What the 1869 MIT Entrance Exam Reveals About Math Today

[@BleuSquid]

For everyone saying how easy these questions are, some back story: MIT had only been established a few years prior, in 1861, followed a couple years later by the first actual classes in 1865. Up until 1869, there was no entrance exam. The requirements for admission that year were simply that "the student must have attained the age of sixteen years, and must pass a satisfactory examination in arithmetic, so much of algebra as precedes equations of the second degree, plane geometry, English grammar, and geography." In the 1869-1870 school year, the student body was composed of 89 first-years, 37 second-years, 56 third-years, and 24 fourth years. (Course Catalogue of the Massachusetts Institute of Technology 1869 - 1870, from the online MIT Archives) These questions aren't meant to be particularly difficult, nor did they need to be. Unlike today, they weren't trying to find the top 10% of applicants, but rather ensure a minimum level of knowledge.

[@michaeledwards2251]

An additional factor is the US civil war which disrupted education. It would have taken until 1869 before the candidates could be expected to have a continuous education.

[@PhrontDoor]

Excellent point there.

[@18890426]

😮

[@18890426]

How about Harvard?

[@xkillmaster3995]

Wish i had known about this when learning factoring back in the day. This visual representation truly is beauty thank you.

[@Blu3B33r]

It's my first time seeing this method and I love it! Thank you for sharing

[@kusanagi533d]

Never have I ever seen the squaring method, thank it was enlightening

[@RAFAELSILVA-by6dy]

One key point is that we can check that (a+b) is a factor of the second expression by setting b = -a and checking that we get zero. And you can see immediately that (a+b) is not a factor of the first expression.

[@polymathematic]

very nice!

[@Cau_No]

This is what I always think of when people citicize the 'new math' - there is no 'new' or 'old', it is always the same mathematics. Just the representation is different. Alas, they don't see it, because they never really understood it in the first place. Thinking the older, written algorithm is better because it is faster cannot be more wrong. You just do the same things, but adding part sums in your head instead of visualizing it. But that's plain memorizing the algorithm. With visualizing, you'll be able to actually understand it. (You even don't have to memorize the binomics formulas anymore, just recognize and or derive them.) And as this is a teaching tool, no one is kept from learning the fasttracker later on. Sadly, many people can't see that connection, because they actually suck at math without knowing it. Mathematics is so much more than crunching numbers.

[@jessejordache1869]

"New Math" is short for "heavily revamped curriculum because Sputnik". It's called a neologism.

[@pedroteran5885]

Tbh, after watching this I'm unsurprised that the US is falling apart. If you tell people to solve x^2+bx+c=0 by solving the system -x-y=b, xy=c, which is what this person presents as an example of how this is supposedly useful, soon enough there will be no running water in that country.

[@Thagrynor]

For me, it isn't so much the area method that annoys me when I see some of these things on social media, it is that the student will get it marked wrong because they ask for, say, 3 x 5 and they put, for example, the 3 on top and 5 on the side and then are told it is wrong because the 3 should be on the side and 5 on top. It is like they are intentionally using a method that should be totally fine but then grading it in a way to try to deny the commutative properties of simply multiplication. Denying portions of mathematic truth in any method makes the teaching of that method questionable, not because the method is questionable but because the teaching method is not applied correctly or with any care to ensure the students learn actual math or mathematical truth. With the algorithm method, there would be no difference with what number was on the first line vs the second line. With the area method, you end up with a rectangle with transposed dimensions and thus, teachers who do not know what they are actually teaching will say it is wrong despite it being absolutely correct.

[@polymathematic]

i definitely get the frustration with requiring a certain method. i do not require particular methods in my class, but i can see how that'd be annoying for a student to complete a problem correctly, but then get dinged 'cause they did it differently from the teacher.

[@Thagrynor]

@polymathematic I had wanted to be a math teacher when I was younger (still do, to be honest, but at 41 and having obligations that preclude going back to school to finish what got derailed due to life emergencies when I was in school, that doesn't seem likely lol) and used to think of creative ideas for things like extra credit, just to make the classroom more fun, like you'd get a few extra points if you opted to use an abacus for a test instead of a calculator. By the same token, like you, I would not have marked a student wrong if they used a slightly varied method (or combined a few steps into one where it was obviously done correctly, such as commuting things mid step) just because it didn't follow the exact method the book showed. If it was clear they understood what they were doing and could prove that via the steps they did show, I felt that was sufficient to prove they got the material and would correctly answer any similar problem presented. It just makes me sad that there are so many teachers who don't understand the material enough or don't care/aren't passionate about the material/teaching to do more than robotically grade something based on what "the book says". It is one reason I love your channel, because I can tell (and you remind me in this way of one of my favorite and most inspiring teachers throughout all of school) that you care about teaching well and care about the material. Good teachers like you seem to be a rarer and rarer commodity these days. And your students are lucky to have you. 🙂

[@geraldsomerville1280]

@@Thagrynor I am perhaps going off at a tangent, but your talk of creative ideas to make classroom work more fun reminded me of how I would write up the results of Physics experiments in secondary school. I would usually know what the results of an experiment should be, but the actual observations would, of course, all be approximate, unlike in Pure Mathematics, so I would estimate the maximum amount of error in every measurement, and calculate an upper bound for the amount of error in each intermediate result alongside calculating the intermediate result itself. It made the whole exercise more interesting mathematically, and also brought out the importance of the order in which results are calculated when the calculations are based on approximate data and not exact data like in a mathematical problem. (The use of calculators or computer programs gives rise to a similar issue of approximation, even if the data on which the calculations are based is exact.)

