He didn't explain it like the asker was 5 though. He should've started by teaching him the alphabet, then taught him basic arithmetic, then probably ended the video because a five-year-old has no chance of learning algebra.
I don't believe I have learned these polynomials, although I have seen them in some maths books and tried them for fun many yrs after graduating. That's probably why I prefer the long division rather than the synthetic version.
I always replace the division with multiplication (I guess it's more or less another way of doing the synthetic division...). You basically take the (x - k) and then multiply with something to get back to the original polynomial. I find this to be easy and quick. To give a step-by-step of the example above (in real life you can of course do it in a matter of seconds): (x - 3)(x^2 - because x(x^2) gets you the x^3 term. The result is: x^3 - 3x^2 - but you don't want -3x^2, you want -x^2, so you have to add 2x^2 by adding 2x to the q(x): (x - 3)(x^2 + 2x - this gives you (x^3 - 3x^2) + (2x^2 - 6x) = x^3 - x^2 -6x. But you want -12x so you must subtract another 6x by subtracting 6 in q(x): (x - 3)(x^2 + 2x - 6) - this gives you (x^3 - 3x^2) + (2x^2 - 6x) + (-6x + 18) = x^3 - x^2 - 12x + 18. Now you can just subtract remainder 5 to get 13 as the constant.
Or said otherwise, we are looking for a, b and r such that (x-3)(x²+ax+b) + r = x³ - x² - 12x + 13 After developing the left side: x³+ (a-3)x² + (b-3a)x - 3b + r = x³ - x² - 12x + 13 Matching the terms of equal powers gives a system of 3 equations: [1] a-3 = -1 [2] b-3a = -12 [3] -3b + r = 13 The resolution is pretty simple. Using [1] we get a = -1+3 = 2 then using [2] b = -12+3a = -12+6 = -6 and finally using [3] r = 13 + 3b = 13 - 18 = -5 So the final answer is f(x) = (x-3)(x²+2x-6) - 5
Funny thing, I had a bit of a blindspot for long division at school, UNTIL I did polynomial (we called it algebraic) long division, and then it became automatic. Something about seeing just numbers hid where they were coming from, but algebraic expressions look like what they are if that makes sense.
Yep. The connection becomes really obvious in hindsight when you realise that our number system is literally just writing the coefficients of a polynomial where x=10 :)
I would start the slow way. First solve for f(x)=(x-3)x^2 +r_0(x) We get r_0(x) is 2x^2 -12x +13 We then do r_0(x)= (x-3)2x - r_1(x) r_1 is -6x+13 Then r_1 = (x-3)(-6) + r_2, r_2=-5 We then unroll it. f(x)=(x-3)x^2 + r_0 = (x-3)x^2 + (x-3) 2x + r_1 = (x-3)x^2 + (x-3) 2x + (x-3) (-6) + (-5) = (x-3)(x^2+2x-6) -5 The tricks and shorthand don't give the person any chance to understand WHAT is going on. And without knowing what is going on, how are they supposed to know if the trick is nonsense? Or if there is a mistake in it? Teaching tricks without foundation is obsolete in this era of caclulators. Teach the foundation, not how to solve some algorithm that we'd hire peasants to do in the 18th century.
I will recommend just do the long devision instead of this method in this video. Long devision works perfectly, and only slower about 20 seconds or less. The benefit is you don’t memory other method to do devision. I said that because I always misplace the -3 instead of 3 on divisor part or confusing other part with long devision while doing exam. Anyways, love your video, keep doing more :)
This. ^ Synthetic division has limited applicability, and requires you to memorize a technique / remember its idiosyncrasies. (More mental load.) Long division works in all cases (even when the divisor isn't so simple), and only requires you to know how division works. Every step is logical, and it's not really as time-consuming as some teachers make it out to be.
ngl, I'm a mechanical engineer that eats, breathes, and shits 2nd-order ODEs and I've never seen an algebra problem worded like this in my entire life. I had no clue what they were even asking me to do.
Hello u done it good - no question - but it must be a math genius kid who will get this with five... No sorry none "normal" educated who is five years old will understand this. Even not a 40 year old one, who never have that at school or simply forget ("the personal math and testing hell has ended long ago, so let this be quickly suppressed" is not really seldom experience and pain from the past ) will understand this - sorry. But is not your fault, it simply impossible to do it in 7 minutes, and even in two hours.
