I'm in college and teachers don't teach at all anymore. Which is why a lot of us are on here trying to learn. Thank you as you are the only one I have run across that makes it interesting and understandable.
11:40 The way I see it, is (taking your example): 340473 / 582 = 585.(decimal) *340473* */* *582* *=* *585* *+* *(R* */* *582)* => as the reminder is to be divided to get the decimal number 585.(decimal) - 585 = 0.(decimal) 0.(something) = (R / 582) R = 0.(decimal) * 582 So on a calculator I just do this: 340473 / 582 = 585.(decimal) Previous answer - 585 = 0.(decimal) Previous answer * 582 = R
Daniel Major calculated through your equation the answer is not 3. The decimal part is the portion of divisor (the 581 you said which should be 582). So you should multiply decimal part with divisor get the residual.
Your boldened line should be corrected to 340473 / 582 = 585 + (R / 582). This is a restatement of the quotient-remainder expression Mr. Woo has used throughout the video; multiplying both sides of this equation by the divisor (i.e. 582), we can see this, as we then have 340473 = 585 × 582 + R, as he writes at the very end of the video. 🙂
For the exercise I chose x to be 1 (so I could do it in my head) x + 1 = 2 x^2 + 5x + 7 = 13 13 / 2 = 6 Remainder 1 Then I tried x to be 2 x + 2 = 3 x^2 + 5x + 7 = 31 31 / 3 = 10 Remainder 1 Neither of these results satisfy the pattern of 'remainder 3.' Did I make an error, or is there something different about these values that it fails to satisfy the mentioned pattern?
After dividing the polynomial, you end up with x + 4 + 3/(x + 1), with the 3 being the remainder and the x + 1 being the divisor. If you choose x to be 1 or 2, the divisor would be 2 and 3 respectively, which is small enough that the remainder can still be divided into it, so the actual remainder is less than 3.
You made a calculation error with x=2. x^2 + 5x + 7 = 21, so the remainders is 0 after division by 3. I think it has something do to with the divisor being smaller than the remainder of the algebraic polynomial division. If you divide 3 by the divisor of your first example, it does give a remainder of 1 and indeed a remainder of 0 in your second example.
I chose the number 8 x + 1 = 9 x^2 +5x+7 8^2+5(8)+7=111 after doing the calculations on paper, I end up with a remainder of 7 however the gap from my quotient and divisor was 4. Was there an error?
Eddie great stuff as always! I do think there is a mistake though. My remainder was 7, and using your figures 340473 / 581, i also get a remainder of 7, with a difference of 5 not 4. Your videos are a great help. Thanks.
When you choose 2 your divisor becomes 2 + 1 = 3, which means you can't possibly have a remainder of 3. X needs to be restricted to positive integers greater than 2, cause it also doesn't work for 1 or 0.
You showed where the remainder comes from in decimal, but you didn't explain to them HOW to get the remainder into an integer form......I know you want them to work it out, but after describing where something comes from, it's probably a good idea to show the steps so they can see how to get the integer out of a calculator (some people memorize rather than rationalize better, so teaching both methods at the same time captures more brains). meaning.....tell them to simply subtract the 585 to get rid of the quotient, then multiply the leftover by 581 to get the remainder.
He did, at the very end. He just didn't actually do the calculation. The equation at the end can be solved for R, which gives the remainder in integer form. Your solution is the way I figured out in primary school, and it's pretty easy as well.
@@Whizzer .... he gave the formula and pretty much told the kids "figure it out"....I'm saying that some kids cannot fully grasp concepts very well and it's best to just tell them straight out and they'll memorize it better. I work with a woman who still figures out her paycheck 1 day at a time.....even though she makes the same amount per hour....and no, she does not add the hours up, she literally multiplies the "8 hours in a day" by her hourly rate, for EACH day....and then adds up all the subtotals for a final total. People Are Dumb
StackExchange, Reddit, and Quora all have mathematics sections. Perhaps submit your questions there: • math.stackexchange.com • www.reddit.com/r/math • www.quora.com/topic/Mathematics
@@nahanpodlejski7605; though he says that as a reference to when the students first explored linear expressions, not arbitrary polynomials such as quadratics and cubics. The content in this video is usually taught to 15-16 year olds (year 10) in the UK, and I imagine it's the same in Australia, where this video was recorded.
