = 6 If you multiply A and C you will obtain 72, as Mr H did. Then you find the two numbers that multiplied will give us 72 and added up will be equal to the coefficient B, which is 17. These numbers are 8 and 9, as shown in the video. Finaly, rewrite the expression using these numbers and finish up factoring by grouping: 12x^2 + 8x + 9x + 6 = 4x(3x + 2) + 3(3x + 2) = (3x + 2)(4x + 3) For the curious, a simple demonstration of why this works: Given the following product, let's use the distributive law: (ax + b)(cx + d) = ac(x^2) + adx + bcx + bd Now we combine the terms that contain x to obtain a quadraditc equation in the standard form: = ac(x^2) + (ad + bc)x + bd The coefficients A and B and the constant term C are as follows: A = ac B = ad + bc C = bd We know that the product of the coefficient A = "ac" and the constant C = "bd" is equal to "acbd", which can be rewritten as "(ad)(bc)" for better visualization. This means that by multiplying A and C we get the product of the numbers that add up to form the coefficient B. So if you find what are the numbers that multiplied will result in AC and added up will result in B, you can rewrite the expression using them and finish off factoring by grouping: ac(x^2) + (ad + bc)x + bd = ac(x^2) + adx + bcx + bd = ax(cx + d) + b(cx + d) = (ax + b)(cx + d)
@@disunitednation2782: It's an interesting method of factoring 2nd degree trinomials and it's always nice to know variations when enlightening students.
In Asia it's common we were thought this in 10th if you don't use this method then what method do you guys used? We were only thought this method in our schools .
who else is taking an online math class? I am over here trying to teach myself because the software that is supposed to be teaching me, makes no sense at all. I have been watching so many videos on youtube and none have helped. I stumbled across this one, a 4 MINUTE VIDEO, and this taught me so much in such little time. I am so grateful this video is on the internet. keep making great content MR H!! Thank you
I'm a Senior, everytime my granddaughter ask me about Math/Algebra, I always turned to your channel to make sure I did it right.. Thank You for the very clear and easy catch up on details of solving the problems.. Always refresh my past memory of Math Solving.
You single handedly managed to teach me how to do this in 2 minutes, I've spend my WHOLE day researching this topic, rewatching the materials my teachers have given me, I've been attempting their methods for so long to no avail. All I can say is thank you, this is a major weight off my shoulders, especially since I'm no good at math, sometimes I wish I could be taught anywhere else besides the US, so many things feel needlessly complicated, and it leads to me giving up hope sometimes.
Needlessly complicated. Quite a phrase. I sure do understand. Please keep with it. Electrical Engineering was hard for me. Professors will say: Of course, it should be! It’s complicated! . They are right, as I found out in real life. Now retired, I do a little studying and always discover new complications that somehow didn’t get included. Oh, most of my math professors got their education at Cambridge. The math department at Waterloo supplied them for the Engineering faculty back then. Quite the experience. Please keep at it, you will be rewarded for it.
bravo. solved 2 equations in under 5 minutes while giving a thorough explanation. thank you for not saying "imagine that we split the equation into two parts" like many other online tutors do. you gave a logical reason for why you multiplied and where the numbers went.
I am 47 years old and I've done a bunch of mathematics in high school and Calulus level mathematics later in college but I have never ever seen in my life that way of factoring. The world and especially the kids need more teachers and educators like you Sir.
@@hqs9585You don't watch his videos without understanding the basics of algebra. He didn't need to explain why it works because if you are at this level you should understand the concepts.
Mr. H thank you for your contributions to the United States School system, you have inspired many struggling highschool students who never would've otherwise graduated.
Never learned this method in my school, college, or engineering. Today encountered a similar equation with a>0, when I was teaching factorization to my secondary 2 daughter (she has math exam tomorrow) and watched this video. Beautiful method. Learned something new. Thanks a lot sir. You saved my day.
I, too never saw this method mentioned anywhere by anybody I've ever known but thanks for presenting it and expanding our minds a little bit. I'm mildly pleased to see quite a few people into math. It sure is good to keep the juices flowing and our minds working! Thank you!
As an engineer I have never seen this method before .till one of my students told me about it.but really, it is now clear for me and very simple .thanks a bunch
Im a college student who has taken a real appreciation for maths and physics in my last couple years of highschool. this is absolutely awesome, thank you Mr H
My savior. I learnt this method in school and become so confident I thought I could never forget it. 2 years later I did and was faced with the prospect of a lifetime of trail and error. it took me half an hour to find this but I am so gald I did, thank you
I never comment on videos, but this one truly deserved one. Studying for a test a couple days away and this was by far the best video that I have watched. Extremely helpful. thank you!
Very interesting and you made me think. 44 yrs teaching and these are still easy and quick with the cross method for most school kids, but I still like your method. The numbers are simpler cases than yours in Australian school level exams, which certainly helps as well. 😊
Accidentally watched this video, I found it great! I have never learned factoring like this. So much easy to solve the final answer! I really wish I could have learned this skill back to the 70s when I was studying in secondary school. 👍👍👍
I have done maths way beyond this level and could have solved it using the traditional(?) method but I have never seen this technique before. It is brilliant.
You make mathematics so easy. This is why Chinese are so good in mathematics. Africans have also got their own tricky way of making calculations easy. Thank you so much I have learned something amazing today.