x^3 - x^2 - 12x + 13 step 1: write (x-3) gap (x-3) gap (x-3) step 2: write coefficient for first as: x^2(x-3) gap (x-3) gap (x-3) step 3: write second coefficient as: x^2(x-3) + 2x(x-3) gap (x-3) because from first and second term -3x^2 + 2x^2 = -x^2 (second term of original equation) step 4: write third coefficient as: x^2(x-3) + 2x(x-3) -6(x-3) because from second and third term -6x -6x = -12x (third term of original equation) step 5: we write : x^2(x-3) + 2x(x-3) -6(x-3) - 5 -5 at the end because we have +18 from the third term but we have 13 in original equation so +18-5 = +13. finally: take (x-3) common (x-3)(x^2 + 2x - 6) - 5 -->ANS
I ain't never seen a question like that, and I know why, it used q and r notation which is from English so obviously we would've skipped that, it also feels kinda useless
You should have explained the case for another way around as well. Like in the form of (x2 + 2x -6) k(x) + r Or, even in the form of (x + x1 ) (x + x2 ) m(x) + r etc. I think, the nightmare of most highschoolers was, they had been explained to find the solution of one case (the teacher had explained to them like to a baby). But then the question of the exam/test they had been given was pretty much different. Like, there is no such general solution for the same topic of various problems.
That's fairly slick, something I don't ever recall doing in math ever, and I went further than most any non-math major. That said, I'm still lost of the purpose of doing it, from the explanation given at the end it seems like he's saying order of operations is hard, so do polynomial division and it makes it easy.
Polynomial division could be useful to simplify a high order polynomial to a lower order one so you can apply a formula to find all the zeros, given you know some of the zeros already. :D
@@zombieherobrian5384 That make's sense, I guess this question was problem set up to show that you can find zeros and the last bit was just to get the student to do a check to verify they did exactly that.
Just a reminder to everyone, synthetic division is actually possible with non-integer rational numbers also. You have to divide all coefficients of the "dividend" by the denominator, first IIRC.
This is included in IGCSE Additional Mathematics (4037) course [Factors of polynomials] I use 3rd line rule as my teacher doesn't like synthetic (which i fcking hate)
Hey i was wondering why this video has so few views, so i checked your videos tab to compare and realized its not there? Is this unlisted or maybe a youtube bug? if it's unintentional you shoukd probably either try to fix or reupload it.
5:50 Finding f(3) this way seems simpler than converting to the form (x-k)q(x)+r and then solving. Sure it is simpler to solve once you already have converted to that form, but the conversion process is far more tedious than just plugging in 3 to the f(x) form. I can do the f(x) form in my head, but the (x-k)q(x)+r form requires pencil and paper. I simply don't understand how that is easier. Also, it would be helpful to explain why you "go ahead and put a 1 down here" at 3:19 as those unfamiliar with this process may not understand why that is a given. Is it because the coefficient of x in (x-3) is 1, or is it because the x^3 coefficient in the original equation is 1? I assume it is the former, but being unfamiliar with this process until today, I can not be certain it isn't the latter? I am pretty sure a 5 year old would have a similar confusion. :)
it is the latter. You're diving the coefficients. Even without pencil and paper multiplying 2 by 3 and subtracting 6 is as easy as having a third power of 3 involved. Finding f(3) is just a matter of finding the remainder, since you know x-3 is 0 -> 1*3 = 3 -1 = 2 *3 = 6 -12 = -6 *3 = -18 +13 = -5 you're done. and all the coefficients are already written down
@@TheLifeLaVita Thank you for your response, however I was unable to follow the logic you were explaining. It still seems much simpler to me to just plug in 3 for x in the original form of the equation: 3*3*3 (or 27) - 3*3 (or 9) - 12*3 (or 36) + 13 = -5 and I'm done. I don't understand why this is thought to be harder to do? I don't need the pencil and paper to perform the calculation in the synthetic division method, I need the pencil and paper to set up the matrix to keep the coefficients straight. But you did teach me something, I believe. if the equation were f(x) = 2x^3 - x^2 -12x + 13, then the first step would be 2*3 = 6... instead of 1*3 = 3...
@@lyrensyn you don't need paper to set up anything. It's already written on the equation. The Ruffini method (what was used for polinomial division) is very simple, and can find you the 0s of a function. Roughly speaking, the method just tells you to spot just by looking at it a zero and then divide by it, which is by far the fastest way of finding a zero, it just isn't that easy most of the times. He drew arrows to show exactly the multiplication but it's just a simple multiplication inside a division. You multiply by the divisor, you subtract until you have a remainder. You start with the first coefficient 1 and follow the same thought all over, as easy as doing 2*6. Now it's not thought to be harder to just do it, that's what I've been telling you, it's just as easy to do both in your head. You're just fixated on "having to write everything" because it's a new thing for you and you didn't understand what was going on completely, it would be the same as me telling you I have to write down 3*3*3 because I never did it before. In a way you can say the division method is easier since you do all the operations in a row without putting on hold any number, as you do with just doing it all directly.
@@lyrensyn Well this is an example of the method. What if the k had been a large number say 127, then x^3 is larger and harder to do in your head. What if f(x) was an equation with the significant polynomial of x^7. For both of these examples solving f(x) by plugging in the value for x would be a daunting task. Then this method of solving for q(x) and r is a shortcut.
