Factoring Trinomials

Factoring a trinomial that has a leading coefficient other than 1 can be annoying and time taking.
In the video, I show a method that is not taught in most high schools in the USA.
The lesson also includes how to factor trinomials with leading coefficient of 1.

[@EdwardKilner]

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!

[@nytrex2001]

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.

[@Lemurai]

Really??? They teach this method in the community college system, it’s called bottoms up factoring.

[@EdwardKilner]

@@Lemurai I’m 78, today, and way back then, this technique was not taught. I’m glad to hear progress has been made.

[@truthkmgmailcom]

H

[@buthainaabbas3885]

. This is cool,much easier this method.We need these ideas to be taught in our schools.Many thanks

[@parkerboyce4906]

Holy crap... I'm here the night before the test, and this guy is my lord and savior. Thank you sir.

[@SonicWillBoopU]

Did you pass?

[@ProGamer-li1zm]

relax man this topic covers at least 5 % of the whole syllabus of the maths topic in total u aint gonna pass like that

[@Y-it123]

same here man this guy is a life saver 🤣🤣

[@Desiresworld9]

Jesus Christ is Lord and Savior

[@Burgerisgood-tf3jp]

Dude same I have an EOC torrmorw

[@sherriclowe1229]

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.

[@mrhtutoring]

Thank you for the comment. And for subscribing.

[@robyn2791]

Really? I taught this method a few times but English kids just don't get it so I use a different method now.

[@marklyles3335]

I've been tutoring for 15 years. NEVER came across this.

[@alfonsorodriguez2739]

@@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.

[@crashpro1694]

it’s called the swish method they teach it at the private christian school i attend

[@sovietunion3721]

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

[@luisclementeortegasegovia8603]

Four minutes of great explanation! Thanks professor 👍

[@mrhtutoring]

You are welcome!

[@hqs9585]

well it was not really explained why it works, just a procedure. SO where is the great explanation?

[@IlTjaylI]

​@@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.

[@hqs9585]

@@IlTjaylI this is so idiotic. I am college Math professor, my background is of course there, you are stupid by becoming personal

[@DavidRobinson-rj2sp]

I'm an Engineer and have never seen factoring quadratic trinomials in this way. BRILLIANT and in the toolbox.

[@badmanskill1112]

Alright Dave ya nerd. 😂

[@disunitednation2782]

What’s application of this in your field

[@DavidRobinson-rj2sp]

@@disunitednation2782: It's an interesting method of factoring 2nd degree trinomials and it's always nice to know variations when enlightening students.

[@RAYYANSyed-vx6oh]

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 .

[@MILKY-it3by]

​@@RAYYANSyed-vx6ohi am from india , i first got introduced to this when i was in 7th grade

[@DiegoSita]

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)

[@Savethirdrock]

Thank you Diegosita!

[@esolo114]

I learned it this way, but I find his way faster and easier.

[@oddling2547]

Thanks Will Hunting

[@DiegoSita]

@@oddling2547 people also call me Mr beast

[@richardwaters2742]

this is a simplified version of factoring by grouping

[@Kassidy_quo]

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

[@SharpeCreations]

Highschool Algebra Teacher here and my kids are going to love me when I show them this! Thank you!

[@mr.x1510]

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

[@mrhtutoring]

Thanks again!

[@jaggisaram4914]

4x+3)(3x+2)

[@aurelian771]

@@jaggisaram4914 use foil method

[@ProGamer-li1zm]

@@mrhtutoring So I can factorise anything without even using the quadratic formula?

[@mrhtutoring]

Yes, but depending on the question, one method maybe be better than the other.

[@ItzASecretBoi]

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.

[@mrhtutoring]

Great to hear!

[@EdwardKilner]

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.

[@armandelven-dechoux5290]

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

[@seabass5484]

This is an incredible method. Don’t know why I was never taught this.

[@luiscamacho1996]

Because this is just another method of factoring. High IQ mathematicians and physicists can factor inmediately just using a little thinking.

[@mrhtutoring]

I highly doubt mathematicians and physicist watch my channel.

[@andreasioannides5905]

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.

[@Monique_Johnson]

Say it again.

[@MathWithGenie]

I teach this method in my algebra courses! We call it 'Slide & Divide'. It's a game-changer!

[@aram5642]

When I open YT to watch a specific video, and I see a recommendation from Mr H channel, there's no way that I can just pass by.

[@mksarav75]

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.

[@looalex8011]

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. 👍👍👍

[@Kenneth-gn8nk]

Thanks!!! I wish you (and your programs) were around 55 years ago, it scares me what level of knowledge in math I could have acquired!!!