I am 68.I was MathsPhobia in my Study Career.My subject was NaturalScience fearing Maths.After watching your teaching I realised that I missed many things in my life by not enjoying Maths.However I am happy that I could guide students fearing maths to turn them into MathsLovers.
first one can be done with a trick of spliting the middle term into two others like this 12x^2 + 17x + 6 12x^2 + 8x + 9x + 6 4x*( 3x + 2 ) + 3*( 3x + 2 ) ( 4x + 3 )( 3x + 2 ) this technique doesn't always work but if spotted it can safe time great video sir :))
His method is faster and would be easier for a beginner to remember. I say this after teaching the "grouping method" (also known as the AC-method at my college) for thirty years now. And by the way, the method you show always works assuming the trinomial is in the proper form and factorable.
Wow, this method is so amazing! I love quadratic equations!! I first learned them 23 years ago as a student and now I teach them full-time, thank you Jesus! Math is so fun!!!
@@SomethingaboutScreens These things vary from country to country, district to district in other cases. Reason why most folk are on such varied levels of education.
Good for you. I taught high school math for 30 years, and most ot the home-schooled kids that were in my classes arrived at this point in algebra. Most Moms and Dads were not up to the teaching of factoring polynomials.
Omg this is the first time a math method made 100% sense to me holy crap. Thank you dude wish you were my math teacher. Life would be way easier. If i get an A on my unit test tomorrow thats on you
@@luiscamacho1996 Thank you Captain Obvious! Of COURSE it's another method of factoring! The question is whether it is more useful or less useful for the typical person factoring a trinomial in an algebra class. A mathematical genius has no need for any of this. But not everyone is Gauss or Newton... thus this method is a useful alternative to the methods generally taught in an algebra class (for non-geniuses). By the way, I've met many "high IQ mathematicians" who are brilliant when it comes to abstract mathematical proofs, but can't do practical computation worth beans. High IQ mathematicians are generally not the same people as "human calculators." Those are two different "gifts" that are most often not in the same person.
This method is a variation on setting the trinomial to zero and solving the equation (at least the last couple steps are). Just my opinion, but I think a typical algebra student would find this method much easier to follow and remember, but perhaps not understanding the principles that make it possible. But, that's not really so important at the beginning level; ease of getting the right answer consistently via whatever shortcut/method is the name of the game. Leave the theory to higher math classes.
I learned factoring in U.S. school system. I had trouble with it then and I still do today. This method should have been taught to the U.S. population. Why wasn’t it taught to us? Simply put, this is another fabulous little tool that is useful in certain circumstances, which deserves a place in our mathematics tool bag!
Nice method. It is effectively reducing the leading coefficient to 1(i.e. monic) and working with fractions that we can eliminate later like this: First example: 12x^2 + 17x + 6 = 12*(x^2 + 17x/12 + 72/12^2). -- note that we have to write the constant term as a fraction with the denominator = 12^2. Now x^2 + 17x + 72 factorises as (x+8)(x+9), so we know that the expression above factorises similarly with fractions having 12 in the denominator: 12*(x^2 + 17x/12 + 72/12^2) = 12*(x + 8/12)(x+9/12) = 12*(x + 2/3)(x + 3/4) = (3x + 2)(4x + 3). Now you should be able to see how the method in the video works. Second example: 6x^2 - 5x - 4 = 6*(x^2 - 5x/6 - 24/6^2) -- compare with x^2 - 5x - 24 = (x + 3)(x - 8): 6*(x^2 - 5x/6 - 24/6^2) = 6*(x + 3/6)(x - 8/6) = 6*(x + 1/2)(x - 4/3) = (2x + 1)(3x - 4) As an aside, a factorisable quadratic expression can be viewed as (ax + c)(bx + d) = abx^2 + (ad + bc)x + cd. In the examples given, the coefficient of x is odd, which allows us to surmise -- without loss of generality -- that ad is odd and bc is even, in other words, both a and d are odd. This can often dramatically shorten the factorisation if done by trial and error. In the first example, 12x^2 + 17x + 6, the odd factors of 12 are 1 and 3 and the odd factors of 6 are 1 and 3, so for (ad+bc) = 17, we should use 3*3+4*2, giving (3x+2)(4x+3). In the second example, 6x^2 - 5x - 4, the odd factors of 6 are 1 and 3 and the odd factor of 4 is 1 (so d = ±1 and c = ∓4), leading to ad+bc= 3*1+2*(-4), giving (3x-4)(2x+1) .
I'm messaging as you seem to understand why the method should work. But I don't see why the final step should work in general. I have my monic factorisation (x-s1/a)(x-s1/a) where s1/a and s2/a are solutions to the original polynomial. To get the factorisation of the original polynomial I should just multiply everything by a. His final 'multiply everything by the reduced denominator trick' only does this if s1 and s2 are coprime - which in general they don't have to be (try his method on 3*x^2 - 9*x + 6 for example). Am I missing something?
@@prettyigirl1 No, you're not missing anything. You have to work with expressions, f(x), that don't have a common factor, otherwise the solutions to f(x) = 0 are actually the solutions to f(x)/k = 0, where k is the common factor, and your s1 and s2 are missing a factor of k. Taking your example, 3x^2 - 9x + 6, the first step has to be to take out the common factor of 3. Then the factorise the reduced expression and put the 3 back at the end, like this: 3x^2 - 9x + 6 = 3 * (x^2 - 3x + 2) which is now trivial, since x^2 - 3x + 2 is monic and can be seen to be equal to (x-2)(x-1). So 3x^2 - 9x + 6 = 3(x-2)(x-1). Perhaps a more instructive example might be 36x^2 + 51x + 18, where we also have a common factor of 3. The method in the video fails. as you realised, if you try working with 36 as the leading coefficient. But if we take out the common factor of 3 first, then we can factorise 12x^2 + 17x + 6, just as I did in my original post, by considering 12*(x^2 + 17x/12 + 72/12^2), where we know the factorisation gives 12*(x+8/12)(x+9/12) = 12*(x+2/3)(x+3/4) = (3x+2)(4x+3). So 36x^2 + 51x + 18 = 3(12x^2 + 17x + 6) = 3(3x+2)(4x+3), but you have that first step that he didn't mention in the video whenever there is a common factor. Hope that helps.