@@TheLifeLaVita You may not need paper in pencil to set up the matrix, but I do because of the way my memory operates. It's not the same as yours. Perhaps after using this method enough times it would open up pathways in my mind that would allow me to perform this just as easily as it is for you, but as someone new to this concept, I simply don't process it in the same manner as you do.
I was taught synthetic division as a way to evaluate a polynomial, and as a way to factor roots out of a polynomial, but never quite made this connection with Euclidean division
In my distance learning to catch up on my degree, I only had a polynomial equation in my math notebook, and in 2 lines, the result was provided without intermediate steps for explanation and no solution given. I had to figure out on my own how to arrive at the result.
I dont think puting it in that form is faster than just pluging in 3 from the beginning. Could be wrong though. It could also be easier as problems get more complex.
that q(x) is pretty confusing because i thought it was going to be another function like f(x) at first lol wait it is, i got double confused and thought it was going to be a coefficient instead
This is 11th grade? I have an A-level in maths (UK) and never faced anything like this. Just thinking about dividing polynomials is giving me a headache.
Lol, the request is funny 11th grade math: please explain like i'm five There's a reason why it's 17yo math and not 5yo math Unless you're a genius 5yo, in which case that person won't be needing to be helped out in the first place.
Can you explain this triple integral? I can not calculate it since the result is dependent on x. 2x+2y+z=10, 2x+2y+z=5, x=2y^2, x=sqrt(4y) find the integration of 15z. Thank you beforehand
I think you are giving polynomial long division a bit of a bad rap. Compared to synthetic division, I think it is easier for novice (and possibly struggling) students to learn and remember - especially ones in grade 11. Polynomial long division more closely matches long division of numbers ( which they already know) and does really not take that much more time . A few seconds maybe if you are really good at it. I used to teach synthetic division in high school but eventually decided it was not worth the time.
I'm going to be "crying on the inside" for a long time...if all math teachers could teach like you, American kids would grow up to be major contributors to the economy and their own families because they understand math. Have you thought of traveling and holding teaching seminars for math teachers?
This is a 11th question? I learnt this in 7th/8th. We divide to get answer. I though we would have to derive the value of k using differential equations but this was easy? Like is this truly 11 th grade question I want to know just put of curiosity
a^1025 / a = 2^1024 * 2 a^1024 = 2^1025 a = plus or minus 1024th root of 2^1025 (or 2^(1025/1024), if you want) This can be slightly simplified as: a = plus or minus 2 * 2^(1/1024)
I thought so too until I did so much synthetic division that I just end up writing the coefficients and doing it all very easily in my head. I do like to fall back to long division to be assured of my answers so generally speaking you’re absolutely right.
I am asking you,that how to become a pro in maths for olimpiad as I am in starting point. I have a doubt that is that possible by lots of practice or not.
would the remainder -5 not be written as -5 / (x-3) since we tried to divide with it but it is not divisible? this is what we learned in school.... but maybe it was wrong
If it was inside the parentheses, yes, so you'd get (x-3)(-5/(x-3)), but that simplifies to just -5 and makes the first term 0 so you can cancel it out easier.
How do you define the square root of a complex number ? And I'm not asking because I don't understand complex number, I'm asking because I do understand and there are two second degree roots of z², so unless the setter specifically mention that by sqrt he means the principal root, or alternatively that by 'x=' he means the set of solutions corresponding to all the relevant roots, that equation isn't actually well defined. Still, if you need a hint: |z| = sqrt (a² + b²) (notice that a² and b² are positive real numbers, so this is well defined), and sqrt (z²) will solve to either a+bi or (-a) + (-b)i (whichever has a positive value in front of i), or both (if you want all the solutions for all the roots). Don't forget to put in the 5's and 12's where they belong, add it all up, and you'll have the solution(s).
Additional note: it is possible, and I think even likely, that the setter is trying to trick you into incorrectly answering that sqrt (z²) = |z|, which would be true for a real number but not a complex one, as I'm sure you've noticed or you wouldn't be calling for help. It would make solving the equation so much easier, but it's wrong, and the setter very carefully stated b ≠ 0 to make sure it was never the right answer.
2³/2³ = 8/8 = 1 2³/2³ = (2/2)³ = 1³ = 1 2³/2³ = 2^(3-3) = 2⁰ Since you start with the same expression, all 3 methods must give the same final answer, which means 2⁰ and 1 are the same value. This reasoning works for any base except 0, as that puts you in the 0/0 situation.