Wait wait wait, these students know how to factor polynomials without knowing how to divide them? How? That means that they just learned the formulas without knowing how and why they work, that's weird.
factorising is not dividing, as some polynomials dont have real roots, as the graphs have only 1 or 2 instead of 3 intercepts, so if you put it in your calculator, youll get a number like i+1.234 or something like that, meaning that it is an imaginary number (i is basically the square root of a negative 2) and therefore indivisible, as you cant root a negative number.
I'm in college and teachers don't teach at all anymore. Which is why a lot of us are on here trying to learn. Thank you as you are the only one I have run across that makes it interesting and understandable.
Math is my least favorite subject and I’m watching this voluntarily... Mr.Woo is quite the teacher!
11:40 The way I see it, is (taking your example):
340473 / 582 = 585.(decimal)
*340473* */* *582* *=* *585* *+* *(R* */* *582)* => as the reminder is to be divided to get the decimal number
585.(decimal) - 585 = 0.(decimal)
0.(something) = (R / 582)
R = 0.(decimal) * 582
So on a calculator I just do this:
340473 / 582 = 585.(decimal)
Previous answer - 585 = 0.(decimal)
Previous answer * 582 = R
Daniel Major calculated through your equation the answer is not 3.
The decimal part is the portion of divisor (the 581 you said which should be 582). So you should multiply decimal part with divisor get the residual.
@@yin78105 Oh yes, you're right. I was confuse while writting my comment.
I corrected it.
@@danielmajor3777 all good bro. we are all here for learning math :D
Your boldened line should be corrected to
340473 / 582 = 585 + (R / 582).
This is a restatement of the quotient-remainder expression Mr. Woo has used throughout the video; multiplying both sides of this equation by the divisor (i.e. 582), we can see this, as we then have
340473 = 585 × 582 + R,
as he writes at the very end of the video. 🙂
All of you should know
Its 581 not 582
For the exercise I chose x to be 1 (so I could do it in my head)
x + 1 = 2
x^2 + 5x + 7 = 13
13 / 2 = 6 Remainder 1
Then I tried x to be 2
x + 2 = 3
x^2 + 5x + 7 = 31
31 / 3 = 10 Remainder 1
Neither of these results satisfy the pattern of 'remainder 3.' Did I make an error, or is there something different about these values that it fails to satisfy the mentioned pattern?
After dividing the polynomial, you end up with x + 4 + 3/(x + 1), with the 3 being the remainder and the x + 1 being the divisor. If you choose x to be 1 or 2, the divisor would be 2 and 3 respectively, which is small enough that the remainder can still be divided into it, so the actual remainder is less than 3.
You made a calculation error with x=2. x^2 + 5x + 7 = 21, so the remainders is 0 after division by 3. I think it has something do to with the divisor being smaller than the remainder of the algebraic polynomial division. If you divide 3 by the divisor of your first example, it does give a remainder of 1 and indeed a remainder of 0 in your second example.
Oh, absolutely right. Yeah, I think what you're saying is what @Teravortryx was saying as well.
You should divide it in such a way that the quotient is x+4 i.e 6 in your case (when x=2)
Eddie, I would need an intensive personal class of Math from you to pass my next final...
Divide by 582, not 581
Harry Morris I was wondering the same thing🤔
@@yin78105 x+1=582 x=581
12:47 Not gonna lie but I am completely mindblown by these.
I chose the number 8
x + 1 = 9
x^2 +5x+7
8^2+5(8)+7=111
after doing the calculations on paper, I end up with a remainder of 7 however the gap from my quotient and divisor was 4. Was there an error?
