Factoring Quadratics WITHOUT Guessing Product & Sum

[@saturday9032]

Okay, wow. This method should've been taught in schools. Incredible! Edit: I am actually lost for words. Sounds like glazing but it isn't

[@wernerviehhauser94]

It was. Do I use it? No. I'm faster with the quadratic formula. This method provides some insight to the attentive student, but no gain in speed over the quadratic formula for real-world numbers

[@farhanrejwan]

this proves how we try to overcomplicate things in attempts to solve them, while all we needed was a simple thinking process.

[@saftheartist6137]

@@wernerviehhauser94I agree the quadratic formula technically works for all these problems and provides even more. The need to guess and check just gets eliminated in all it’s entirety.

[@kevinstreeter6943]

@@wernerviehhauser94 I have taught math. I recommend using the quadratic formula as a last resort. They are too many ways you can make a mistake.

[@kevinstreeter6943]

@@saftheartist6137 I have taught math. I recommend using the quadratic formula as a last resort. They are too many ways you can make a mistake.

[@claude77573]

My only suggestion is that "c" is used to define two different things: The third "element" of the quadradic equation and the "constant value" added to or subtracted from the second element of the quadratic, b. To help prevent confusion, I would suggest calling the second constant value "k", so we don't have two c's with different meanings.

[@thomasdemilio6164]

Right, mathematical rigor must be provided

[@shabirmir9597]

k= sqrt{(b/2)-(c)}; m = (b/2)+(k); n =(b/2)-(k). Final factors: (x +/- m)(x +/- n). k must be within bracket to take care of sign e.g. x**2 + 5 x - 14 provides k = 4.5, m=2.5+4.5=7, n=2.5-4.5= (-2). Factors are (x+7)(x-2)

[@geoffstrickler]

Simpler to remember than the quadratic formula, arguably easier to calculate. But as others pointed out, it’s essentially the same method.

[@KRYPTOS_K5]

The same suggestion of gpt4 who doesn't know this universal method to my amazement!!! Brasil

[@robappleby583]

@@thomasdemilio6164clarity.

[@fgvcosmic6752]

This is pretty much just how the quadratic formula works. This is how the quadratic formula should be taught, in my opinion; by starting with this as motivation

[@gregorymorse8423]

The quadratic formula is derived from the completing the square method. Rookie mistake.

[@notcraig3204]

@@gregorymorse8423 You trolling? This method is the same steps as completing the square.

[@DarIsANoob]

@@gregorymorse8423formulas can be derived from a multiple of methods. Rookie mistake.

[@prefabrication]

​@@gregorymorse8423 This is literally completing the square. He's doing so by finding c in 16² - c² = 192.

[@mikechrissan]

That is nothing like completing the square

[@danielnegussie4894]

It is called the Poh Shen Loh method. A math professor named Poh Shen Loh discovered this method. The math professor himself said he found this method buried in a very very old Math book. Here is the link kzbin.info/www/bejne/jnyliWOoaK9gb7ssi=LiskkjE6oaHBAPZY

[@wernerviehhauser94]

i've heard it called Hindi Method or Sridhara method. We used to learn it as an application of Vieta. Nonetheless, all I use is the quadratic formula - it's faster than everything else for real-world numbers in physics problems.

[@ramunasstulga8264]

​@wernerviehhauser94 this is why I don't give a shit about how a formula is named because it's only about politics

[@maanasdongre1581]

Well it is poh Shen loh method but it is explained better

[@glennschexnayder3720]

It’s the ancient Babylonians’ method. They were very advanced in math and astronomy. If NJ Wildberger ever gets his channel restored, you’ll be able to see his demonstration of a real, Babylonian worked problem using this method.

[@chocolateangel8743]

@@glennschexnayder3720 Professor Loh basically condensed the method and made it easier to understand. This is especially true when explained with a graph.

[@angelviloria4966]

I came here to refresh my memory on factoring so I can begin teaching my girls (8th & 6th grades). I love this method because I hate ever having to say “guess and test”. I always look for methods/process or creat one. This is excellent. I can’t wait to teach it.