I don't think this is explaining like the person is 5. You didn't teach them counting, addition, subtraction, multiplication, division, or exponents. And you haven't covered algebra.
no. 7 divided by 3 is 2 remainder 1. so, 7 = (3)*(2) + 1 if you work with a remainder you don't divide it further. that's what you'd do to get decimal numbers (7/3 = 2.3333...) then you don't have any remainder.
lol i just to make a lot of effort to understand the problem it was very dificult for me know i am calc3 level and i cant solve school olympia problemes and some times integration bee i cant belive my progress
No (x-3) is only dividing f(x), the resulting value for q(x) and r are found from there. Picture this 31 divided by 3, you’ll get 10 as the quotient and 1 as the remainder, not 1/3.
@@sorryboss8550 yea but then you'd write it as 10 remainder 1 not 10 + 1? cause he's adding it at the end? if you add it to the end wouldn't it be 10 + 1/3?
@@turtleboi3919 the equation asks it to be rewritten as (x-k)q(x)+r so the plus r in the equation. So when we calculate the division, we get the quotient and then plug it into the q(x) part and then take the remainder and plug it into the r part, next up we already know x and we already know k so the new expression becomes (3-3)(x^2+2x-6)+5, and bedmas dictates that we do brackets before addition so we end up with +5 remaining, so essentially the problem has 2 parts the beginning where we get the quotient and the remainder and the end when we plug that into the expression
@@sorryboss8550 I have absolute no brain when it comes to math. The only reason I watch these videos is seeing him fill an entire whiteboard with, what is in my mind, gibberish, and then he enthusiastically writes more gibberish and proudly proclaims it's the answer. 😁
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"You don't have to beg me, I would love to explain this to you". I love this teacher, I appreciate so much your work!
He didn't explain it like the asker was 5 though. He should've started by teaching him the alphabet, then taught him basic arithmetic, then probably ended the video because a five-year-old has no chance of learning algebra.
@@chitlitlah bruh
@@chitlitlah you're a funny guy. thanks for the laugh. I needed it
@@chitlitlah trueeeee actually
I wish my teachers were like him back in high school!
You're one of the best teachers i've seen. Love watching how you explain.
Thank you! 😃
I agree, I absolutely love watching these explanation videos
I don't think I've ever learned about polynomial division at school. Or if I did, I've totally forgot about it.
Same
I learnt it.
Australia Year 11 Extension 1 Maths
I didn learn it during my time.. but new curriculum has polynomial..
I don't believe I have learned these polynomials, although I have seen them in some maths books and tried them for fun many yrs after graduating. That's probably why I prefer the long division rather than the synthetic version.
I don't even understand how it works
You're one of the most dedicated and enthusiastic math professors out there. I hope you'll achieve everything
Thank you!!
I always replace the division with multiplication (I guess it's more or less another way of doing the synthetic division...). You basically take the (x - k) and then multiply with something to get back to the original polynomial. I find this to be easy and quick. To give a step-by-step of the example above (in real life you can of course do it in a matter of seconds):
(x - 3)(x^2 - because x(x^2) gets you the x^3 term. The result is:
x^3 - 3x^2 - but you don't want -3x^2, you want -x^2, so you have to add 2x^2 by adding 2x to the q(x):
(x - 3)(x^2 + 2x - this gives you (x^3 - 3x^2) + (2x^2 - 6x) = x^3 - x^2 -6x. But you want -12x so you must subtract another 6x by subtracting 6 in q(x):
(x - 3)(x^2 + 2x - 6) - this gives you (x^3 - 3x^2) + (2x^2 - 6x) + (-6x + 18) = x^3 - x^2 - 12x + 18. Now you can just subtract remainder 5 to get 13 as the constant.
It's basically what I do too. It seems complicated, but it's actually pretty simple!
I do this with the box method. I’m a high school math teacher.
Or said otherwise, we are looking for a, b and r such that
(x-3)(x²+ax+b) + r = x³ - x² - 12x + 13
After developing the left side:
x³+ (a-3)x² + (b-3a)x - 3b + r = x³ - x² - 12x + 13
Matching the terms of equal powers gives a system of 3 equations:
[1] a-3 = -1
[2] b-3a = -12
[3] -3b + r = 13
The resolution is pretty simple. Using [1] we get
a = -1+3 = 2
then using [2]
b = -12+3a = -12+6 = -6
and finally using [3]
r = 13 + 3b = 13 - 18 = -5
So the final answer is f(x) = (x-3)(x²+2x-6) - 5
Funny thing, I had a bit of a blindspot for long division at school, UNTIL I did polynomial (we called it algebraic) long division, and then it became automatic. Something about seeing just numbers hid where they were coming from, but algebraic expressions look like what they are if that makes sense.
Yep. The connection becomes really obvious in hindsight when you realise that our number system is literally just writing the coefficients of a polynomial where x=10 :)
Try this trig equation th-cam.com/video/Skl236rfDYs/w-d-xo.htmlsi=6X4lYQoiV6iysk6d
I would start the slow way.