edit: added NOT before coprime on the 2nd line of the 2nd paragraph. @@prettyigirl1Good catch! Your polynomial factors to 3(x - 1)(x - 2), but Mr. H's method gives just (x - 1)(x - 2) without the coefficient of 3. I did some scratch work, and agree with you about coprime factors: in the step of reducing (x - 3/3)(x - 6/3) to (x - 1)(x - 2), we're effectively "dividing by 3" twice. We should do it once: (x - 1)(3x - 6) or (3x - 3)(x - 2). So for this specific example, I think we can all agree to factor out the 3 to get x^2 - 3*x + 2 first. Unfortunately, I don't know how to prove anything about the general case here, but I think the method works even if the pair (s1 and s2) are NOT coprime, as long as the trio (s1, s2, and a) are coprime. (That is, both numerators and the denominator must share a factor in order for this not to work.) example: 3*x^2 - 10*x + 8. The method eventually gives us (x - 4/3)(x - 6/3). s1 = 4, s2 = 6, and a = 3. Even though s1 and s2 are NOT coprime (they both have 2 as a factor), a doesn't have 2 as a factor. So the trio is coprime and the method works!
Nice explanation on WHY this method works. But there is no way i'm even gonna try explaining this in class. This is why I don't like magical algorithms, yea it makes things easier, but then students think Math is just a bunch of random rules that you just have to memorize, not understand.
Why didn't I have you as a math teacher back in high school!? So much simpler and clearer than when I was taught. Now I'm preparing for computer science in college, this is helpful
@mrhtutoring thanks. I recently got on the presidents list in first semester. Took programming courses and kept a 4.0 GPA. Here's hoping I can keep the momentum
The method is simple, but what's not explained is *why* it works. Understanding "why" is more important than being able to blindly follow an algorithm. Maybe that's why you weren't taught this in high school.
I was never shown this as a student. But as a tutor, students have shown me this method from their professors in college. I believe it was 6/7 years ago. "Slide and Divide" is the name given for this method.
When was the first time you were introduced to it? I don’t know why it started showing up, after what it seems from the comments, so many of us not bring in school for a long time. I’m wondering if it was always available but teachers opted to never do it until one day some said “hmm this seems to be the universal way of factoring any type of quadratic, so I’m going to use it.”
It’s like the AC method. The AC method doesn’t do any dividing. In the example from the video, it would the first step, multiplying the a and c. But after that is different. You’ll find the product of 72 and sum of 17. Then replace the b term in the original quadratic with those 2 values. Finally you would factor by grouping to get the factored form.
This way is sooo much easier than the way my professor taught me thank you so much this unit started out as a struggle and I had a feeling I was not going to do good with the method we were shown but after this video I am so much more comfortable
OMG OMG OMG!!!! This is just brilliant! Thank you so much for making this video. Yeah, I grew up in the US and I never learned about factoring this way. I’m in my late 70s now and doing math problems to keep my brain in shape. The workbook I’m using has some hard problems and now I have an algorithm for figuring them out. YES!!!!
I am also 77 and I try to keep my brain active by learning calculus (and adv algebra) on RUclips. Wish this was available when I was trying to become an engineer. Instead I went to easy route by majoring in accounting.
12x^2+17x+6 , factor: first multiply 12=a and c=6 togeather, you get 72. now look for factors of 72 which add up to 17 ie.8 and 9 now write 17x in terms of 8 and 9 so as to get 17x= 9x+8x. insert into equation; 12x^2 +8x +9x +6 =12x^2+9x+8x+6 factor; 3x(4x+3)+2(4x+3) (4x+3)(3x+2);
im middle schooler, and they taught us this in the whole semester, but in the most hardest way. and i cant believe i was able to understand it in 4 mins! THANKSSSS Subscribing!
You have the gift of clarity exposing matters that in themselves have complications; my congratulations on it. As for mathematics, a subject to which I have been linked all my life, they tend to be slippery in allowing their secrets to be revealed. In my opinion, the most relevant aspect of this video is that you do reveal some of those secrets, which are nothing more than manifestations of the relationships among numbers; by the way, exploring such relationships is what mathematicians have set as the goal of their profession.
@@kevinstreeter6943 Yours is a very valid point of view, as is mine, the one to which you respond. Probably, both coincide in responding to the expectations that their respective issuers have on the subject. I look at the half-full glass.
In India we are taught 3 methods to solve a quadratic equation 1) splitting middle term ( the same as taught by you) 2) Completing the square 3) quadratic formula.
Oh my goodness thank you so much for providing such a clear and concise explanation. I'm going back to school for engineering, and I'm going over all different math problems in preparation.
One tip - to find out the pair of numbers that adds up to 72 ( or any no.) You can prime factorize that no.if it is large 😊 , also i'm a high school student in india and we've been taught this method when quadratic polynomial has even both negative or either negative or positive sign in their linear or constant term in our 8th grade only 😅 this is both positive so kinda easy. ( Both negative too )
In India, I guess we have learnt to factorise Non- Monic Quadratic polynomial by the help of "middle term" method. With extreme practise, we gain the ability to factorise a Monic- Quadratic polynomial mentally (when coeff. of x² is 1) but we get trouble while solving a Non- Monic Quadratic polynomial. Sir Dr. H has tried to convert a polynomial to a Monic- Quadratic, therefore now we can factorise this mentally too. Thank you, Sir
if you can't remember the special rules of this method, do the fraction factoring instead: 12x^2+17x+6 12(x^2+17/12 x+72/144) 12(x+8/12)(x+9/12) (3*4)(x+2/3)(x+3/4) (3x+2)(4x+3) 😊
If the goal was procedural computation, sure. But that is not the goal, especially for high school students. The conceptual underpinnings of factorization are much more important.