[@angelviloria4966]

I bet they aren’t teaching this in their school.

[@biswanathbera1260]

For small numbers this method is sheer time consuming. But for bigger numbers it is helpful.

[@dmitrmax]

Dude, teach them the generic way to solve quadratic equations, not this magic shit

[@ranjittyagi9354]

​@@angelviloria4966I would assume the same. However, just as others advised, please don't skip the formula taught in school. Remember the roots and (-b +- sqrt (b²-4ac)) / 2a

[@GeoCalifornian]

⁠@@ranjittyagi9354 How did you square the “b”?? …Thank you in advance!!

[@zulfiqarali4888]

Love this. So much of boring guessing work of stone age in school. m = sum/2 + sqrt((sum/2)^2 - product) n = sum/2 - sqrt((sum/2)^2 - product)

[@Masteroogway-e4i]

that's basically quadratic formula

[@bjornfeuerbacher5514]

You can also do it in an another way, which in the end is essentially equivalent to the one in the video: first complete the square, x² + 32x + 192 = x² + 32x + 16² - 16² + 192 = (x + 16)² - 64. Then write this as a difference of squares: (x + 16)² - 64 = (x + 16)² - 8² = ( (x + 16) + 8 ) ( (x + 16) - 8 ) = (x + 24) (x + 8).

[@Reginald425]

It always comes back to completing the square

[@50Steaks68]

You lost me at completing the square lmao I hate doing that. Guess and check with calculator is by far the fastest and easiest way to solve these

[@lookiii1]

Do it with with x² + 4x + 10

[@samhardy6030]

​@@lookiii1It works well. You complete the square, then add and subtract the square root of the completing adjustment (X+2+√6)(X+2-√6)

[@ranjittyagi9354]

Your equation furnishes complex roots. Here they are: ​-2 + 2.4494897427832i and -2 - 2.4494897427832i Please check.

[@ffggddss]

In actuality, this method is exactly equivalent to a variation of completing the square. • Divide by the leading coefficient if it isn't 1, to get x² + bx + c = 0 • Square half the linear coeff., (½b)² = B, and add & subtract that to the LHS, to get x² + bx + B + c - B = (x + ½b)² - [B - c] • Now treat the LHS as a diff of squares, which then factors into the product of (x + ½b ± √[B - c]) If you apply this to the examples in the video, I think you'll see that it amounts to the same process. In fact, all quad. soln. methods are essentially equivalent. That said, the method in the video is quite good. Fred

[@frankiej9642]

yes indeed

[@viasciencemath]

Another way to think of this is that the solutions to x^2 + b x + c = 0 are x = - b/2 +/- sqrt( (b/2)^2 - c ) which is just a rewriting of the quadratic formula.

[@Chemest_a]

Yall this is literally the quadratic formula broken down into steps. Fire video tho!

[@crazydoc3770]

I saw this method once, but after time went by, I completely forgot it existed, and when I wanted to search it, I couldn’t find it anywhere, so thanks for bringing it up again

[@rainerzufall42]

This is just a manual execution of the abc or pq formula (there are many names for it). Still impressive. For simplification I use a = 1 => pq formula for x² + px + q = 0 => x_{1/2} = - p/2 +/- sqrt( (p/2)² - q ) (x - x_1)(x - x_2) = 0 => m = -x_1 and n = -x_2. That's p/2 -/+ sqrt( (p/2)² - q ).

[@alemswazzu]

So completing the square is also called PQ method. Extra bonus is giving a result in vertex form.

[@Parasuraman-ey4wo]

Very nice. Someone was smart enough to think if this and formalize, removing guess work. I would still factorize mentally in terms of sum and product, and in the case of large numbers that are difficult for me to break down that way, I would use this procedure. Appreciate this new and simple way.

[@IAMVenos]

Sending this to every math teacher I’ve ever had 🔥

[@yassertariqvideo]

Excellent. When 'c' is a large number then it really helps solving this way. Thanks for jogging my memory, after a hiatus of 53 years.

[@tomthinlaishram4695]

The method we use at our schools are very lengthy and consumes a lot of time. Good that I got to learn this method! Thanks a lot!