First solve for f(x)=(x-3)x^2 +r_0(x)
We get r_0(x) is 2x^2 -12x +13
We then do r_0(x)= (x-3)2x - r_1(x)
r_1 is -6x+13
Then r_1 = (x-3)(-6) + r_2, r_2=-5
We then unroll it.
f(x)=(x-3)x^2 + r_0 = (x-3)x^2 + (x-3) 2x + r_1 = (x-3)x^2 + (x-3) 2x + (x-3) (-6) + (-5) = (x-3)(x^2+2x-6) -5
The tricks and shorthand don't give the person any chance to understand WHAT is going on. And without knowing what is going on, how are they supposed to know if the trick is nonsense? Or if there is a mistake in it?
Teaching tricks without foundation is obsolete in this era of caclulators. Teach the foundation, not how to solve some algorithm that we'd hire peasants to do in the 18th century.
I will recommend just do the long devision instead of this method in this video. Long devision works perfectly, and only slower about 20 seconds or less. The benefit is you don’t memory other method to do devision. I said that because I always misplace the -3 instead of 3 on divisor part or confusing other part with long devision while doing exam.
Anyways, love your video, keep doing more :)
This. ^
Synthetic division has limited applicability, and requires you to memorize a technique / remember its idiosyncrasies. (More mental load.)
Long division works in all cases (even when the divisor isn't so simple), and only requires you to know how division works. Every step is logical, and it's not really as time-consuming as some teachers make it out to be.
I never teach synthetic division for this reason. I use the box method or long division when I teach algebra 2.
wow i hate math and dont even have it as a subject but still watched it cause of ur amazing explanation
ngl, I'm a mechanical engineer that eats, breathes, and shits 2nd-order ODEs and I've never seen an algebra problem worded like this in my entire life. I had no clue what they were even asking me to do.
That's why engineers are not mathematicians.
I love synthetic division. It makes the work so much easier if you meet the correct initial conditions.
I dont know why i got recommended this channel, but im learning more than I did in high school and College
The problem was trivial but so fun! thank you for your beautiful solution bprp!
I have but what a way to explain it. Truly a god at teaching.
I'm early so I would like to thank you for all your work ❤
Hello
u done it good - no question - but it must be a math genius kid who will get this with five...
No sorry none "normal" educated who is five years old will understand this.
Even not a 40 year old one, who never have that at school or simply forget ("the personal math and testing hell has ended long ago, so let this be quickly suppressed" is not really seldom experience and pain from the past ) will understand this - sorry.
But is not your fault, it simply impossible to do it in 7 minutes, and even in two hours.
Just work it in reverse: plug in 3, calc answer mod 3 (which is 1) subtract 1, then factor (x-1) out that poly to get (x-1)(x^2-12)
3:02 I'd the coefficient isn't 1, you can just factor out the coefficient out to make it 1, remember to divide the coefficient after doing synthetic
I love your enthusiasm!
As a five year-old, I can confrim that I totally understand this
As a newborn, I can confirm that I totally understand this
👏So interesting to learn about something I've never even stumbled upon even back in high school
x^3 - x^2 - 12x + 13
step 1: write (x-3) gap (x-3) gap (x-3)
step 2: write coefficient for first as: x^2(x-3) gap (x-3) gap (x-3)
step 3: write second coefficient as: x^2(x-3) + 2x(x-3) gap (x-3)
because from first and second term -3x^2 + 2x^2 = -x^2 (second term of original equation)
step 4: write third coefficient as: x^2(x-3) + 2x(x-3) -6(x-3)
because from second and third term -6x -6x = -12x (third term of original equation)
step 5: we write : x^2(x-3) + 2x(x-3) -6(x-3) - 5
-5 at the end because we have +18 from the third term but we have 13 in original equation so +18-5 = +13.
finally: take (x-3) common
(x-3)(x^2 + 2x - 6) - 5 -->ANS
*You could also just rewrite x³ -x² -12 +13 as x³ -3x² +2x² -6x² -6x² +18 -5 and then factor out the x-3 from the first six terms.*
Loving these videos. Thank you!
I ain't never seen a question like that, and I know why, it used q and r notation which is from English so obviously we would've skipped that, it also feels kinda useless
I have never been happier to remember polynomial long division.
You should have explained the case for another way around as well.
Like in the form of (x2 + 2x -6) k(x) + r
Or, even in the form of (x + x1 ) (x + x2 ) m(x) + r
etc.
I think, the nightmare of most highschoolers was, they had been explained to find the solution of one case (the teacher had explained to them like to a baby).
But then the question of the exam/test they had been given was pretty much different.
Like, there is no such general solution for the same topic of various problems.
amazing!
thank you for exposing me to this!
i did the polynomial long division...
thank you for exposing me to synthetic division!
That's fairly slick, something I don't ever recall doing in math ever, and I went further than most any non-math major. That said, I'm still lost of the purpose of doing it, from the explanation given at the end it seems like he's saying order of operations is hard, so do polynomial division and it makes it easy.