I showed my math teacher this method in class after watching this video and he was absolutely baffled at how it kept working. I think he teaches factoring differently now.
Mr H, you are a true life saver. Tomorrow I'm going to be writing Business Mathematics and Statistics and I'm really confident after watching this video and the other's,you have made life easier. May blessings follow you.❤
My teacher means well but her explanations are very confusing. However, after looking at this video and copying down all the expressions and formulas, I finally understand much more about factoring. Thank you so much.
I’m 78 and never learned this in high school nor engineering school. Don’t have a strong need to know this method, but it brought a big smile to my face. Thank you!
Yes, it seems like a much easier method that what I was taught in school too. Simple to remember and use. I really like it.
Really??? They teach this method in the community college system, it’s called bottoms up factoring.
@@Lemurai I’m 78, today, and way back then, this technique was not taught. I’m glad to hear progress has been made.
H
. This is cool,much easier this method.We need these ideas to be taught in our schools.Many thanks
No shit talks, gets right to the point, very easy to understand and much easier than what my school had taught me, Thank you very much professor
I already use this method
Yeah, I know another youtuber math person who blabs for 20 minutes for a single problem. None of it pertains to the math problem itself, just random stuff that doesn't help with the problem.
I've been teaching algebra for 38 years and I have never seen this method before!! I am using it in my lesson on Friday. Thank you, Professor H. I am now a subscriber.
Thank you for the comment.
And for subscribing.
Really? I taught this method a few times but English kids just don't get it so I use a different method now.
I've been tutoring for 15 years. NEVER came across this.
@@marklyles3335that is not a real method. It is voodoo math and should be avoided. What he is doing is just a u-substitution in disguise. You are better off teaching the u-substitution and get the benefits later.
it’s called the swish method they teach it at the private christian school i attend
Highschool Algebra Teacher here and my kids are going to love me when I show them this! Thank you!
I teach this method in my algebra courses! We call it 'Slide & Divide'. It's a game-changer!
Holy crap... I'm here the night before the test, and this guy is my lord and savior. Thank you sir.
Did you pass?
same here man this guy is a life saver 🤣🤣
Jesus Christ is Lord and Savior
ME RN
SAME!!!
I'm 69 years old and still learning. Very grateful for your videos. Please keep doing it, sir.
Nice 😁
Me too. Almost 70 years old and still learning elementary school math. Great help for dementia prevention.
جيد جداً.
Nice one.
But I prefer the good old sum and product method, followed by the factoring by grouping method, which is faster:
12x^2 + 17x + 6
The coefficients A and B and the constant term C are as follows:
A = 12
B = 17
C = 6
If you multiply A and C you will obtain 72, as Mr H did. Then you find the two numbers that multiplied will give us 72 and added up will be equal to the coefficient B, which is 17. These numbers are 8 and 9, as shown in the video.
Finaly, rewrite the expression using these numbers and finish up factoring by grouping:
12x^2 + 8x + 9x + 6
= 4x(3x + 2) + 3(3x + 2)
= (3x + 2)(4x + 3)
For the curious, a simple demonstration of why this works:
Given the following product, let's use the distributive law:
(ax + b)(cx + d)
= ac(x^2) + adx + bcx + bd
Now we combine the terms that contain x to obtain a quadraditc equation in the standard form:
= ac(x^2) + (ad + bc)x + bd
The coefficients A and B and the constant term C are as follows:
A = ac
B = ad + bc
C = bd
We know that the product of the coefficient A = "ac" and the constant C = "bd" is equal to "acbd", which can be rewritten as "(ad)(bc)" for better visualization. This means that by multiplying A and C we get the product of the numbers that add up to form the coefficient B.
So if you find what are the numbers that multiplied will result in AC and added up will result in B, you can rewrite the expression using them and finish off factoring by grouping:
ac(x^2) + (ad + bc)x + bd
= ac(x^2) + adx + bcx + bd
= ax(cx + d) + b(cx + d)
= (ax + b)(cx + d)
Thank you Diegosita!
I learned it this way, but I find his way faster and easier.
Thanks Will Hunting
@@oddling2547 people also call me Mr beast
this is a simplified version of factoring by grouping
I'm an Engineer and have never seen factoring quadratic trinomials in this way.
BRILLIANT and in the toolbox.
Alright Dave ya nerd. 😂
What’s application of this in your field
@@disunitednation2782: It's an interesting method of factoring 2nd degree trinomials and it's always nice to know variations when enlightening students.
In Asia it's common we were thought this in 10th if you don't use this method then what method do you guys used? We were only thought this method in our schools .
@@RAYYANSyed-vx6ohi am from india , i first got introduced to this when i was in 7th grade
who else is taking an online math class? I am over here trying to teach myself because the software that is supposed to be teaching me, makes no sense at all. I have been watching so many videos on youtube and none have helped. I stumbled across this one, a 4 MINUTE VIDEO, and this taught me so much in such little time. I am so grateful this video is on the internet. keep making great content MR H!! Thank you
Need to say WOW again, you are correct that I've never seen it quite like this. YOU MAKE MATHEMATICS ENJOYABLE AGAIN. Thanks Dr. H
Thanks again!
4x+3)(3x+2)
@@jaggisaram4914 use foil method
Yes, but depending on the question, one method maybe be better than the other.
Well son of a gun. Chat, this is a math teacher. Pure music to my ears.