[@edobolo-h2q]

i've had thoughts of this but never found a way to make them useful to my math classes, well until i found this video i could actually do something with it. THanks!

[@unknwonboy-fm5ei]

we are literally solving a quadratic , to solve another quadratic . Although the approach is nice.

[@airking2883]

Well it's always a single term quadratic which is much easier than multi term quadratic equation

[@fahrenheit2101]

​@@airking2883yes, but completing the square is not only already familiar, but is equivalent.

[@uncannyroaches5933]

Well the difference is.. x²=k type quadratic isn't exactly solving something.

[@DaniCanales10]

I haven’t even been taught factorisation of quadratic formulas yet, but found this video flabbergastingly interesting and really easy to understand.

[@bushrahossain2209]

Bro is 6 years old

[@ADN0Infinity]

Great job! The method is powerful, try its generalization. For instance, - -if you have x^3+bx^2+cx+d, and x1, x2, x3 are solutions, then x1+x2+x3=b, x1.x2 + x1.x3 + x2.x3 =c and x1.x2.x3=c. -if you have x^4+bx^3+cx^2+dx+e and x1, x2, x3, x4 are solutions, then x1+x2+x3+x4=b, x1.x2 + x1.x3 + x1.x4 + x2.x3 + x2.x4 + x3.x4=c and x1.x2.x3 + x1.x2.x4 + x2.x3.x4=c and x1.x2.x3.x4 = e💙💜💙💜💙💜💙

[@mr.cabbage4428]

This is a fun video, I found the general form for this method by not plugging anything in where x^2+bx+c=(x+((b/2)+sqrt(((b/2)^2)-c))*(x+((b/2)-sqrt(((b/2)^2)-c)) If there is a coefficient on x^2 divide it by every term and put it on the outside of the formula a(x^2+(b/a)x+(c/a)) We can treat b/a as “b” and c/a as “c” for the formula

[@matthewmanzanares6798]

This method is what I figured out in deriving the quadratic formula which I thought was that's how you actually derive it instead of completing the square. Although this ensures that you get its factors, I don't think it's the method I'm going to use mainly in my factoring, because to factor quickly, a method that allows you to being able to do it mentally or at least minimal writing would be my bet. But this method would require me a lot of thinking, writing, and/or storing numbers in my head. This method actually gave me an idea. instead of just using the quadratic formula, just use the formula derived from this method directly because comparing it to the quadratic formula, it is easier to compute the square root. when a = 1 m = b/2 + √( (b/2)² - c ) n = b/2 - √( (b/2)² - c ) it is essentially the same as the method, but for me using the formula is faster than doing all the work

[@kdaviper]

Or perhaps solving one equation for a variable and plugging it into the other equation

[@DaneBrooke]

Wonderful video. Since this amounts to using the quadratic formula for writing the factors of the quadratic, you could have just gone for "How to write the factors of a quadratic by using the quadratic formula". And *that* would have been a wonderful video. And nobody would have guessed or checked anything.

[@soilsurvivor]

Great method, and VERY well-presented! Just wordy enough without feeling like you're dragging out minor details. Concise, but not sparse. Thank you!

[@Adityasharma-00]

In this question 15:37 2x²-7x-4=0 it will written as *X²-7x-8* =0 after finding Root (x-8)(x+1)=0 then divide 2 both side (x-4)(x+1/2)=0 This also easy by their method *Maine isse aur aasan bna diya*

[@HotNoBeans]

Babe wake up new factoring method just dropped

[@starlogonova]

Brilliant. I am sure my 8th grade math teacher in my school will find this interesting.

[@ajnagarofficial]

This is the best trick i have ever seen.thankss

[@jakubjiricek7806]

I don't know about the other parts of the world, but here in the Czech republic, some 35 years ago, we were not taught to guess, but we learned the standard formula for the roots, to which this "new" approach is equivalent.

[@Arima_Kousei0_0]

1:50 he did not considered 8 and 24 because that's the answer

[@cakespreader]

probably because he wants to show this method as “easier”

[@Dsonsee]

Thanks so much for this method! I teach in highschool and even though I've explored alternatives, I never thought of something so elegant. I especially like that it takes away the arbitrariness of using the quadratic roots equation, or the frustrating guesswork of assigning numbers and seeing if they work.