Polynomial division could be useful to simplify a high order polynomial to a lower order one so you can apply a formula to find all the zeros, given you know some of the zeros already. :D
@@zombieherobrian5384 That make's sense, I guess this question was problem set up to show that you can find zeros and the last bit was just to get the student to do a check to verify they did exactly that.
I am in grade 9th (India) CBSE and this question is nothing. Literally every good or moderate student can solve this question
Just a reminder to everyone, synthetic division is actually possible with non-integer rational numbers also.
You have to divide all coefficients of the "dividend" by the denominator, first IIRC.
Why they didnt teach this way to me in middle school. I wouldn't have struggled so much..
In my country this is one of the first few chapters of additional math syllabus (taught for grade 9-10 equivalent)
lemme guess Indian ? asian ?
@@shreysinghrawat5469 for me it was Brazil, but it was a private school and I didn’t pay attention…
Surprisingly, I managed to solve it
This is included in IGCSE Additional Mathematics (4037) course [Factors of polynomials]
I use 3rd line rule as my teacher doesn't like synthetic (which i fcking hate)
omg this is super useful thank you so much
Hey i was wondering why this video has so few views, so i checked your videos tab to compare and realized its not there? Is this unlisted or maybe a youtube bug? if it's unintentional you shoukd probably either try to fix or reupload it.
Yea could be a YT bug I too didn't find it in the list
@@epicgams5406 How did you comment one day ago? the video is 20 minutes old.
@@charlesmichirugallolmao ya and @epicgams5406 too 1 day ago
judging by the new replies and all the comments now on the vid i guess he finally fixed it so it's public now
(x-1)(x²-12)+1
could also be one of the possibilities
5:50 Finding f(3) this way seems simpler than converting to the form (x-k)q(x)+r and then solving. Sure it is simpler to solve once you already have converted to that form, but the conversion process is far more tedious than just plugging in 3 to the f(x) form. I can do the f(x) form in my head, but the (x-k)q(x)+r form requires pencil and paper. I simply don't understand how that is easier.
Also, it would be helpful to explain why you "go ahead and put a 1 down here" at 3:19 as those unfamiliar with this process may not understand why that is a given. Is it because the coefficient of x in (x-3) is 1, or is it because the x^3 coefficient in the original equation is 1? I assume it is the former, but being unfamiliar with this process until today, I can not be certain it isn't the latter? I am pretty sure a 5 year old would have a similar confusion. :)
it is the latter. You're diving the coefficients. Even without pencil and paper multiplying 2 by 3 and subtracting 6 is as easy as having a third power of 3 involved.
Finding f(3) is just a matter of finding the remainder, since you know x-3 is 0 -> 1*3 = 3 -1 = 2 *3 = 6 -12 = -6 *3 = -18 +13 = -5 you're done. and all the coefficients are already written down
@@TheLifeLaVita Thank you for your response, however I was unable to follow the logic you were explaining. It still seems much simpler to me to just plug in 3 for x in the original form of the equation: 3*3*3 (or 27) - 3*3 (or 9) - 12*3 (or 36) + 13 = -5 and I'm done. I don't understand why this is thought to be harder to do?
I don't need the pencil and paper to perform the calculation in the synthetic division method, I need the pencil and paper to set up the matrix to keep the coefficients straight. But you did teach me something, I believe. if the equation were f(x) = 2x^3 - x^2 -12x + 13, then the first step would be 2*3 = 6... instead of 1*3 = 3...
@@lyrensyn you don't need paper to set up anything. It's already written on the equation. The Ruffini method (what was used for polinomial division) is very simple, and can find you the 0s of a function. Roughly speaking, the method just tells you to spot just by looking at it a zero and then divide by it, which is by far the fastest way of finding a zero, it just isn't that easy most of the times. He drew arrows to show exactly the multiplication but it's just a simple multiplication inside a division. You multiply by the divisor, you subtract until you have a remainder. You start with the first coefficient 1 and follow the same thought all over, as easy as doing 2*6.
Now it's not thought to be harder to just do it, that's what I've been telling you, it's just as easy to do both in your head. You're just fixated on "having to write everything" because it's a new thing for you and you didn't understand what was going on completely, it would be the same as me telling you I have to write down 3*3*3 because I never did it before. In a way you can say the division method is easier since you do all the operations in a row without putting on hold any number, as you do with just doing it all directly.
@@lyrensyn Well this is an example of the method. What if the k had been a large number say 127, then x^3 is larger and harder to do in your head. What if f(x) was an equation with the significant polynomial of x^7. For both of these examples solving f(x) by plugging in the value for x would be a daunting task. Then this method of solving for q(x) and r is a shortcut.