[@jakemccoy]

Wow, I wish you were my math teacher 40 years ago. I understand your lessons immediately. My math teacher was like, “Just kinda guess until you get the correct factors.”

[@jasoncook2715]

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]

Glad hear that. Good luck in college!

[@jasoncook2715]

@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

[@mojota6938]

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.

[@eugeneassoahmorcher2042]

I'm also finding this helpful and also offering Computer science at the University .

[@eugeneassoahmorcher2042]

​@@jasoncook2715how is the course going for you ?

[@21ruevictorhugo]

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!!!!

[@mrhtutoring]

Glad you enjoyed it!

[@user-rk9wd8wt2z]

In india every 9 standard student know this method😂😂

[@buthainaabbas3885]

@@user-rk9wd8wt2z Indians are genius in math and physics .l admit that.

[@stanroelker7443]

I am also 77 and I try to keep my brain active by learning calculus (and adv algebra) on KZbin. Wish this was available when I was trying to become an engineer. Instead I went to easy route by majoring in accounting.

[@aap223]

Most of the Indians are liars like their prime minister gautamdas modani.

[@Max2chess]

I am 54 years old. I just love learning and I love your videos. Great teaching stile together with excellent topics. Thank you.

[@wiz4rdx440]

This method might be a little longer but it's so much easier to understand for myself when doing difficult trinomials. Thanks so much!

[@konradyearwood5845]

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.

[@atharvyadav-us4wn]

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);

[@azimuth4850]

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.

[@alfonsorodriguez2739]

@@azimuth4850a u-substitution is probably the easiest

[@grahammcfadyenhill9555]

This method is called "decomposition". Good students learn it, struggling students do not. All can learn "slide and divide".

[@marklyles3335]

I can't articulate how much I appreciate your videos. All I can say is that you are one of God's many gifts.

[@taranmehta6356]

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

[@mrhtutoring]

Best of luck!

[@formulanova12]

Mate what did I just see!? This is so easy, quick and useful!

[@mrhtutoring]

Glad to hear it!

[@MichaelMaysMusicMinistries]

HOW DID I NOT LEARN THIS?!!! Thank you so much for this awesome technique.

[@mrhtutoring]

You're so welcome!

[@CydVillegas]

I have a test tomorrow and this helped me A LOT. The method they teach in our school is slower and much harder to grasp. So thank you for teaching a more faster and efficient way to do this. I'm now a subscriber:)

[@mrhtutoring]

No problem. Thanks for subscribing!

[@NameIsWario]

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.

[@krishgounder5116]

Excellent. I am 60 years old, helping my granddaughter with her grade 10 maths. This is a very easy method 😂🎉

[@anjaninverma3553]

I am 55 years old. I wish that you would have my teacher in my school day.

[@alejandroramirez3944]

This guy made me understand this topic in 4 minutes. Thank you sir, please keep teaching and stay safe.

[@sunana2702]

Your math’s skill is fantastic. I love the way you teach.

[@mrhtutoring]

I appreciate that!

[@hkmmar1]

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.

[@grahammcfadyenhill9555]

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.

[@hkmmar1]

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.”

[@dooziejohn6165]

In my grade school, I struggled with the guessing game. Glad to know this systematic method. Thank you.

[@Monique_Johnson]

I instantly subscribed before the first problem was completed. The simplicity was amazing! Thank you Mr H.

[@mrhtutoring]

Thank you for subscribing.

[@Neolith_]

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.

[@mrhtutoring]

You're very welcome! Good luck in college!

[@SomethingaboutScreens]

Can you explain how education system works there? Coz this is taught in grade 6 in india. when i was 13 maybe

[@Neolith_]

@@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.

[@SomethingaboutScreens]

@@Neolith_ ohh

[@janastraubhar3546]

Thank you, Mr. H! You managed to remove my anxiety by learning this simple factoring example!

[@lilghost1794]

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!

[@mrhtutoring]

Appreciate the nice comment. 😊

[@elmoreglidingclub3030]

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!

[@tesskansas]

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.

[@grahammcfadyenhill9555]

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.

[@All_8_days_of_a_week]

U wil unemploy my teacher

[@rainbzsploop]

WOW ! i have been takign many classes but i could not quite understand it but this small video literally explained me better then 67 hours of study and 60$ spent!

[@RexxSchneider]

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) .

[@swsennel]

This is a fantastic explanation of why the method works.

[@prettyigirl1]

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?

[@RexxSchneider]

@@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.

[@swsennel]

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!

[@eugene188]

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.