[@geraldsomerville1280]

I think that the fundamental error here is in insisting that there is only one method. Everyone, whether teacher or student, should rather be thinking of trying out as many and varied methods of solving the same problem as possible. Only by approaching the same problem from different angles can you be sure you understand the situation.

[@kevinpaulus4483]

@@geraldsomerville1280 Our teacher in what you call junior and senior year didn't give a fuck what your method was you could just write answers if they were correct you got the points otherwise you didn't. Sometimes that made it kinda easy cause most of "solve for x" kinda problems have inputs in the range of [-4,+4]. So a little brute force and some guesstimation often could get you the result very rapidly. He wasn't as great of a teacher as I believe he could have been, he was hard to understand speech wise and had narcolepsy, but the ability to use whatever method you had to solve the problems was so liberating and he commanded a decent amount of respect in a mostly stupid male practical schooling classroom which says a lot.

[@louml6233]

I love your explanation. It's easier to see and to follow. Thank you

[@dougnettleton5326]

I got the idea of factoring the second trinomial, and I did that, but I don't see how the box really helped. You just needed two terms that multiply to give the third term and add to give the second. Once you have the two factors, you can divide the (a+b). Then again, I don't really see an advantage "using the box", you just write the trinomial above the binomial and do the "long multiplication" and add up the like terms. It's the exact same number of multiplication and addition operations. That said, I enjoyed the video. Thanks.

[@polymathematic]

It's definitely a visual way to organize the process and maybe gives a bit more insight into how factoring works.

[@Alfi-rp6il]

@@polymathematic The best way to learn how factoring works is to apply long division. But the 'insight' has do exist BEFORE doing so: When it is possible to devide the whole numarator by a + b, it must be possible to devide one of the factors in the numarator by (a + b). In fact, it is the second one: (a^2 - 2ab - 3b^2) : (a + b) = a - 3b => (a^2 - 2ab - 3b^2) = (a + b) * (a - 3b) By carrying out this multiplication we regain the term (a^2 - ...). This second step is important to understand that the same 4 multiplications (! cf. the box!) occure in the long division as well as in the multiplication of the factors. THIS creates the 'insight'.

[@mrsoikawa]

Awesome video, thank you

[@disgruntledtoons]

The thing is, with basic arithmetic students should soon memorize the answer to the simpler calculations so that they can tackle more difficult problems later on. For instance, by the end of first grade a child should not have to count out single-digit addition on the fingers. If the answer is not memorized by that point something has gone wrong and needs to be addressed.

[@HansLemurson]

One thing I like to do when I see polynomials is to imagine X as 10, and then see what they become as base 10 numbers. So long as there's no carrying, it works out pretty well. When you do A as 10 and B as 1 in the example polynomial, you quickly see that although the first trinomial evaluates to 309, the the second trinomial evaluates to 77, which is clearly divisible by 11.

[@JavedAlam24]

Hey that's a cool idea!

[@pwmiles56]

Funny story: In the library of my college in Cambridge, UK, I once discovered some Maths Tripos papers from around 1802. These were the exams for a bachelor's degree. I was struck by one question: if you have some rectangular boxes with side lengths of such-and-such, how many would fit into a warehouse of such-and-such length, width and height. All the numbers were given; it amounted to knowing the formula for volume and doing some basic arithmetic. It's possibly explained by the fact that, due to the influence of Newton, for a long time ALL Cambridge students were required to pass in mathematics. As a sometime maths lecturer, I developed a horror of these so-called intuitive, visual methods. Not everyone thinks that way, or needs to. The "obvious" (to me) method is to investigate factorisations, by way of the quadratic formula. I used to say to the students "Don't try to guess the factors! In an exam, you don't want to be guessing. Use the formula!"

[@Skank_and_Gutterboy]

That problem is not as simple as a person might think. You can't cram a full box of merchandise into a gap where there's only remaining room for 1/4 of a box (unless you're FedEx and don't care about your customers).

[@highpath4776]

yes, thats a problem, when numbers can be easily brute forced its straightforward, working out the best "algorithm" for the general cases to solve for any variables is more tricky. I think also the question is what Jeff Bezos worked out for amazon box fixing for not only optimum sizing also for optimum mail costs ( ready a Kellys Post Office Directory for the time and look at the Royal Mail schedule of rates and its bewildering (still is for business bulk rates )

[@Skank_and_Gutterboy]

@@highpath4776 No doubt. I remember in an undergrad engineering class discussing how many ping-pong balls you can put into a cylindrical tank, packing factors, etc. It can get complicated quick.

[@JavedAlam24]

Well if the degree of the polynomial is any greater than two, you can't use the quadratic formula anymore! You basically have to make educated guesses and then check.

[@serbanudrea9429]

When I was in school the area model was called distributivity. Maybe the area model can help in understanding this concept, but I think that the concept by itself is important and it shouldn't be forgotten. Besides, to the best of my knowledge, if a question is formulated like no. 3, it is expected that the examinee first performs the multiplication and then the division. Point being, maybe, to make the question harder to answer.