@@TheLifeLaVita You may not need paper in pencil to set up the matrix, but I do because of the way my memory operates. It's not the same as yours. Perhaps after using this method enough times it would open up pathways in my mind that would allow me to perform this just as easily as it is for you, but as someone new to this concept, I simply don't process it in the same manner as you do.
I was taught synthetic division as a way to evaluate a polynomial, and as a way to factor roots out of a polynomial, but never quite made this connection with Euclidean division
interesting. I was taught it as an extension of Euclidean division but not quite as a way to evaluate or factor out the roots of a polynomial.
If this was the type of math I had to do in my 11th grade, I am 100% sure I would get 100 in every math test.
I'm in 4th year of college, and I remember vividly learning this in high school wow
In my distance learning to catch up on my degree, I only had a polynomial equation in my math notebook, and in 2 lines, the result was provided without intermediate steps for explanation and no solution given. I had to figure out on my own how to arrive at the result.
I dont think puting it in that form is faster than just pluging in 3 from the beginning. Could be wrong though. It could also be easier as problems get more complex.
What 5? Like '5' years old?
My thought on first seeing synthetic division was "...wait, that works?" 😂 It seems almost too good to be true.
It's basically the same as long division, you just remove the x part to make it visually simpler.
Wow that's a lot of dry erase markers in the background! Most people out here buying gold or crypto, but he's got his priorities straight!
I have a digital image processing exam tomorrow. I don’t know why I’m watching this?
Its 2am wtf am i watching, i already joined the workforce 3 years ago haha.
I learned polynomial long division, but never synthetic division, I don't think.
I had completely forgotten about polynomial division damn
MInd blown. No teacher ever told me to use long division for _evaluating_ a function.
The method you used at 3:20 is called Ruffini Method? I learnt it that way in school
that q(x) is pretty confusing because i thought it was going to be another function like f(x) at first lol
wait it is, i got double confused and thought it was going to be a coefficient instead
In Italy we call that Ruffini's rule/law
wait cant you think of certain numbers as separate function in of themselves? like infinity and zero
This is 11th grade? I have an A-level in maths (UK) and never faced anything like this. Just thinking about dividing polynomials is giving me a headache.
Lol, the request is funny
11th grade math: please explain like i'm five
There's a reason why it's 17yo math and not 5yo math
Unless you're a genius 5yo, in which case that person won't be needing to be helped out in the first place.
I haven’t used long or synthetic division in so long.
bro please make a video on (arctanx divided by x) integration?
I learned polynomial division... but nothing like this
Can you explain this triple integral? I can not calculate it since the result is dependent on x. 2x+2y+z=10, 2x+2y+z=5, x=2y^2, x=sqrt(4y) find the integration of 15z. Thank you beforehand
Thank you so much!
The answer i want is what is the practicle application for this ....
I think you are giving polynomial long division a bit of a bad rap. Compared to synthetic division, I think it is easier for novice (and possibly struggling) students to learn and remember - especially ones in grade 11. Polynomial long division more closely matches long division of numbers ( which they already know) and does really not take that much more time . A few seconds maybe if you are really good at it. I used to teach synthetic division in high school but eventually decided it was not worth the time.
Thank you for the revision its just that i have a question what grade are your videos generally intended for? Thanks in advance
Below 11th grade I'm guessing since I'm learning Calculus in 11th
@@xadxtya aight thanks
I'm going to be "crying on the inside" for a long time...if all math teachers could teach like you, American kids would grow up to be major contributors to the economy and their own families because they understand math.
Have you thought of traveling and holding teaching seminars for math teachers?
This is a 11th question? I learnt this in 7th/8th. We divide to get answer. I though we would have to derive the value of k using differential equations but this was easy? Like is this truly 11 th grade question I want to know just put of curiosity
Anyways nice work bro
Why didn’t you divide (r) by (x-3)? Since it’s the remainder don’t you have to divide by the divisor?
Never mind, it’s because it’s outside the parenthesis.
Hey man can you help me with this problem because there is so much results in this problem I hope you will solve it.
Problem:
(a^1025)/(2a)=2^1024
a^1025 / a = 2^1024 * 2
a^1024 = 2^1025
a = plus or minus 1024th root of 2^1025
(or 2^(1025/1024), if you want)
This can be slightly simplified as:
a = plus or minus 2 * 2^(1/1024)
Why isn’t the remainder r shown being divided by the divisor (x-k) ?
Why don't you subtract r before dividing by x-3?
Personally , I think that long division is much easier and simpler than synthetic one. (No hate to the video just giving my opinion)
I thought so too until I did so much synthetic division that I just end up writing the coefficients and doing it all very easily in my head. I do like to fall back to long division to be assured of my answers so generally speaking you’re absolutely right.
@@sorryboss8550 yeah that's true. It just requires getting into it. Still using long division as an authentication is great.
i passed calc 2 last year and i dont remember any of this shit lmao
Me neither lmao.