[@parinose6163]

I just wanted to let you know that I never learned that before! Many thanks. It's another pain in the back removed with a quadratic polynomial!

[@jonathanimanuel5237]

It's just blew up my mind! I can't think of a word that describes this method, I think I'll use it for the rest of my life. By Guessing, hello new technique.

[@Study_Sensei_Class10th]

just use the quadriatic formula

[@IRVINMILLER]

Very clever! x^2+bx+c=0 x=-b/2(+/-)(b^2-4c)^.5/2 Someone noticed that the roots where b/2 +/- the same number. Then it becomes obvious how you developed your approach. Telling the students your thought processes is the way math was supposed to be taught. It was not intuitively obvious to me until I looked at the quadratic formula.

[@u_ufc587]

Bro u are an absolute life saver the amount of time i can save instead of guessing is astronomical

[@Swanicorn]

Why can't you just use the quadratic roots formula? It doesn't need any guesswork. It will give 2 roots values (m,n) which you can write as (x-m)(x-n). This approach is literally that quadratic formula manipulated. (-b +- D)/2a. As you can see, the same number D (discriminant) is added and subtracted from -b. D is the playing the same role as "c" in this approach.

[@lapse9continuum765]

Like others have said, this is really just a version of the quadratic formula. The two factors could also be found by this formula: [b+/- sqrt(b² - 4ac)] / 2a It's just the quadratic formula but without the negative on the first "b." Great video!

[@neitoxotien2258]

Hmmm... I love this method, since back in the day I love Factoring than Quadratic formula. This method gives me a new ways. Thank you.

[@ASINGH-li8eq]

You explained so good , I am astonished to see this ,we never learnt this in school .it is really amazing . thanks .

[@lastchance8142]

My daughter is a high school math teacher. She likes this method, but can't change the approved curriculum. This is why things don't get better in US schools.

[@jamesharmon4994]

IMO, factoring is a waste of time.

[@davidwright8432]

Surely there's nothing against teaching it after teaching the 'approved' method? It could be done easily, in a 40 min class time!

[@scottekoontz]

I taught HS math, and always found time for additional niceties like this. Doesn't matter if it is approved, although she may want to avoid quizzing students on it if it really is a backward administration. But for the test that wraps up all methods, i.e., solve using any method, students can use this method if they wish.

[@JavedAlam24]

​@@jamesharmon4994 you've clearly never had to integrate anything requiring partial fraction deconposition

[@pianoman5259]

​@@jamesharmon4994respectfully, your opinion is wrong unless you are implying there is a faster way

[@rgmolpus]

FANTASTIC! I've suspected there was a formula or algorithm for finding the factors, but never found one. Thank you for the explanation!

[@HellGuyRj]

Earlier in my days, we used to do prime factorisation of the constant which simplifies the guessing game with massive amount. But yes, this is amazing.

[@matthiasklein9608]

And this is exactly what is called the p-q formula (at least here in Germany) for solving quadratic equations: x^2 + px + q = 0 x = -p/2 +/- sqrt((p/2)^2 -q). It’s just not displayed as a single formula but broken down into two steps.

[@uncannyroaches5933]

I mean instead of complicating things just use the standard quadratic formula. But sure maybe you guys have your own tricks. I always like to keep things simple.

[@Study_Sensei_Class10th]

@@uncannyroaches5933 yeah fr the quadratic formulae is so simple why no one uses it

[@karikosonen3047]

I havent seen this before. Absolutely brilliant! I wish i knew this in the 90´s !

[@skylerharris3914]

Just completing the square and using the solutions of the function equaling zero as the factors which is what they are.

[@raizen_u]

Ive been looking for a way to solve factoring quadratics without guessing for months now so thank you !