[@JGlaister]

I clicked on this video to see how to solve the problem on the thumbnail. I sometimes try to figure them out even though I flunked algebra in 1972. I picked up a little when my son was in school. Now that my granddaughter comes to me for help, I've been a bit baffled by those boxes she keeps drawing. This helped.

[@polymathematic]

so glad to hear it was helpful!

[@samanthahellawell8141]

And the rest of you much more advanced and experienced maths’ people can say how easy it is but I am just totally thrilled that with a bit of working through myself as he did I can understand a MIT entrance exam question!

[@magisterparsons]

I appreciate your clear and cogent explanation of the pedagogy. Great video!

[@polymathematic]

thanks for watching!

[@dougnettleton5326]

I love the word pedagogy. I have an acquaintance who uses it incessantly.

[@Jack_Callcott_AU]

I'm surprised how simple many of those questions are. Question 6 and 7 are just high school level. Question 6 is a linear equation in X. Question 7 is simple set of simultaneous equations. Question 2 looks very easy also. I thought it was supposed to be difficult to get into Harvard; maybe in 1869 it was easy.

[@codescience1061]

Sorry to correct you, but this is a MIT admission test. I agree with the rest of your comment.

[@Jack_Callcott_AU]

@@codescience1061 How embarrassing. I forgot to check. BTW I don't mean to be disparaging about Harvard or MIT. America has great universities, and these 2 are at the top of the list. Thanks for the correction. 👍

[@camelopardalis84]

Time might have been an important factor. "Can you do this? And quickly so?" They didn't have calculators back then.

[@Jack_Callcott_AU]

@@camelopardalis84 Good point. 👍

[@jconcah]

High school level math would seem perfectly placed here for an entrance exam, which is sat by high schoolers? I'd guess this isn't only for math students either, but for all students including those studying literature or history

[@georget8008]

you should know that in the mid 19th century, the precollege schooling lasted 9 years (instead of 12 today). So, this question is absolutely normal for the level of the students those days

[@MrSummitville]

Since modern day students have 12 years, instead of only 9 years, they should know this. Oh, but wait ... Every student in the Chicago Public Schools *failed* the State Math exam.

[@clemenswinklmaier5824]

Vieta's Formular is such a great tool! With CAS at hand it's just one of the cool things you leave out to make time for other new stuff. I was hoping you'd point out that if the product is divisible by (a+b) at least one factor has to be divisible by (a+b), thus reversing the order of division and multiplication will yield "more easy" computations

[@HenrikMyrhaug]

It frustrates me as a university student, whenever I see people say stuff like "if it aint broke, don't fix it" or "math doesn't change". Of course, the goal of math is to have consistent results, but there are infinite different methods to obtaining those results, and some methods are eaiser to learn, compute the result faster, or give a better understanding. To someone who has never seen a new method, your child showing it to you might make it seem difficult. But that's because children don't make good teachers, and adults who don't teach math aren't usually good at figuring out what they are looking at when they see an unfamiliar method. Honestly thank god we aren't learning math like we used to in the past! I was taught so many inefficient, hard to memorize, and poorly explained methods, that I had to invent my own methods to be able to solve many problems, and to this day I still occasionally discover better methods of doing maths. I'll be glad if my kids learn math in a different way from what I do, if that helps them understand easier!

[@travcollier]

A lot of those 'weird' techniques aren't about being easier. They are about trying to get the kids to better understand what the maths is actually about. So you start off with stuff like very cumbersome tally mark methods, but you can just see exactly what is going on. Then you move onto more efficient ways. This one is both that and the other thing parents seem to get freaked out about "new math"... It is preparing them for a more complex thing those parents don't even remember or maybe never learned. Multiplying polynomials using the area method isn't really any better than the standard way (distributive method isn't exactly difficult), but when you are factoring and solving for roots it gets really useful. "Completing the square" Honestly, my big complaint with maths teaching (grade school level) is that it isn't very well motivated by real world problems. Really should be taught alongside physics and maybe some basic engineering. That would be good for physics too, since it is normally taught in a way which avoids anything but the simplest calculations.

[@Loots1]

as soon as i saw the question my intuition was the 2nd one was factorable, ive never heard of this box technique before though, pretty cool

[@antonemberbroque4452]

I remember learning both the standard algorithm and area method, but I had no idea the area method could be applied to polynomials. It reminds me of doing trig sub in calc: the way that works for me is to draw out the triangle labeling each side with come combination of x, a and sqrt.

[@samanthahellawell8141]

Well I am a beginner who can just about understand this and will be writing it all down and going through it. I think it’s so cool and thank you for showing me!

[@aroundandround]

Thumbnail was misleading and the reason I clicked after staring at it for a few seconds.

[@polymathematic]

In what sense?

[@aroundandround]

@@polymathematic It only has the part of the numerator indivisible by (a+b).

[@JavedAlam24]

Isn't that the whole purpose of thumbnails? To pique your interest? I wouldn't call it misleading

[@Dominic_Berry]

It's even easier than that. If you know one of the factors needs to be a+b, then the only other factor that can give the a^2 and -3b^2 terms is a-3b. Then you just need to check that multiplying gives the correct -2ab term.

[@TimRobertsen]

That was wild! I subscribed

[@polymathematic]

thank you!

[@RadicalCaveman]

I always multiply using trapezoids.