I am asking you,that how to become a pro in maths for olimpiad as I am in starting point. I have a doubt that is that possible by lots of practice or not.
This still makes zero sense
thats's the Ruffini division
would the remainder -5 not be written as -5 / (x-3)
since we tried to divide with it but it is not divisible?
this is what we learned in school.... but maybe it was wrong
If it was inside the parentheses, yes, so you'd get (x-3)(-5/(x-3)), but that simplifies to just -5 and makes the first term 0 so you can cancel it out easier.
i just watched a math video for entertainment
Could you help me with this problem please? How do you solve x = (5|z| + sqrt(z^2))/12 where z is a complex number in the form of a + bi where b ≠ 0
How do you define the square root of a complex number ? And I'm not asking because I don't understand complex number, I'm asking because I do understand and there are two second degree roots of z², so unless the setter specifically mention that by sqrt he means the principal root, or alternatively that by 'x=' he means the set of solutions corresponding to all the relevant roots, that equation isn't actually well defined.
Still, if you need a hint: |z| = sqrt (a² + b²) (notice that a² and b² are positive real numbers, so this is well defined), and sqrt (z²) will solve to either a+bi or (-a) + (-b)i (whichever has a positive value in front of i), or both (if you want all the solutions for all the roots). Don't forget to put in the 5's and 12's where they belong, add it all up, and you'll have the solution(s).
Additional note: it is possible, and I think even likely, that the setter is trying to trick you into incorrectly answering that sqrt (z²) = |z|, which would be true for a real number but not a complex one, as I'm sure you've noticed or you wouldn't be calling for help. It would make solving the equation so much easier, but it's wrong, and the setter very carefully stated b ≠ 0 to make sure it was never the right answer.
Can you explain,how to become 1, any number power of zero , can explain with reason?
2³/2³ = 8/8 = 1
2³/2³ = (2/2)³ = 1³ = 1
2³/2³ = 2^(3-3) = 2⁰
Since you start with the same expression, all 3 methods must give the same final answer, which means 2⁰ and 1 are the same value.
This reasoning works for any base except 0, as that puts you in the 0/0 situation.
an eleventh grader struggling to solve this problem.
me who learnt it in 9th grade on the very first day
I don't think this is explaining like the person is 5. You didn't teach them counting, addition, subtraction, multiplication, division, or exponents. And you haven't covered algebra.
In that case, show me the long division !
That must be a really smart 5 year old 😂
NOT THE WEBASSIGNS!!!!!
I used reverse deduction mentally.
YOU WRITE THE HORIZONTAL LINE ON THE F BEFORE THE VERTICAL ONE?!?!?!??!
I saw this question in 10th class pre boards💀
“I am five “💀
Put the team on his back !
please explain trigonometry
Where the hell does q(x) and r come from?
im trying to pay attention but all i see is
"demonstrate that fkr"
Wouldn’t the remainder be (-5)/(x-3)
no. 7 divided by 3 is 2 remainder 1.
so, 7 = (3)*(2) + 1
if you work with a remainder you don't divide it further. that's what you'd do to get decimal numbers (7/3 = 2.3333...) then you don't have any remainder.
You're thinking of the form a/b = q + r/b. In this case, both sides are multiplied by b: a = bq + r.
Thankyou
lol
i just to make a lot of effort to understand the problem
it was very dificult for me
know i am calc3 level and i cant solve school olympia problemes and some times integration bee
i cant belive my progress
isnt it +(-5)/(x-3) for the remainder?
No (x-3) is only dividing f(x), the resulting value for q(x) and r are found from there. Picture this 31 divided by 3, you’ll get 10 as the quotient and 1 as the remainder, not 1/3.
@@sorryboss8550 yea but then you'd write it as 10 remainder 1 not 10 + 1? cause he's adding it at the end? if you add it to the end wouldn't it be 10 + 1/3?
@@turtleboi3919 the equation asks it to be rewritten as (x-k)q(x)+r so the plus r in the equation. So when we calculate the division, we get the quotient and then plug it into the q(x) part and then take the remainder and plug it into the r part, next up we already know x and we already know k so the new expression becomes (3-3)(x^2+2x-6)+5, and bedmas dictates that we do brackets before addition so we end up with +5 remaining, so essentially the problem has 2 parts the beginning where we get the quotient and the remainder and the end when we plug that into the expression
I don't know why I keep watching these videos. I understand nothing of what is being said.
Check out the unit on polynomial division on Kahn academy if interested you could learn all these problems in little time😃
@@sorryboss8550 I have absolute no brain when it comes to math. The only reason I watch these videos is seeing him fill an entire whiteboard with, what is in my mind, gibberish, and then he enthusiastically writes more gibberish and proudly proclaims it's the answer. 😁
@@Gremriel 😂😂😂😂