[@animewarrior7]

Your videos are so easy to understand and are so helpful. Thank you so much sir 😊

[@mrhtutoring]

You are most welcome

[@Orillians]

@@mrhtutoring Could you please tell how polynomials and trinomials etc are different? Im very confused

[@georgekroll2901]

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.

[@Man-idk-et5wp]

Explained this in 30 seconds while my teacher had to take an hour

[@gappzzzyt903]

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

[@hameedamathtuber]

Very interesting to watch your lectures. Awesome teaching

[@rleroygordon]

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.

[@nuggetz760]

THIS JUST SAVED MY LIFE PROFESSOR. TYSM. In school they taught us a difficult and long method and i didnt understand a thing but with ur help i understand now and im sure ill pass easily my exam :D

[@mrhtutoring]

Glad to hear it.

[@vinaykumar9002]

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.

[@user-tb7pq2sz9i]

Yeah.. exactly I learned this in school 😅

[@waqasjamilkhan5323]

Vinay bhai... basically he is using the first method

[@alfonsorodriguez2739]

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.

[@lolman1758]

Absolutely love your mathematics skills!

[@azzaeldessoky5870]

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

[@majora424]

I’m in algebra and was confused on how to factor but this method made it way easier.

[@Geers7Omoorstay]

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.

[@Geers7Omoorstay]

@@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.

[@murarirajagopalan3771]

What a wonderful teacher! Kudos!

[@mrhtutoring]

Wow, thank you!

[@crystalclarity6766]

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!

[@shantelvengesai]

may God bless this man 😭 im in g11 and i had forgotten how to factories this is the best method ever

[@MariahBellaQuizzagan-ex3ti]

Thank you Professor, this is way quicker and easier way to understand 👍

[@nethvegz3465]

Very clear, short cut method and accurate explanation,thank u sir❤❤❤

[@mrhtutoring]

You are most welcome!~

[@theLordsboy]

This guy is a terrific teacher. His students are blessed to have him giving this kind of instruction.

[@mrhtutoring]

Thank you for the nice comment.

[@luthermcfarlane9501]

I am retired engineer and I never saw this method before. It is great to know as many methods as possible.

[@chocolateangel8743]

This is most commonly called the "Bottom's-Up" method (for factoring trinomials). It works by compensation. I was born in the 80's. It wasn't something I was taught, either. Ever since the introduction of Common Core math (more conceptual teaching approach), there has been a big push to teach kids multiple ways of solving any problem.

[@alfonsorodriguez2739]

This is not a method. This is voodoo math

[@phlogistanjones2722]

THANK YOU! Super clear and well explained. FULLY comprehend the process. KUDOS!

[@drdrizzl9756]

Thank you, this method is easy and useful. Very good explanation, great video!

[@mrhtutoring]

Glad you enjoyed it!

[@MadScientyst]

This is AMAZING & I have 2 Math Degrees, wouldn't have thought to go that route! Thank u Mr H!! 🤩🤩

[@mrhtutoring]

Glad to help!

[@wileynicholson1682]

66 years old and first time learning this. Awesome!

[@dzeksonmakgajwer3010]

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 :))

[@JDMathematicsAndDataScience]

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.

[@ravisankar9669]

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.

[@cuongcao5781]

THANK YOU! THIS IS THE ONLY WAY I COULD UNDERSTAND HOW TO FACTOR A QUADRATIC EXPRESSION WITHOUT BEING CONFUSED!!!!!!

[@tanishk1538]

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 )

[@user-je6ff5dk1n]

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

[@bytemark6508]

I suppose it's a method that works, but I like to do this in a different way. The numbers (if they exist) can be deducted from this little trick : From the aX2 + bX + c, the numbers n1 and n2 can be obtained by n1*n2 = a*c and n1+n2 = b. The two numbers n1 and n2 are then used simply by writing the term bX as two terms n1X + n2X, and then taking the factors apropriately. In the first case, n1*n2 = 12*6 = 72, and n1+n2 = 17, the only numbers that satisfy this are 8 and 9 For the second one, a=6, b=-5, c=-4, thus, n1*n2 = 6*(-4) = -24, and n1+n2 = -5. So only +3 and -5 work here. The first exercise: 12x*x + 17x + 6 = 12x*x + 8x + 9x + 6 = 4x(3x +2) + 3(3x +2 ) = (3x+2)(4x+3). The second was a little easier. 6x*x - 5x - 4 = 6x*x +3x - 8x - 4 = 3x(2x +1) - 4 (2x+1) = (2x+1) (3x-4).... I suppose the way you show it is a little different, but it's the same idea. By using my trick, I am simply looking for 2 numbers whose sum is equal with "b" and product is equal with "a*c". On small enough numbers, that is a very simple visual task

[@salhakim2322]

Thank you so much for this method of factoring trinomials. It is so much easier than the way we are taught in school!