[@Steven-v6l]

*OR* you could just do a polynomial division as we learned back in the 1970's; before "area multiplication" was imagined. (a^2 - 2ab - 3b^2) / (a+b ) = (a - 3b) Hence: (3a^2 + ab - b^2) * (a - 3b) You don't need "area multiplication" here either. But I guess some differently abled math major might identify with "area multiplication".

[@polymathematic]

It's true that there are other methods to solve that kind of problem.

[@midnighttrain-jz2my]

a^2 -2ab -3b^2 = a^2 -3ab+ab -3b^2 = (a^2-3ab) + (ab-3b^2) = a(a-3b) + b(a-3b) = (a+b)(a-3b)

[@Rend-gi3sq]

I've never seen these before. It's impressive thanks for sharing. 👍

[@ralphhardie7492]

🎉🎉🎉 Enjoyed that 🎉🎉🎉 Good teaching makes maths fun 😊

[@polymathematic]

thank you very much!

[@NeoAF10]

This is so cool!

[@user-cq8xh5vl1p]

I disagree. The instruction was to multiply first and then divide.So he got it wrong.

[@polymathematic]

nah, I got it right.

[@tassiedevil2200]

@@polymathematic Arguably you got the right answer to your own question - on your YT title slide which isn't quite the same - and yes I already saw elsewhere in the Comments the semantics about it not saying "and then" divide.

[@JavedAlam24]

@user-cq8xh5vl1p you'd make a nasty teacher

[@scarletevans4474]

I actually made a silly mistake here, as I first factored (a+b) out of both expressions (receiving non-divisible a^2 in one of them), then later down the line, while multiplying -4b by a^2/(a+b) i forgot the square in a^2 and ended up with "-4a^2b-4ab" instead of "-8a^2b" in the final result 😀😀 I really like 5th problem! Set x=a+b, y=a-b, simplify to: "(x^2+y^2)/(x^2-y^2)", then to: "1+(2y^2)/{(x+y)(x-y)}", then as x+y=2a and x-y=2b you quickly get: 1 +a/(2b) +b/(2a) -(2ab)/(2ab) = (1/2)×(a/b+b/a) .

[@polymathematic]

very nice substitution for that fifth problem!

[@cosmosgato]

Amazing video!

[@polymathematic]

thank you! glad you enjoyed it :)

[@JohnJones-w7i]

That was fun. Thanks.

[@MsAcpaul]

That area method is a great way to teach students to expand brackets. I still find teachers who hate it though. My teaching rounds I used areas to teach completing the square and my supervisors comment afterwards was I was babying the students too much with that method. My lecturer said she would have done the same lesson plan. All maths stems from geometry.... this method was probably one of the first methods when algebra was first thought of.

[@rickm409]

Nice! never thought of it that way!!!

[@spottymaldoon]

Apart from computer aptitude, it's hard to think of stuff today's math students can do that math students of the same age back then couldn't.

[@stevenrn6640]

The box method: lets makes something easy confusing as F. Underwhelming.

[@polymathematic]

what confused you about it?

[@stevenrn6640]

​@@polymathematic I am thinking from the standpoint of (as you said) a 2nd grader. The box method (and I have helped kids that have had their math skill destroyed by teaching this "method"), while interesting, makes a simple problem (such as 13X14) difficult as kids divide smaller and smaller section into little boxes and count. There is a place for banked learning and simple rote memory and this is a perfect example. I am seeing engineers coming into the work place that can't do basic math in their head and count on their fingers like a 2nd or 3rd grader. It is sad beyond belief how educators have failed 1-2 generations of kids. It does have a side benefit of teaching how to calculate the area of a square but it overly complicates a simple procedure into a complex operation.

[@AlexD02721]

A bit click-baity. The thumbnail didn’t mention multiplying 2 quadratics. Tried to solve & couldn’t without that knowing about the 2nd quadratic.

[@polymathematic]

Seriously?

[@Theo0x89]

That's the only reason I clicked on the video. I couldn't solve the thumbnail with the factorization method explained in video. Watched the whole video, didn't really learn anything knew. Well clickbaited.

[@jlmassir]

​@@Theo0x89 Me too.

[@polymathematic]

@@Theo0x89 i think it's a bit silly to expect the exact problem statement to be in the thumbnail. but i'm sorry you feel mislead. hope you'll stick around for other vids!

[@julianbruns7459]

​@@polymathematicthen why even write a problem in the thumbnail if its "silly" to expect the correct one?

[@eamonnwalker4512]

4:23 "We didn't split the 13 into 10+3." Yeah, it's a good thing we all memorized those 13 times tables back in the fifth grade.

[@albertogarcia4177]

Let g(a)=a²-2ab-3b², we can see g(a) as a polynomial in "the variable a", notice g(-b)=0, so by the Ruffini rule g(a) is divisible by a+b. Using long division we find the quotient a-3b and rest 0. Now is easy proceed the result.

[@wasimohammad4280]

This question is to check the multiplication and division ability of students than to get the answer.

[@JavedAlam24]

How would you know? Did you write it?

[@2ndavenuesw481]

If multiplication were taught left to right, as Menninger recommends in Calculator's Cunning, students would struggle less with multiplying polynomials Foil and multiplication of polynomials generally would be more "automatic."

[@MichaelFJ1969]

I instead substituted b = x-a (and hence x = a+b) and then simplified the second factor to get: a^2 - 2ab - 3b^2 = 4ax - 3x^2. From here it's obvious that there is a common factor of x. So we get: x*(4a-3x). Then substitute back x = a+b gives: (a+b)*(a-3b). This is purely mechanical and requires no guesswork.