[@VuNam_MCVN]

6:26 m= 32 - n Replace m in the product Check if n and m is negative or not

[@bfelten1]

Absolutely beautiful. 🥰 No need to memorize any formula, not even the (a+b)(a+c) or the (a+b)+(a-b) or (a+b)(a-b), because you can easily complete them on the next line. What are the odds that you remember the quadratic formula some 30-40 years from now? I'm 76 years old, and even though I graduated from university with maths as one of my majors, I don't (because I almost never use it).

[@Study_Sensei_Class10th]

by ur logic what are the odds that you have to SOLVE a quadratic formulae 30 years from now ?? this is litrally basic maths

[@SM-ev3pv]

This is an AMAZING method? You should be nominated for the Fields Medal.

[@Matthewmellow2k]

this is absolutely freaking amazing! thank you! i have always found the guessing method to be flawed. this is an awesome exact method.

[@simonekentish7491]

In my experience, most students who struggle with multiplication struggle with fractions even more. It’s probably more efficient to stick to finding factors of ac first, and for more challenging problems, make an organized list of the factor pairs. What does help from this method is to recognize that larger b values (compared to ac) mean that the numbers are further apart. This understanding helps with guess and check.

[@emayteetee73]

I agree, but then we are teachers. Teaching this method to the masses (especially Foundation Tier) would baffle them. Higher Tier GCSE students with strong algebra skills who enjoy generalising results could take to this way. They also appreciate being shown multiple methods and are able to select the most appropriate or efficient method when necessary.

[@catburner1896]

This actually really cool and helpful, I might use this next time instead of the quadratic equation

[@NichaelCramer]

Other commenters have mentioned that this technique is basically (just) how the quadratic equation works. While, strictly speaking, this may be true I still think that this video is extremely valuable in the way that it views the QE from rather different direction. In short, you can never too many tools in your toolbox. ;-) (More to the point I think that in some cases this technique could be quicker/handier/less cumbersome than using the QE. For example, if your goal were to factor the equation rather than simply obtaining its roots -say as a sub-step in dealing with a larger problem.)

[@notnyw]

Always wondered if there was a mathematical way of determining the factors and was told I had to guess the two numbers. Never knew about this method. Very well presented. Many thanks Jensen.

[@Neekalos]

Why not just use the quadratic formula?

[@andrewkarsten5268]

@@Neekalosa lot of American school teach the guess and check method before the quadratic formula. They force you to spend at least a month on guess and check before teaching you complete the square or quadratic equation

[@ahmadafgan6591]

I am very grateful for your method. It helps me a lot, during my exams and worksheets. I think this should be known further, you are an inventor.

[@megumarutakashishi3003]

I've seen OutlierOrg but different approach but same principle. Thank you so much. You earn a subscribe

[@musaadebisi9105]

Same here

[@NaHBrO733]

So completing the square? This is the method used to prove the quadratic formula. In the end, you either guess and check, or use the formula as it is. Once you know the formula, these steps are redundant. I only use this to solve integer counting problems There are many different methods, which can be easily proven equivalent to each other. I am good at math, so I can understand all those methods easily and generalize them to the formula. For some students, maybe one method is easier to understand than the other. *Suppose you want to factorize ax^2+bx+c, just solve it x=x1,x=x2 using the formula. Then (x-x1)(x-x2)=0 is the factorization.

[@paulcooper8818]

Nice, this process is easier to follow than using the formula. Really good as long as the formula is also presented and understood for use in purely algebraic manipulations

[@user-gr5tx6rd4h]

In fact I invented this method (or very similar) some decades ago (now retired from high school). But often with much less work. Take x^2 - 6x - 91 = 0. Only integer factoring of 91 is 7 x 13, so I write (x - 7) (x - 13), yes - both with a minus. Now it is easy to see if any - must be made into a + (which is "easier" than the opposite). Since 7 - 13 = -6, we must have + on the 7. Thus: (x + 7) (x - 13).

[@negativeM]

This puts the quadratic formula to shame. Thanks for sharing this! I can also use more knowledge in the mathematics language. After all the years of not knowing this or ever thinking about a way to make such a thing viable. And finally I came across this. Props.

[@mathe-info]

I really like about this approach that you immediately get to s = -b/(2a) where s is the x-coordinate of the minimum/maximum of the parabola. With this you can calculate the y-coordinate by plucking x=s into the formula and you get the location of the minimum/maximum.