[@marcbennet4346]

One way to look at humanity is by splitting it into 2 categories: those whose knowledge of a complex subject is such that they make it seem simple & easy (this is a tiny minority); and those whose ignorance of a simple subject is such that they make it seem complex & difficult (this is the overwhelming majority). Congrats, and thank you Professor ! !

[@user-xi4ye4qq9y]

Very helpful for my summer math packet. Thank you for making this less confusing for me :)

[@mrhtutoring]

You are so welcome!

[@luisclementeortegasegovia8603]

Substitution rules are very important, and are used a lot for algebra problems. They are a big help for derivatives, integrals, etc. Thanks professor 👍

[@maherhaddad6455]

Nice way to solve. I am an engineer and I never heard of it. Thank you for the information.

[@mrhtutoring]

👍

[@robdunsmuir732]

You just changed my life. Thank you for this

[@larrydickenson8922]

The way I explained it to my kids is to list the prime factors of the first and last number vertically beneath them. Then group them to find the coefficient of the middle term. Thus the 12 becomes 2,2,3 and the 6 becomes 2,3. The 3 pairs with the 3 and the 2,2 pairs with the 2 resulting in 9+8=17. Kind of like his method but without moving numbers all over the place.

[@mrhtutoring]

I've tried that method with my students also. But many of them had difficulty doing it that way. Especially, if the c value was something like 12 or 24.

[@alittax]

Superb, thank you!!!

[@jmich7]

100 per cent thumbs up! I also started thinking of the transformation of 12, 17, 6 into 3, 2, 4, and 3! Amazing

[@mrhtutoring]

🙏

[@user-us4gd7yx2l]

I had a test tomorrow morning you just saved my life

[@rcb3921]

The whole benefit of writing lines when you simply an expression is that each line is equivalent to the one before. This method breaks that rule ... (6xx) - (5x) - (4) (1xx) - (5x) - (24) at the given roots... [-1/2, 3/4]. How would your students defend these operations if challenged, except to say "it seems to work."? Better to multiply the whole expression by the leading Coefficient so it can be seen how (36xx) - (30x) - (24) then factors to the form (6x + ?)(6x - ?). Determine the unknowns according to FOIL. Reduce to the preferred form.

[@RexxSchneider]

It works by considering a quadratic like ax^2 + bx + c = a*(x^2 + bx/a + ac/a^2) Then if you can factorise x^2 + bx + ac = (x + m)(x + n), you can also factorise x^2 + bx/a + ac/a^2 = (x + m/a)(x + n/a) where m/a and n/a will reduce. So you now have ax^2 + bx + c = a*(x^2 + bx/a + ac/a^2) = a*(x + m/a)(x + n/a) and you'll always be able to distribute the initial 'a' to clear the fractions. 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).

[@rcb3921]

@@RexxSchneider Yes, this is more or less what I have said. The difference between us is that I have multiplied the whole expression by the leading coefficient to make it a perfect square for easy factorization, and you have divided the whole expression by the leading coefficient to make it easy to factorize. I prefer the method that you have shown, but the one that I described seemed like an appropriate correction to what's shown in the video. What you have done makes sense. But, it is not what is shown in the video, which is a sloppy shortcut that produces misleading internal steps.

[@rcb3921]

@@angelofelisimolacruz1462 I like what you've done here, I have no problem with it. But this working out does not contain the statement x² + 17x + 72. The video does, and it is invalid in its context. That is my complaint.

[@AngeloLaCruz]

@@rcb3921 Correct, In the video of @mrhtutoring the transition from 12x² + 17x +6 to x² + 17x + 72 has an implied substitution. Namely X = 12x. Thus, 12x² + 17x +6 becomes X² + 17X + 72 and with a note ... DO NOT FORGET THE FACTOR 12. I would not choose to present this method to My Students.

[@geminwanzo]

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) 😊

[@mistymouse6840]

I prefer this method as each expression is equal to the previous expression, and so one can see why it is valid.

[@stellarnucleosynthesis]

I was reading a complicated version of factoring trinomials but this was honestly such a better method - thank you so much Mr. H.

[@alihoussney7870]

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!!!

[@zantonio1310]

I really love this channel ❤ Now, because of your channel, my love for maths increases 📈📈📈📈. Thank you so much. And keep it up ❤

[@mrhtutoring]

Wow, thank you!