[@tassiedevil2200]

While I can see this box method is useful for book-keeping all the terms in expanding out brackets (the distributive law at work) I do find the "area" motivation somewhat strange when stretched beyond grade school level. Really, it relies for its two dimensionality on the application to treating the product of just two sums at a time. For natural numbers it seems OK as arranging "things" - like counting 2D arrays of beads or blocks, although why wouldn't the product of three quantities deserve a volume interpretation, but even going to products of real positive numbers seems to attribute dimension to the quantities, like combining lengths to get areas - which seems unwieldy once variables are involved.

[@JeffSpieker]

1) if e=8, then e − [(e + 1)+ 2] + (e − ³√(e)) * √(e − 4) = 8 - 5 + 12 = 15 2) simplify 3a − [b + (2a − b) − (a − b)] = 3a-b-a = 2a-b 3) Multiply (3a² + ab − b²) by (a² − 2ab − 3b²), and divide the product by (a + b). = (3a²+ab-b²) (a-3b) (a+b) / (a+b) = 3a³-9a²b+a²b-3ab²-ab²+3b³ = 3a³-8a²b-4ab²+3b³ 4) Reduce the following fraction to its lowest terms: (x⁶ + a²x³y) / (x⁶ − a⁴y²) . = (x⁶ + x³ a² y) / (x⁶ - a⁴ y2) = x³ (x³ + a² y ) / (x³ +a²y)(x³-a²y) = x³ / (x³ - a²y) = x³ / (x³ - a²y) 5) Simplify ((a + b)/(a − b) + (a − b)/(a + b)) ÷ ((a + b)/(a − b) − (a − b)/(a + b)) =((a+b)² + (a-b)²)/(a-b)(a+b) ÷ ((a + b)/(a − b) − (a − b)/(a + b)) =a²+2ab+b²+a²-2ab+b² / (a-b)(a+b) ÷ ((a + b)/(a − b) − (a − b)/(a + b)) = 2a²+2b² / (a-b)(a+b) ÷ ((a + b)/(a − b) − (a − b)/(a + b)) = 2a²+2b² / (a-b)(a+b) ÷ ((a+b)² - (a-b)²)/(a-b)(a+b) = 2a²+2b² / (a-b)(a+b) ÷ a²+2ab+b² - a²+2ab-b² / (a-b)(a+b) =2a²+2b² / 4ab =(a²+b²)/ 2ab 6) (3x − 4)/2 − (6x − 5)/8 = (3x − 1)/16 24x-32 / 16 - 12x-10/16 = 3x-1/16 24x-32 - 12x + 10 = 3x -1 9x = 21 x = 7/3 = 2.3333... 7) 7x-5y=24, 4x-3y=11 x = (24-5y)/7 (96-20y)/7 - 3y = 11 96-20y-21y=77 y = 19, x = 17

[@davidwright5719]

I didn’t know it was possible to turn “try factorizing first” into a 15m video.

[@polymathematic]

Well, now you do

[@jimiwills]

The question literally asks you to divide the *product* by a+b, so wouldn't you get in trouble for dividing before finding the product?

[@polymathematic]

I don’t see why. One of the beautiful things about division is that whether you divide a product before you multiply or after, it’s the same either way.

[@victorzurkowski2388]

If a question ask to multiply 2 polynomias, and then divide the result by another, the first part of the question has not been answered here. Had I marked this answer, I would give partial credit for it.

[@polymathematic]

@@victorzurkowski2388 well, i guess i'm lucky you were not marking it then! personally, i would certainly give full credit for a student who cleverly saved themselves some trouble and still came up with the correct result.

[@ArchmageOfTheCahling]

​@@victorzurkowski2388 the question did not use the word "then".

[@Jamesmacdermid21000]

@@victorzurkowski2388 writing the two polynomials in (---)(---) form IS the product. You haven't expanded it yet, but it literally is the product, in factored form. So canceling out the factor is following it word for word, you took the product (in factored form) first, and divided. If it said "Find the product in expanded, simplified form then divide by a+b" then you may have an argument

[@nasalimbu3078]

(a^2-b^2)/a+b (A+b) (A-b)/ A+b (A-b)^0 = 1

[@jmguevarajordan]

That question has a 'bad' intention behind it. It is quite clear that (a+b) is a factor of the second factor, by the residue theorem, so you can factorize or divide it by a+b and get the (a-3b) and (3a^2+ab-b^2)(a-3b) is a correct answer. However, the problem says 'multiply', so if You do not perform the multiplication in the above expression then they take points away from your correct answer. Terrible!.

[@WaitingforGodel]

Right, someone taking the test has to guess what the teacher wants and how they would grade the answer

[@Brigadier_Beau]

From that long ago, they probably expected polynomial long-division. I learned how to do that ages ago, but I would have to find a tutorial to remember at all. I prefer more modern techniques.

[@jmguevarajordan]

@Brigadier_Beau yes, You are right. The guy who wrote that question was thinking that the students would multiply first the two factors, each one with three terms, followed by the división.👍

[@pedroteran5885]

If you don't do what you're being asked to do, how is your answer 'correct'?

[@jmguevarajordan]

@pedroteran5885 yes it is true your statement and that is exactly the 'bad' intention in this question. In fact, You can divide first and multiply after that, which is better and shorter, and there is a chance that You don't get full crédit in your answer.