[@gagikantonyan]

I am stunned. Perfect.

[@Kounomura]

Analytically, the extreme value (minimum or maximum) of the function f(x)= ax^2+bx+c is at 2ax+b=0, i.e. at x=-b/2a. The solutions of the equation are to the left and right of this, at an equal distance from this. In fact, the solving formula also expresses this.

[@Quroxify]

Haha i remember one day one student I was tutoring bitching about where the quadratic formula came from. So I derived it on the blackboard for him. I would suggest as an exercise (or a future video in this series) that you can do it too. Thanks for this . I like it just as much as the quadratic formula because it's cute, it's easy and it's novel. One thing this could have embellished a bit is the form of the graph and it's roots.

[@ammaradil3796]

This method should go viral, we have to make children's lives easier. My school years would've been so not depressing if this method was around my time.

[@ranjittyagi9354]

Most methods were there in books outside of the ones we were told to follow. However, the teachers who could think of explaining how to approach a given problem from different angles weren't around in some respectable number. They followed a particular book and stopped at that. There has been a very asked cha ge in how children learn these days. So many follow good channels on KZbin as well. No wonder then, the depressing days had to be endured by us.

[@malluzenitsu]

x = [-b±√(b²-4ac)]/2a

[@scottekoontz]

This is introduced in many math classes, but alas too many parents will claim this is more difficult or "I never learned it that way" and blame this on Common Core.

[@johnsolari4933]

Great method. I think it's easier to understand this method than to memorise and remember the quadratic equation.

[@gendo1123]

What i have done for similar problems is right m=b-n then n(b-n)=c so n^2-bn+c=o then use x=(-b±squareroot(b^2-4ac))/(2a) for n^2-bn+c as it would allow me to right ax^2+bx+c=0 in form of (x+n)(x+m)=0

[@alanthayer8797]

NIIIIIICE WRK MY BOY! Simplicity is a Gift

[@mastermichael0751]

I’ve been taught this in school since seventh grade. Seeing other comments, it appears to me that I was apparently very lucky to have known this method from as early as I have

[@baileyradel3104]

Compelling method. Reminds me of "completing the square" since you're taking half the b value and factoring out an a>1. A "benefit" of using the c value and listing factors is that students can generate a table of y = c/x in their graphing calculators and look for integer coordinate pairs. I love the link to multiplying conjugate pairs in this method though! I'll stick with the AC method for a>1 for now.

[@شعرکوتاه-ع7ظ]

Mathematics is the basis of all sciences and you explained it well, thank you

[@iikitsunex-x]

Mind blowing method yet so intuitive and simple! I wish this was taught in my classes!! Thank you 😊

[@notcraig3204]

Played this on x2.6 speed and quadratic formula got the answer first with plenty to spare. While this certainly gives a cool perspective that seems to speak to students, we should probably spend more time pondering what to do about the fact that the formula (by definition the easiest way to do something) is so intimidating to students.

[@sphereguanzon2217]

This is scarily accurate. Even with irrational radicals

[@cerisaac8149]

Wish I knew this wayy earlier. Thanks to this video I just realized that you can actually also use the Quadratic Formula for factoring! (by replacing -b to positive)

[@cerisaac8149]

Oh and you can derive the process/formula used in the video from the Quadratic Formula(where -b is positive) and vice versa, which makes sense since both ultimately gives the same answer

[@yvg2009]

This method already taught to me 🔥🔥🔥

[@naveenpokala6442]

Excellent and simple method, thanks for bringing it up

[@younesdz6567]

1st solution:( b/2)- √((b/2)^2 -c) 2nd solution:(b/2)+ √((b/2)^2 -c) I think this is it

[@pietergeerkens6324]