I'm a Senior, everytime my granddaughter ask me about Math/Algebra, I always turned to your channel to make sure I did it right.. Thank You for the very clear and easy catch up on details of solving the problems.. Always refresh my past memory of Math Solving.
You single handedly managed to teach me how to do this in 2 minutes, I've spend my WHOLE day researching this topic, rewatching the materials my teachers have given me, I've been attempting their methods for so long to no avail.
All I can say is thank you, this is a major weight off my shoulders, especially since I'm no good at math, sometimes I wish I could be taught anywhere else besides the US, so many things feel needlessly complicated, and it leads to me giving up hope sometimes.
Great to hear!
Needlessly complicated. Quite a phrase. I sure do understand. Please keep with it. Electrical Engineering was hard for me. Professors will say: Of course, it should be! It’s complicated! . They are right, as I found out in real life. Now retired, I do a little studying and always discover new complications that somehow didn’t get included. Oh, most of my math professors got their education at Cambridge. The math department at Waterloo supplied them for the Engineering faculty back then. Quite the experience. Please keep at it, you will be rewarded for it.
bravo. solved 2 equations in under 5 minutes while giving a thorough explanation. thank you for not saying "imagine that we split the equation into two parts" like many other online tutors do. you gave a logical reason for why you multiplied and where the numbers went.
I am 47 years old and I've done a bunch of mathematics in high school and Calulus level mathematics later in college but I have never ever seen in my life that way of factoring. The world and especially the kids need more teachers and educators like you Sir.
Say it again.
I can't articulate how much I appreciate your videos. All I can say is that you are one of God's many gifts.
Four minutes of great explanation! Thanks professor 👍
You are welcome!
well it was not really explained why it works, just a procedure. SO where is the great explanation?
@@hqs9585You don't watch his videos without understanding the basics of algebra. He didn't need to explain why it works because if you are at this level you should understand the concepts.
@@IlTjaylI this is so idiotic. I am college Math professor, my background is of course there, you are stupid by becoming personal
Some people were just born to do certain things in life. You truly are an amazing educator
This guy is a terrific teacher. His students are blessed to have him giving this kind of instruction.
Thank you for the nice comment.
Mr. H thank you for your contributions to the United States School system, you have inspired many struggling highschool students who never would've otherwise graduated.
Never learned this method in my school, college, or engineering. Today encountered a similar equation with a>0, when I was teaching factorization to my secondary 2 daughter (she has math exam tomorrow) and watched this video. Beautiful method. Learned something new. Thanks a lot sir. You saved my day.
I, too never saw this method mentioned anywhere by anybody I've ever known but thanks for presenting it and expanding our minds a little bit. I'm mildly pleased to see quite a few people into math. It sure is good to keep the juices flowing and our minds working! Thank you!
This particular type of Trinomial was giving me grief recently. I am glad I found this video.
As an engineer I have never seen this method before .till one of my students told me about it.but really, it is now clear for me and very simple .thanks a bunch
Im a college student who has taken a real appreciation for maths and physics in my last couple years of highschool. this is absolutely awesome, thank you Mr H
Best of luck!
My savior. I learnt this method in school and become so confident I thought I could never forget it. 2 years later I did and was faced with the prospect of a lifetime of trail and error. it took me half an hour to find this but I am so gald I did,
thank you
I was struggling really hard with factoring and really wasn't sure if I was doing it right but I have to say this guy explains it phenomenally
46 and still excited to learn and thankful! God bless! ❤
This guy made me understand this topic in 4 minutes. Thank you sir, please keep teaching and stay safe.
I’m a student currently taking alegebra and this technique is a complete game changer. I couldn’t thank you enough.
That was the most enlightening 4 minutes of my mathematical life! You are wonderful Professor H!
I Agreeeeee !!!
I am 55 years old. I wish that you would have my teacher in my school day.
I never comment on videos, but this one truly deserved one. Studying for a test a couple days away and this was by far the best video that I have watched. Extremely helpful. thank you!
Appreciate the nice comment. 😊
Very interesting and you made me think. 44 yrs teaching and these are still easy and quick with the cross method for most school kids, but I still like your method. The numbers are simpler cases than yours in Australian school level exams, which certainly helps as well. 😊
This is amazing, I am 58, and i dont recall that none of my advanced math was taught in this way, at least not to me.
Very nice!
Accidentally watched this video, I found it great! I have never learned factoring like this. So much easy to solve the final answer! I really wish I could have learned this skill back to the 70s when I was studying in secondary school. 👍👍👍
I taught Mathematics in high school and university some years back. I never came across this method. Thank you for such an amazing method.
I have done maths way beyond this level and could have solved it using the traditional(?) method but I have never seen this technique before. It is brilliant.
You make mathematics so easy. This is why Chinese are so good in mathematics. Africans have also got their own tricky way of making calculations easy.
Thank you so much I have learned something amazing today.
I am 68.I was MathsPhobia in my Study Career.My subject was NaturalScience fearing Maths.After watching your teaching I realised that I missed many things in my life by not enjoying Maths.However I am happy that I could guide students fearing maths to turn them into MathsLovers.
first one can be done with a trick of spliting the middle term into two others like this
12x^2 + 17x + 6
12x^2 + 8x + 9x + 6
4x*( 3x + 2 ) + 3*( 3x + 2 )
( 4x + 3 )( 3x + 2 )
this technique doesn't always work but if spotted it can safe time
great video sir :))
His method is faster and would be easier for a beginner to remember. I say this after teaching the "grouping method" (also known as the AC-method at my college) for thirty years now. And by the way, the method you show always works assuming the trinomial is in the proper form and factorable.
Your math’s skill is fantastic. I love the way you teach.
I appreciate that!