[@williamprice1844]

Well I know a pastor that went to Dallas theological seminary that they made answer his Greek Grammer in latin,so we really don't require that level of efficiency anymore.

[@ralphmelvin1046]

Seems really hard to me I went to high school back in the seventies. Back then algebra was not even a mandatory class Algebra 1 was not a mandatory class to take.. it's hard for me to learn now at my age 61 but I'm trying

[@RealQinnMalloryu4]

{3a^2+3a^2 ➖}+{ab+ab ➖ } ➖ b^2={6a^4+ab^2} ➖ b^2 =6ab^6 ➖(b^2)^2={6ab^6 ➖ b^4}=6ab^2 2a^2 +ab ➖ b^2 ÷a^2+2ab ➖ 3b^2=2ab ➖ 3b

[@boysen01]

So that is what "completing the square" is? If so, very good explanation. Thank you

[@KatyaOrlova-jp1gy]

That's actually an amazing way to visualize math. If I was taught it this way I would definitely understand it better.

[@dan-florinchereches4892]

Well i would just try to consider expressions as polynomial in a an check if they divide by a+b this means -b is a root. We easily see the second polinomial is b^2+2b^2-3b^3=0 so we can divide it ... Factoring is too hairy for me in this problem Still i am wondering how muhc they would have to study in university to catch up to calculus and everything they needed

[@mathcanbeeasy]

If I formulate a statement "multiply six by three and divide the result by two", I think it is quite clear that an objective of the exercise is to see if the student knows both the multiplication and division. The same thing is here. If the statement says to multiply that two expressions, I don't think it's right to skip that requirement. Ok, after he multiplies them, he can start from the beginning for decomposition and simplification, but I think, in your solution, a requirement is skipped. Anyway, the exercise can be solved very easily by multiplying the expressions, writing the result as an expression in a and do the division by a+b exactly like dividing polynomials. 3a^3-8a^2b-4ab^2+3b^2 is obtained.

[@polymathematic]

i doubt it, but it's certainly possible! one reason that standardized tests have probably moved toward multiple choice questions where that kind of ambiguity isn't possible.

[@JavedAlam24]

That's not true at all. There are often situations occuring in mathematics questions where you can use your knowledge to finish the problem much faster than you otherwise could have. This is often intentional, and I feel that you may not realise that the order of operations when multiplying and dividing does not affect the result. There is no 'skipping', rather the operations are done in a more optimal order.

[@mathcanbeeasy]

@@JavedAlam24 Read again what I wrote. It is very true. If the statement clearly says "...divide the result ..." you have to get that result first. If the exercise stated by me, above, is scored with 2p, then 1p is for the multiplication result and 1p is for the final result. We are not talking about the grid test. A student must know clearly how to read a statement, first of all that is what it is about. The statement is not "calculate 6*3:2". If that was the case, he could do whatever calculations he wanted. If the statement is clear "first do the multiplication and then divide the RESULT by two", be sure that, as a teacher, I am interested if he knows how to do both multiplication and division and if he knows how to read a statement correctly.

[@ericslavich4297]

This was my thought, too; the question calls for two separate calculations with two distinct results. However, I wondered if conventions were different at the time such that skipping the intermediate step was expected. If it were me, I would show the multiplication and then separately show the division and comment that if only the final quotient was required the initial multiplication would have been unnecessary.

[@mathcanbeeasy]

@@ericslavich4297 In that period, the difficulty of the subjects was very low. The multiplication of some expressions in two unknowns was actually an exam subject. The more difficult item would have been the division, which can be done as the division of two simple polynomials in the unknown a and b as a parameter. For example, what is: (6a^3-2b^3+2a^3b+ab^3-7a^2b+5ab^2-a^2b^2)/(3a+ab-2b)?! This is pretty hard, and no chance for sum and product. But if we write 2(3+b)a^3-b(7+b)a^2+b^2(b+5)a-2b^2 and (3+b)a-2b as two polynomials in a and we apply the division algorithm we can easily get 2a^2-ab+b^2.

[@geoninja8971]

Well, I'm not getting into MIT any time soon! I do recommend a grid for larger polynomial products to my high school son....

[@umutgokce3185]

I did (a²-2ab-3b²) = (a²-2ab+(b²-b²)-3b²) = (a²-2ab+b²) - (4b²) = (a-b)²-(2b)² = (a-b-2b)(a-b+2b) = (a-3b)(a+b)

[@picrust314]

When you used the areamethod backwards, should you not in THIS case start with a rectangle with side a+b ? Since that is what you want

[@bethhentges]

I first saw the area model in 1984. It’s certainly not “new.”

[@polymathematic]

It may not be new, but it’s taken on particular salience in the last 15 years, stemming basically from its implementation as part of a number of common core compliant curricula.

[@windBlasta]

@@polymathematic Was taught the box method in 1982 - 9th grade - then taught the X multiplication factorisation method - and finally the ever reliable -b+/- etc.... They are all reliable and all correct - different people learn in different ways - some are visual - some not - anyway - great demonstration of the power of this method. - Keep up the good work

[@aosidh]

I didn't learn long division of polynomials until 400-level algebra 🙈

[@aisolutionsindia7138]

simplification wasnt asked, wonder if someone writing the expression as-is would have been awarded a full score

[@imagseer]

First time I've seen that, and I can see how conceptually useful it might be. I often wonder if Math is easier for people who speak French because they already see the numbers broken up? Cent quatre-vingt-dix-huit = (198)

[@potentialofprotonis3]

why are all of these questions so easy?