This is fundamentally Poh Shen Loh factoring, derived differently; but you've missed an opportunity to combine with the Australian (i.e. "down under", referring to the placement of a in the denominator) technique for factoring quadratics with a 1. When b is odd, we encounter an unusual circumstance where the decimal fraction 4.5 may be more convenient than the common fraction 9/2, using this shortcut for squaring a half integer: 1.5² = 0.25 + 100 * (1 * 2) = 2.25 2.5² = 0.25 + 100 * (2 * 3) = 6.25 3.5² = 0.25 + 100 * (3 * 4) = 12.25 4.5² = 0.25 + 100 * (4 * 5) = 20.25 5.5² = 0.25 + 100 * (5 * 6) = 30.25 : 9.5² = 0.25 + 100 * (9 *10) = 90.25 : The Australian method for handling a 1 proceeds by initially ignoring that a 1 as, using your example: 3.5² - c² = (2)(-4) = - 8 becomes c² = 12.25 + 8 = 20.25 = 4.5² and then factoring as [ (2x - 3.5 - 4.5) (2x - 3.5 + 4.5) ] / 2 = [ (2x - 8) (2x + 1) ] / 2 = [ (2x - 8) / 2 ] (2x + 1) = (x - 4) (2x + 1). Notice how the a value of 2 was initially placed in both factors, as well as "down under" in the denominator. When a is composite and must be factored across both final factors, this trick delays having to determine that factoring and distribution until a final step, not interfering with the rest. The factoring of a reduces to identifying the common factor in each numerator factor - much simpler than previous methods.

[@trien30]

Po Shen Loh, not Poh Shen Loh.

[@jerryeubanks491]

Great job where were you 58 years ago when I started seeing factoring this is so much easier

[@ttg4r4taerwgaegrrgea86]

are you the father of Alex Eubanks?

[@Clint52279]

Watching this ticks me off because not only does the more efficient methid make sense, but the OG method makes more sense than what I was taught. My math education sucked noodles. I took calculous in HS, and was fumbling in the dark the whole time.

[@BirhaneGselasie]

Amazing, this method should include in our curriculum .

[@dalhasaid3321]

Great video ❤ deserves millions of viewers

[@SargonofQueens]

When I first watched this video, I thought it was brilliant. Now I know you can just take half of b and square it, the number will be the perfect square of that equation. Add that number to each side and you will get the answer faster. For instance 1x square -6x-91 . We can move -91 to the right. Half of 6 is 3 . 3 square is 9. Add 9 to both sides, Now you get a (x - 3) square. = 91+9 X-3 = root 100. X = +- 10+3.

[@ramonalejano671]

Since I was not taught this, I was using the quadratic formula if I can't guess the factors. Thanks.

[@wmatos]

Very good. Thank you.

[@RexxSchneider]

Isn't the method shown no less effort than factorising ax^2 + bx + c by calculating D = b^2 - 4ac and then writing the factorisation as a.( x + (b + √D)/2a) ).( x + (b - √D)/2a )? That has the advantage of using a formula that's either already familiar, or at least will have further applications later on. If you are worried by big numbers, just calculate d = (b/2)^2 - ac and write the factorisation as a.( x + (b/2 + √d)/a ).( x + (b/2 - √d)/a) which scales thing down by a factor of 4 (like the video does).

[@fahrenheit2101]

Yes. It's just an alternate form for the quadratic formula. This is the famous Po Shen Loh method, and 3B1B has a good video covering it in a little more depth

[@confronta]

True, but in general students that barely understand functions have zero insight in the quadratic formula. I think this method is much closer to their skills.

[@golalislive]

cool, bur useless to say the least for anyone who has even some practice in factorization this method is actually more time consuming than just guessing or checking(if you do it in a right way) and for more complex quadratic both this method and the quadratic formula should take around the same time to compute, I would actually prefer this for complex quadratics with large coefficients as it is more simpler than the quadratic formula. Good Job, but still more of a show method than an actual one.