Mate what did I just see!? This is so easy, quick and useful!
Glad to hear it!
I’m 70 and exploring math. Especially the stuff I learned just to get through school. This is fascinating and very helpful!! Very much appreciated!
Wow, this method is so amazing! I love quadratic equations!! I first learned them 23 years ago as a student and now I teach them full-time, thank you Jesus! Math is so fun!!!
HOW DID I NOT LEARN THIS?!!!
Thank you so much for this awesome technique.
You're so welcome!
Just starting my freshman year of college. Goodness this was so much simpler than what I was instructed to do. Thank you very much, Mr. H.
You're very welcome! Good luck in college!
Can you explain how education system works there?
Coz this is taught in grade 6 in india. when i was 13 maybe
@@SomethingaboutScreens These things vary from country to country, district to district in other cases. Reason why most folk are on such varied levels of education.
@@Neolith_ ohh
This is great. I am a home school mom. My kid is really struggling with trinomials. Thanks so much. You have helped me to help her.
Good for you. I taught high school math for 30 years, and most ot the home-schooled kids that were in my classes arrived at this point in algebra. Most Moms and Dads were not up to the teaching of factoring polynomials.
Omg i am here the night before a test and this was a literal life saver. Thank you sir. You are amazong and idk why i was never taught this in school
Omg this is the first time a math method made 100% sense to me holy crap. Thank you dude wish you were my math teacher. Life would be way easier. If i get an A on my unit test tomorrow thats on you
This is an incredible method. Don’t know why I was never taught this.
Because this is just another method of factoring. High IQ mathematicians and physicists can factor inmediately just using a little thinking.
I highly doubt mathematicians and physicist watch my channel.
@@luiscamacho1996 Thank you Captain Obvious! Of COURSE it's another method of factoring! The question is whether it is more useful or less useful for the typical person factoring a trinomial in an algebra class. A mathematical genius has no need for any of this. But not everyone is Gauss or Newton... thus this method is a useful alternative to the methods generally taught in an algebra class (for non-geniuses).
By the way, I've met many "high IQ mathematicians" who are brilliant when it comes to abstract mathematical proofs, but can't do practical computation worth beans. High IQ mathematicians are generally not the same people as "human calculators." Those are two different "gifts" that are most often not in the same person.
This method is a variation on setting the trinomial to zero and solving the equation (at least the last couple steps are). Just my opinion, but I think a typical algebra student would find this method much easier to follow and remember, but perhaps not understanding the principles that make it possible. But, that's not really so important at the beginning level; ease of getting the right answer consistently via whatever shortcut/method is the name of the game. Leave the theory to higher math classes.
I learned factoring in U.S. school system. I had trouble with it then and I still do today. This method should have been taught to the U.S. population. Why wasn’t it taught to us? Simply put, this is another fabulous little tool that is useful in certain circumstances, which deserves a place in our mathematics tool bag!
Nice method. It is effectively reducing the leading coefficient to 1(i.e. monic) and working with fractions that we can eliminate later like this:
First example: 12x^2 + 17x + 6 = 12*(x^2 + 17x/12 + 72/12^2). -- note that we have to write the constant term as a fraction with the denominator = 12^2.
Now x^2 + 17x + 72 factorises as (x+8)(x+9), so we know that the expression above factorises similarly with fractions having 12 in the denominator:
12*(x^2 + 17x/12 + 72/12^2) = 12*(x + 8/12)(x+9/12) = 12*(x + 2/3)(x + 3/4) = (3x + 2)(4x + 3).
Now you should be able to see how the method in the video works.
Second example:
6x^2 - 5x - 4 = 6*(x^2 - 5x/6 - 24/6^2) -- compare with x^2 - 5x - 24 = (x + 3)(x - 8):
6*(x^2 - 5x/6 - 24/6^2) = 6*(x + 3/6)(x - 8/6) = 6*(x + 1/2)(x - 4/3) = (2x + 1)(3x - 4)
As an aside, a factorisable quadratic expression can be viewed as (ax + c)(bx + d) = abx^2 + (ad + bc)x + cd. In the examples given, the coefficient of x is odd, which allows us to surmise -- without loss of generality -- that ad is odd and bc is even, in other words, both a and d are odd. This can often dramatically shorten the factorisation if done by trial and error.
In the first example, 12x^2 + 17x + 6, the odd factors of 12 are 1 and 3 and the odd factors of 6 are 1 and 3, so for (ad+bc) = 17, we should use 3*3+4*2, giving (3x+2)(4x+3).
In the second example, 6x^2 - 5x - 4, the odd factors of 6 are 1 and 3 and the odd factor of 4 is 1 (so d = ±1 and c = ∓4), leading to ad+bc= 3*1+2*(-4), giving (3x-4)(2x+1) .
This is a fantastic explanation of why the method works.
I'm messaging as you seem to understand why the method should work. But I don't see why the final step should work in general. I have my monic factorisation (x-s1/a)(x-s1/a) where s1/a and s2/a are solutions to the original polynomial. To get the factorisation of the original polynomial I should just multiply everything by a. His final 'multiply everything by the reduced denominator trick' only does this if s1 and s2 are coprime - which in general they don't have to be (try his method on 3*x^2 - 9*x + 6 for example). Am I missing something?
@@prettyigirl1 No, you're not missing anything. You have to work with expressions, f(x), that don't have a common factor, otherwise the solutions to f(x) = 0 are actually the solutions to f(x)/k = 0, where k is the common factor, and your s1 and s2 are missing a factor of k.