[@polymathematic]

It was right after the civil war, so there wasn’t a huge applicant base.

[@potentialofprotonis3]

@@polymathematic ohh i see, thats actually such an interesting aftermath effect of the war that no one would realize

[@jonlittle5032]

uh ... Distribution, any one? Works for numbers and polynomials. Same thing, just three different ways of looking at it.

[@miro.s]

Clear when all areas are positive. But you didn't show how to generalize it for negative areas.

[@polymathematic]

the factoring process or the area model for numerical multiplication?

[@purplerpenguin]

It reveals nothing about "math today", just that, for whatever reason, the exam was trivially easy back then.

[@polymathematic]

huh. I wonder if the "for whatever reason" has any content that might reasonably be explored?

[@MrSummitville]

@purplepeguin - Many students would fail this test. Recently, *EVERY* student, in every Chicago Public School, failed the State Math Exam. 2 years ago, *EVERY* student, in every Baltimore Public School failed the State Math Exam. That is the reality of today's educational system.

[@2ndavenuesw481]

What would John von Neumann say about contemporary students and how they are taught? Would he say it is better or worse than before?

[@allanwrobel6607]

Get a Tardis, go back in time to 1970, and be my math teacher.

[@polymathematic]

would love to :)

[@quigonkenny]

(3a²+ab-b²)(a²-2ab-3b²)/(a+b) (3a²+ab-b²)(a-3b)(a+b)/(a+b) (3a²+ab-b²)(a-3b) 3a³-9a²b+a²b-3ab²-b²a+3b³ 3a³ - 8a²b - 4ab² + 3b³

[@senbonzakurakageyoshi662]

Bro I was a fool XD I try to multiply then tried to divide it by a+b XD

[@attica7980]

What is modern about the area model for multiplication? The area model is how the ancient Greeks did algebra.

[@pwmiles56]

I believe their preferred approach to number was by ratio. Euclid gives both an area proof and a ratio proof of the Pythagorean theorem, but the area proof needs more axioms (to define area itself). Generally areas, to them, meant actual physical areas, rather than algebraic constructs like x^2, xy etc. Only the Arab mathematicians (most were actually Persian), following in the footsteps of the Indians, decisively detached number from geometry. Just my takes!

[@joergbertram4981]

Nice 😊

[@txikitofandango]

I multiplied the polynumber [3 1 1] by [1 -2 -3] and then did a synthetic division by -1. I didn't want to wait around and see if (a+b) divided into either factor. But I still appreciate the video

[@RAFAELSILVA-by6dy]

You can see that (a + b) is a factor of the second expression by checking you get zero by setting b = -a.

[@nihal_farooq_m]

We do this, but without boxes😅

[@prajwalchitrakar-i7u]

But boxes are the way it was first done

[@RB1-n5m]

I really enjoyed this video. But I think I have spotted an error. At the very end, the solution was: 3a^3 * - 8a^2b - 4ab^2 + 3b^3 The 3rd term in this , - 4ab^2 should be +ve It is, after all the the product of: -3ab^2 * -ab^2 is it not? So, the correct solution is: 3a^3 * - 8a^2b + 4ab^2 + 3b^3 Am I right?

[@BruceKoerner]

PROBLEM: The areas model is a great way to UNDERSTAND multiplication, it is a lousy way to DO it.

[@polymathematic]

i agree i wouldn't use it for multiplying numbers, but i love it for multiplying polynomials (and factoring them).

[@tassiedevil2200]

@@polymathematic A matter of taste I suppose, although I find the "application" for factoring of very dubious efficacy. For factoring here it is just the old quadratic guess and check - indeed you'd have a much longer video if you had "happened" to try it on the first quadratic. For polynomial multiplication perhaps it's a way to ensure that one does not lose any terms in applying the distributive law, but hard to see that it is a terrific insight, and it seems to implicitly assume some spurious "dimensionality" to the variables - here everything had uniform degree.

[@DavidLopez-gs1fb]

Tell us the truth: you saved that exam back from your years as a bachelor, didn't you? You presented yourself that exam back in 1869, and you've been thinking about it since then... Confess!!!!

[@polymathematic]

lol, I’m old no doubt, but not quite that old :)

[@spacelemur7955]

As a 70 y.o., I can only fondly reminisce over how young the uploader looks.

[@aqueous3051]

All of these are early high school problems lol

[@KrytenKoro]

3a4+a3b-a2b2-6a3b-2a2b2+2ab3-9a2b2-3ab3+3b4 3a4-5a3b-12a2b2-ab3+3b4 3a3-8a2b-4ab2+3b3

[@nadonadia2521]

If modern techniques are so well, they would teach you that (a²-2ab-3b²) is the beginning square of difference of two numbers when we see -2ab we have to think to (a-b)² so (a²-2ab-3b²) =(a²-2ab+b²-b²-3b²) we add b² in order to find (x-y)² term. (a²-2ab-3b²) =(a²-2ab+b²-b²-3b²), (a²-2ab-3b²) =(a²-2ab+b²)-4b² (a²-2ab-3b²) =(a-b)²-4b², and now another identity X²-Y²=(X-Y)(X+Y) (a²-2ab-3b²) =((a-b)-2b)((a-b)+2b) (a²-2ab-3b²) =(a-3b)(a+b) and these techniques will helps you for university courses, not drawing rectangles.