[@rufescens]

Derivations for the methods you're introducing would be nice. Otherwise, it just feels like we're learning techniques by rote, without understanding them.

[@mrhtutoring]

Some KZbin channels explain how a car runs, how the motor operates, etc. While others show how to drive the car. I'm trying to do the latter.

[@rufescens]

@@mrhtutoring Honestly, that’s disappointing. It promotes a brainless approach to math. Solving math problems shouldn’t be like driving a car. The latter is functional. Math should be about understanding. (And I’m not talking about complex derivations. An explanation of why the trick in this video works, for instance, would keep it from just appearing like a magic trick.)

[@RexxSchneider]

@@rufescens It works by considering a quadratic like ax^2 + bx + c = a*(x^2 + bx/a + ac/a^2) Then if you can factorise x^2 + bx + ac = (x + m)(x + n), you can also factorise x^2 + bx/a + ac/a^2 = (x + m/a)(x + n/a) where m/a and n/a will reduce. So you now have ax^2 + bx + c = a*(x^2 + bx/a + ac/a^2) = a*(x + m/a)(x + n/a) and you'll always be able to distribute the initial 'a' to clear the fractions. 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).

[@rufescens]

@@RexxSchneider Thanks, Rexx! I'm following your derivation for the most part. This is the sort of thing I appreciate--actually understanding how the math works. Actually, I was really on my way to understanding this on my own, in my own less technical way. When we factor an equation in the form of ax^2+bx+c, we're looking for (mx+n)(px+q), where m X p = a, n X q = c, and mq + np = b. So I see how multiplying a X c [12 X 6 in the first example] would give us (m X p) X (n X q), which is also the product of the addends of b, (m X q) X (n X p). In other words, in the first example, the two addends of 17 have to multiply to give us a X c, or 12 X 6, which is 72. That's as far as I went, but I think I would have made sense of the rest of the trick if I thought further. My main point, really, was to express objection to Mr. H's teaching approach on this channel. What's the use of teaching a technique in math without explaining why it works? To allow students to do better on tests or solve problems more quickly without actually understanding the subject? I find that lame!

[@djung650]

It all depends on timing. You wouldn't want to derive the quadratic formula for most middle school students. Still they might need to learn how to factor trinomials with leading coefficient being not one. Using quadric formula is functional as well in most American middle or high school math classrooms given the time allotted. If I have preference, I would rather encourage beginning algebra students to use the method Mr. H showed in the video more than quadratic formula. That way they get to hone factoring numbers so that by practice they get a better number sense. I wonder if Mr. H's presentation is introduced after the boring quadratic formula, it might get many students excited to question how it works and comes the great teaching moment for derivation and understanding. A good teacher knows when to teach what; otherwise mathematical explanation/derivation is still teaching by rote (vs understanding).

[@somatematicaemaisnada4648]

Wow, this video on Factoring Trinomials is so informative! I appreciate the detailed breakdown and the insights you've shared here.

[@Charlie20077]

This blew my mind! It's quicker then any method I've learned in school. Thank you!

[@arithene]

you must be actor😍

[@oscarmedina1597]

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.

[@Strawberrybeaches]

I am literallly dropping my jaw on how amazingly easy this makes factoring equations now. I wish they taught this method in the US.

[@ysngngys7753]

A straight forward method compared to guess and check works best. Thank you for this.

[@Toha-mk4hz]

In 😂 Bangladesh when we are in class 5 we are tought by this method.

[@derwolf7810]

Objection: You missed a step, which results in wrong factorizations for specific terms, such as the following. Example: 20x^2 + 58x + 42 Your process applied: x^2 + 58x + 840 (x + 28/20) (x + 30/20) (x + 7/5) (x + 3/2) (5x + 7) (2x + 3) Missing step: 20/(5*2) (5x + 7) (2x + 3) = 2 (5x + 7) (2x + 3) But the proper factorization is: (5x + 7) (4x + 6) Depending on what you need the factorization for, it might forgivable though (for example in case you solve for x in 20x^2 + 58x + 42 =0).

[@timeonly1401]

Very interesting. So this technique works ONLY IF we first factor out GCF from all the terms. In this case, we would factor out 2 to give 2(10x²+29x+21). Then apply the presenter's technique ONLY on the reduced trinomial to give (5x+7)(2x+3), then the 2 "follows along" (gets coped) to make the final answer of 2(5x+7)(2x+3). Thx!!

[@user-iu8uk5tc9s]

He didn't miss the step because the expressions in both examples have no common factor in the coefficients of the terms. He might have chosen such examples on purpose.