Taking your example, 3x^2 - 9x + 6, the first step has to be to take out the common factor of 3. Then the factorise the reduced expression and put the 3 back at the end, like this:
3x^2 - 9x + 6 = 3 * (x^2 - 3x + 2) which is now trivial, since x^2 - 3x + 2 is monic and can be seen to be equal to (x-2)(x-1). So 3x^2 - 9x + 6 = 3(x-2)(x-1).
Perhaps a more instructive example might be 36x^2 + 51x + 18, where we also have a common factor of 3. The method in the video fails. as you realised, if you try working with 36 as the leading coefficient. But if we take out the common factor of 3 first, then we can factorise 12x^2 + 17x + 6, just as I did in my original post, by considering 12*(x^2 + 17x/12 + 72/12^2), where we know the factorisation gives 12*(x+8/12)(x+9/12) = 12*(x+2/3)(x+3/4) = (3x+2)(4x+3).
So 36x^2 + 51x + 18 = 3(12x^2 + 17x + 6) = 3(3x+2)(4x+3), but you have that first step that he didn't mention in the video whenever there is a common factor. Hope that helps.
edit: added NOT before coprime on the 2nd line of the 2nd paragraph.
@@prettyigirl1Good catch! Your polynomial factors to 3(x - 1)(x - 2), but Mr. H's method gives just (x - 1)(x - 2) without the coefficient of 3. I did some scratch work, and agree with you about coprime factors: in the step of reducing (x - 3/3)(x - 6/3) to (x - 1)(x - 2), we're effectively "dividing by 3" twice. We should do it once: (x - 1)(3x - 6) or (3x - 3)(x - 2). So for this specific example, I think we can all agree to factor out the 3 to get x^2 - 3*x + 2 first.
Unfortunately, I don't know how to prove anything about the general case here, but I think the method works even if the pair (s1 and s2) are NOT coprime, as long as the trio (s1, s2, and a) are coprime. (That is, both numerators and the denominator must share a factor in order for this not to work.)
example: 3*x^2 - 10*x + 8. The method eventually gives us (x - 4/3)(x - 6/3). s1 = 4, s2 = 6, and a = 3. Even though s1 and s2 are NOT coprime (they both have 2 as a factor), a doesn't have 2 as a factor. So the trio is coprime and the method works!
Nice explanation on WHY this method works. But there is no way i'm even gonna try explaining this in class. This is why I don't like magical algorithms, yea it makes things easier, but then students think Math is just a bunch of random rules that you just have to memorize, not understand.
Thank you so much i am a couple nights bwfore the test and you just saved my. God bless
thank you so much bro, this method is 10 times easier than the tutoring I get from maths courses with real money. The best part, it is free.
Why didn't I have you as a math teacher back in high school!? So much simpler and clearer than when I was taught. Now I'm preparing for computer science in college, this is helpful
Glad hear that. Good luck in college!
@mrhtutoring thanks. I recently got on the presidents list in first semester. Took programming courses and kept a 4.0 GPA. Here's hoping I can keep the momentum
The method is simple, but what's not explained is *why* it works. Understanding "why" is more important than being able to blindly follow an algorithm. Maybe that's why you weren't taught this in high school.
I'm also finding this helpful and also offering Computer science at the University .
@@jasoncook2715how is the course going for you ?
I was never shown this as a student. But as a tutor, students have shown me this method from their professors in college. I believe it was 6/7 years ago. "Slide and Divide" is the name given for this method.
In Ireland we called it "bump and run". Bump "a" over to "c", find the factors of "b", run "ac" back to divide those factors.
When was the first time you were introduced to it? I don’t know why it started showing up, after what it seems from the comments, so many of us not bring in school for a long time. I’m wondering if it was always available but teachers opted to never do it until one day some said “hmm this seems to be the universal way of factoring any type of quadratic, so I’m going to use it.”
It has differing names, depending on who you ask but 'ac method' seems to be the most common.
It’s like the AC method. The AC method doesn’t do any dividing. In the example from the video, it would the first step, multiplying the a and c. But after that is different. You’ll find the product of 72 and sum of 17. Then replace the b term in the original quadratic with those 2 values. Finally you would factor by grouping to get the factored form.
This way is sooo much easier than the way my professor taught me thank you so much this unit started out as a struggle and I had a feeling I was not going to do good with the method we were shown but after this video I am so much more comfortable
I’ve known about the ac-method for years, but never saw this particular way of working it. Thanks for showing us another tool for our toolbox.
I was reading a complicated version of factoring trinomials but this was honestly such a better method - thank you so much Mr. H.
OMG OMG OMG!!!! This is just brilliant! Thank you so much for making this video. Yeah, I grew up in the US and I never learned about factoring this way. I’m in my late 70s now and doing math problems to keep my brain in shape. The workbook I’m using has some hard problems and now I have an algorithm for figuring them out. YES!!!!
Glad you enjoyed it!
In india every 9 standard student know this method😂😂
@@शिव-ठ6प Indians are genius in math and physics .l admit that.
I am also 77 and I try to keep my brain active by learning calculus (and adv algebra) on RUclips. Wish this was available when I was trying to become an engineer. Instead I went to easy route by majoring in accounting.
Most of the Indians are liars like their prime minister gautamdas modani.
12x^2+17x+6 , factor:
first multiply 12=a and c=6 togeather,
you get 72.
now look for factors of 72 which add up to 17
ie.8 and 9
now write 17x in terms of 8 and 9 so as to get 17x= 9x+8x.
insert into equation;
12x^2 +8x +9x +6
=12x^2+9x+8x+6
factor;
3x(4x+3)+2(4x+3)
(4x+3)(3x+2);
This is my preferred method, just learned this from an old book from 1930s, it seems to involve the least amount of "hunting" for me.