[@polymathematic]

this is a super silly objection.

[@marzipanhoplite17]

I completely agree with you and add that your technique is much faster than drawing rectangles

[@alvarofernandes8225]

The explanation is good, but I understand that only after otaining the product the division by (a+b) can be done.Therefore, If I was the teacher, I would say it is a clever way to calculate but not what I asked.

[@taresy6789pp]

there was no AI and there IQ higher than 2025 lol🤣

[@AlexandreA-w5c]

Something isn't right, what if a + b = 0 ? You can't simply cancel (a + b) out!

[@AshleyPOliver]

Missed the problem if a=-b though, so you're not coming iitno my maths class

[@simmo5071]

Actually that technique goes back to Babylonian times, watch PoShen Lo

[@Benj-p5w]

I think you try to convince yourself of what you think is your good level in maths. How do you knowbwhat the examiners wanted to prove at the time?

[@nihal_farooq_m]

Nice questions, Completed all 7 of them.

[@KookyPiranha]

why waste ur time on easy ahh problems like these

[@nihal_farooq_m]

@KookyPiranha I don't know, I just feel to do it

[@fredpagniello3267]

Your explanation could have been done in half the time.

[@afterthesmash]

He did. I ran the video at 2x. Others might prefer less pace.

[@fredpagniello3267]

@afterthesmash The tutorial was a Rube Goldberg: lots of steps to explain a fairly simple procedure. So was the approach to the problem; use three steps where just one is sufficient.

[@user-wo8gt1xg5n]

Why not this one: aa-2ab-3bb : a+b = a-3b aa+ab ---------- = -3ab-3bb -3ab-3bb -------------- 0 You can also multiply those 2 poly and divide by a+b after. No geometry needed. Area model works well for low numbers, can you guess 276X^2-272x-128 = (23X+8)(12x-16) ? And yes thumbnail is a click-bait: 3aa+ab-bb : a+b = 3a-2b 3aa+3ab -------------- = -2ab-bb -2ab-3bb ---------------- = 2bb so 3a^2+ab-b^2 = (3a-2b)(a+b) + 2b^2 It certainly cannot be divided

[@polymathematic]

the area model isn't about guessing. if a quadratic is factorable, the area model (or box method as it's more typically called for polynomials) will factor it. in your example, you'd first divide everything by 4 since there's a GCF between terms, and that would give you 69x² - 68x - 32. then you're looking for two numbers which have a product of -2208 (that is, the product matches 69 × -32) and a sum of -68. you do not *guess* at this point, rather you list out the factors of -2208 and see which add up to -272. in this case that ends up being 24 and -92. then you place 24x and -92x in the boxes and factor, giving you 4(23x+8)(3x-4). you'll notice this is in fact precisely the same process one would use for other factoring methods. split the middle term, undoing distribution or FOIL, all these are the same. the box method is just a way to organize the process.

[@user-wo8gt1xg5n]

@@polymathematic The easy way for my example is to solve the equation and write (x-x1)(x-x2) I'm sorry I "can't see" (not guess) when 2 big numbers are related especially if we talk about real numbers not integers. Few years ago I was forced to cleanup my son brain because his math teacher told him 4 versions how to subtract 2 numbers and he was so confused to the point where he was unable to subtract. Teaching a lot of methods is good for skilled kids but also a good way to push others away.

[@polymathematic]

@@user-wo8gt1xg5n sorry, i'm not sure what you mean by "solve the equation" and then re-write the factors. given your example, if you set that quadratic equal to 0 and then solved with the quadratic formula, you would get solutions at -8/23 and 4/3. using your notation, that would give you factors of (x-(-8/3)) and (x-4/3). you can see this is not the same as the factored form of the expression, though it can be manipulated to match. more than that though, it looks like we're talking about two different skills. factoring and solving are obviously related, but they are not the same. in the case of the problem in this video, for example, there's no solving at all. and while you can use the quadratic formula to get "solutions" and then fit those back into factors, this is a bit like baking a cake just to see what ingredients you used.

[@AnwarAdnan-e5w]

In my country a good 15 years old student can solve all of this equations.

[@MrSummitville]

Unfortunately, that country is not the USA.

[@sebastianparamera2424]

It's extremely easy.

[@polymathematic]

did you expect something else from an algebra question written 150 years ago?

[@sebastianparamera2424]

@polymathematic I believed you thought otherwise, as per your description: "Demanding content and unique approach to problem-solving".

[@maxm33]

It seems like an algebra exam, but first he talks about arithmetic. Smth at the low middle school level 😳 I think everyone who is preparing to enter MIT knows the basic formulas and what they correspond to in reality 😂 Waste of time

[@billkillernic]

lol all questions seem so elementary in Greece the math exam to get into a university is a lot harder like all 7 questions maybe could exist in the first section where the "freebies" are (basic theory questions for the people that aren't good at math to get some points instead of 0 lol ) like there are no integrals/ problem solving (as in describe some sort of geometry and ask you to find or solve something or similar) no determinants the above is like junior high stuff or at best freshman high school stuff (so like 4 grade after elementary school)

[@chrislloyd5415]

Set rhs=x and solve. Boring!

[@fatalberti]

meh. very simple. we are dumb and spoiled