[@adityaroy5300]

Very thanks sir i got 5/5 in section of quadratic equations in my bank manager exam. I'm from India and a aspirant of govt bank exams

[@adityaroy5300]

I have to do 35 questions in 20 min which also include 5 questions from quadratic equations

[@TonyBT05]

1:09 product of C and A not only C, And also another method is that you can factorize 192 like we do in HCF and LCM and then multiply them to make 2 numbers that do the job

[@jordanrutledge7943]

This is essentially completing the square, I like it 😊

[@AaryanAbreo]

From the bottom of my heart, thank you

[@currytut455]

Using this method, I took it upon myself to derive a general equation which will help solve for the foctors in a single step. Interestingly, this general solution is just the quadratic equation in disguise😂😂😂. When I was deriving the solution for m and n, I discovered I stumbled upon the quadratic equation, using the method he illustrated for finding m and n without guessing. It's amazing to see the connection between the two ideas. Anyway, here is the solution. foctors = (x + m) , (x + n) factors = (x + [(b + (b2 - 4ac)^1/2)/2a]), (x + [(b - (b2 - 4ac)^1/2)/2a]) using the same idea you can also as well derive the same equation. I would explain how I derived it, but the response would be unnecessarily lengthy.

[@tambuwalmathsclass]

Thanks to Poh shen Loh for making our work easier.

[@trien30]

His name is Po Shen Loh, not Poh Shen Loh.

[@krissyai]

I generalized it with the roots adding up to -b/a and their product being c/a, I did the same thing where one root is essentially the x value for the vertex plus h, the other is the x value for the vertex minus h m = -b/2a + h n = -b/2a - h Their product should be c/a m . n = (-b/2a + h)(-b/2a - h) m . n = b²/4a² - h² b²/4a² - h² = c/a At this point you may have realized that it's building up the main formula to solve for quadratics h² = b²/4a² - c/a We multiply c/a times 4a/4a h² = b²/4a² - 4ac/4a² h² = (b² - 4ac)/4a² h = √(b² - 4ac)/2a Yes, the discriminant shows up here, being b² - 4ac Let's denote the discriminant by using ∆ h = √∆/2a The main formula is x = (-b ± √∆)/2a So if we just replace √∆/2a for h x = -b/2a ± h I should point out that -b/2a is the x value for the vertex in every quadratic, where f(-b/2a) is the y value for this vertex of this same quadratic f(x) This in a way proves that each of the roots is the same distance away from the x value of the vertex, I believe this applies for every polynomial that has an even degree

[@riseofgaming9429]

If you look very very very closely at the procedure, Its just the quadratic formula but in step by step method of solving the determinat to get C and adding and subtracting that to b/2a which will give the two roots

[@biscotty6669]

Very nice video. Small suggestion: use c' in the m/n formulas or another name to avoid any confusion with c in the quadratic equation.

[@davidurman5595]

Yes, a different name would be helpful. Like how about D? The D could be thought of as standing for displacement, that is, the displacement of each of the two factors from B/2.

[@NyteRazor]

You can also factor by grouping 2x^2-7x-4 = 2x^2 -8x+1x -4 = (2x^2-8x) + (1x-4) = 2x*(x-4) + (x-4) -> factor (x-4) from each group making -> (2x+1)(x-4)

[@diwakardas1807]

In the thumbnail it's clearly written a new amazing "factoring method". That is to find the factors..... Not a new method of finding the value of x in the quadratic equation Which we all know the best is the quadratic formula 😊 hope it explains the confusion

[@saeidabbasi7152]

Hi there 🎉good good job If you in put as a(x-m)(x-n)=0 and your factor form as ax^2+bx+c=0 and find the ROOTS as m&n then plug into first equation a(x-m)(x-n)=0 it's much more easier 😊

[@rlouisw]

Good method, and well presented. (The rest of this is not for you, but more for the existence of this method in the first place) Ridiculous though, factoring is not a difficult process, and this method is not only completely unnecessary but much more difficult than just factoring it. I speak to you as a math tutor, and when the kids come to me being confused by yet another factoring method to make things "easier", and I show them how very easy it is to just skip to the end and solve it directly, each one is more shocked than the last at how very easy this all is. And they are truly shocked when I show them how easy it is when a doesn't equal 1. I mean... look at what happened with x^2 + 9x + 20, fractions were introduced to what should have been HORRIBLY SIMPLE! Stop babying these kids with all these "factoring methods", they can handle the guessing and checking (not to mention, it is actually the far easier method). *frustrated rant over, I had a day...*