@@azimuth4850a u-substitution is probably the easiest
This method is called "decomposition". Good students learn it, struggling students do not. All can learn "slide and divide".
we call it middle term splitting@@grahammcfadyenhill9555
Yes,it was taught when I was in 8 standard,in 1968.it is good revision at my 70.
Very interesting to watch your lectures. Awesome teaching
The best explanation so far I have got to understand this concept. Thank you so much sir👏👏👏👏👏👍🌹🌻🌹
This method might be a little longer but it's so much easier to understand for myself when doing difficult trinomials. Thanks so much!
Your videos are so easy to understand and are so helpful. Thank you so much sir 😊
You are most welcome
@@mrhtutoring Could you please tell how polynomials and trinomials etc are different? Im very confused
Absolutely love your mathematics skills!
U wil unemploy my teacher
im middle schooler, and they taught us this in the whole semester, but in the most hardest way. and i cant believe i was able to understand it in 4 mins! THANKSSSS
Subscribing!
I had a test tomorrow morning you just saved my life
You have the gift of clarity exposing matters that in themselves have complications; my congratulations on it. As for mathematics, a subject to which I have been linked all my life, they tend to be slippery in allowing their secrets to be revealed. In my opinion, the most relevant aspect of this video is that you do reveal some of those secrets, which are nothing more than manifestations of the relationships among numbers; by the way, exploring such relationships is what mathematicians have set as the goal of their profession.
@@kevinstreeter6943 Yours is a very valid point of view, as is mine, the one to which you respond. Probably, both coincide in responding to the expectations that their respective issuers have on the subject. I look at the half-full glass.
In India we are taught 3 methods to solve a quadratic equation
1) splitting middle term ( the same as taught by you)
2) Completing the square
3) quadratic formula.
Yeah.. exactly I learned this in school 😅
Vinay bhai... basically he is using the first method
In a decent high school course in the United States we are taught 5 ways. One of the ones you are missing is u-substitution.
Excellent. I am 60 years old, helping my granddaughter with her grade 10 maths. This is a very easy method 😂🎉
Oh my goodness thank you so much for providing such a clear and concise explanation. I'm going back to school for engineering, and I'm going over all different math problems in preparation.
Good luck with your studies!
Decided to learn this the night before the test and wow MR H you have saved me.
Thank you, this method is easy and useful. Very good explanation, great video!
Glad you enjoyed it!
One tip - to find out the pair of numbers that adds up to 72 ( or any no.) You can prime factorize that no.if it is large 😊 , also i'm a high school student in india and we've been taught this method when quadratic polynomial has even both negative or either negative or positive sign in their linear or constant term in our 8th grade only 😅 this is both positive so kinda easy. ( Both negative too )
In India, I guess we have learnt to factorise Non- Monic Quadratic polynomial by the help of "middle term" method. With extreme practise, we gain the ability to factorise a Monic- Quadratic polynomial mentally (when coeff. of x² is 1) but we get trouble while solving a Non- Monic Quadratic polynomial. Sir Dr. H has tried to convert a polynomial to a Monic- Quadratic, therefore now we can factorise this mentally too.
Thank you, Sir
I learned this in the United States before seeing this video. But it’s more useful to expand the linear term and factor by grouping.
Thank you so much for this method of factoring trinomials. It is so much easier than the way we are taught in school!
I am also almost 70 years old and still learning. Amazing trick. Thanks.
Very helpful for my summer math packet. Thank you for making this less confusing for me :)
You are so welcome!
What a wonderful teacher! Kudos!
Wow, thank you!
I really love this channel ❤ Now, because of your channel, my love for maths increases 📈📈📈📈. Thank you so much. And keep it up ❤
Wow, thank you!
wow. it's been 45+ years since i was taught factoring and we NEVER were shown this in Ontario, Canada. This is amazing Mr H.
very, very cool. thx.
Sir, I’d like to thank you for geting me through pre calc. You are a blessing, I along with everyone else here appreciates you.
You're welcome! I am glad to know my videos were helpful.
if you can't remember the special rules of this method, do the fraction factoring instead:
12x^2+17x+6
12(x^2+17/12 x+72/144)
12(x+8/12)(x+9/12)
(3*4)(x+2/3)(x+3/4)
(3x+2)(4x+3)
😊
I prefer this method as each expression is equal to the previous expression, and so one can see why it is valid.
you must be actor😍
If the goal was procedural computation, sure. But that is not the goal, especially for high school students. The conceptual underpinnings of factorization are much more important.
Been learning this in class for 5 days and this man just taught in me 4 minutes
I am literallly dropping my jaw on how amazingly easy this makes factoring equations now. I wish they taught this method in the US.
I showed my math teacher this method in class after watching this video and he was absolutely baffled at how it kept working. I think he teaches factoring differently now.
may God bless this man 😭
im in g11 and i had forgotten how to factories this is the best method ever
I’m in algebra and was confused on how to factor but this method made it way easier.
100 per cent thumbs up! I also started thinking of the transformation of 12, 17, 6 into 3, 2, 4, and 3! Amazing
🙏
I instantly subscribed before the first problem was completed. The simplicity was amazing! Thank you Mr H.
Thank you for subscribing.
Mr H, you are a true life saver. Tomorrow I'm going to be writing Business Mathematics and Statistics and I'm really confident after watching this video and the other's,you have made life easier. May blessings follow you.❤
I'm another senior. (83) Want to tighten up my math a bit to understand basics of physics. This was very cool. and clear. Thanks.
This blew my mind! It's quicker then any method I've learned in school. Thank you!
My teacher means well but her explanations are very confusing. However, after looking at this video and copying down all the expressions and formulas, I finally understand much more about factoring. Thank you so much.
Glad it helped!
You are great. You use simple methods and you explain without talking too much