We did this 2 months ago and I've forgotten already so this is a life's
Plz can anyone tell me how to add missing digrees for the following:
3•X^7•y^5+5x^4•y^3+z^5•y^2
Sir, thank you for your good work
Eddie great stuff as always! I do think there is a mistake though. My remainder was 7, and using your figures 340473 / 581, i also get a remainder of 7, with a difference of 5 not 4. Your videos are a great help. Thanks.
Yeah, he should've written 340473 ÷ 582 instead of 581.
I am wondering what role your textbook plays in the lesson?
You wrote 581 instead of 582 in the division.
Thank you for sharing knowledge
You need to teach teachers mate! :(
Hmmm when I used his chosen number, and the formula written on the board I got 586.01...
Can u explain
Wait what the Australian math syllabus teaches students to integrate before learning about long division?
Kh Bye no, they musy be an accelerated class, you learn poly div a term or 2 before calc
...but is this lesson really an answer to the question "why is it important"????
Sir, Please upload statics and dynamic lecture
I chose 2 and my answer was 7. That’s x+5 with no remainder... I must be missing something
Didn't get you
When you choose 2 your divisor becomes 2 + 1 = 3, which means you can't possibly have a remainder of 3. X needs to be restricted to positive integers greater than 2, cause it also doesn't work for 1 or 0.
In what institute do you teach?
You showed where the remainder comes from in decimal, but you didn't explain to them HOW to get the remainder into an integer form......I know you want them to work it out, but after describing where something comes from, it's probably a good idea to show the steps so they can see how to get the integer out of a calculator (some people memorize rather than rationalize better, so teaching both methods at the same time captures more brains).
meaning.....tell them to simply subtract the 585 to get rid of the quotient, then multiply the leftover by 581 to get the remainder.
He did, at the very end. He just didn't actually do the calculation. The equation at the end can be solved for R, which gives the remainder in integer form.
Your solution is the way I figured out in primary school, and it's pretty easy as well.
@@Whizzer .... he gave the formula and pretty much told the kids "figure it out"....I'm saying that some kids cannot fully grasp concepts very well and it's best to just tell them straight out and they'll memorize it better.
I work with a woman who still figures out her paycheck 1 day at a time.....even though she makes the same amount per hour....and no, she does not add the hours up, she literally multiplies the "8 hours in a day" by her hourly rate, for EACH day....and then adds up all the subtotals for a final total.
People
Are
Dumb
@@Felix-nm8dw ... no
Where can I post my maths doubt ? Is there any email or fb
StackExchange, Reddit, and Quora all have mathematics sections. Perhaps submit your questions there:
• math.stackexchange.com
• www.reddit.com/r/math
• www.quora.com/topic/Mathematics
Hooray!
what year 7 and they know integration??
They were in Year 7 when they learnt how to do long division, which is even more surprising XD.
👍 nice
Hey Eddie here u kinda did an oopsie. The remainder when 340473 is divided by 582 is 3 not 581
10:30
Nice one lol
What class is this? Grade 12?
Yes. This is High school level.
In the first video he says "your 13 years old brains"....
@@nahanpodlejski7605; though he says that as a reference to when the students first explored linear expressions, not arbitrary polynomials such as quadratics and cubics. The content in this video is usually taught to 15-16 year olds (year 10) in the UK, and I imagine it's the same in Australia, where this video was recorded.
in my country, polynomial is thought for year 3 of middle school, or 15 year olds.
@@Meteo_sauce in myn we just don't learn it, until last exam of high school at 17/18 we don't see it, maybe in some university/college but idk
Wait wait wait, these students know how to factor polynomials without knowing how to divide them? How? That means that they just learned the formulas without knowing how and why they work, that's weird.
factorising is not dividing, as some polynomials dont have real roots, as the graphs have only 1 or 2 instead of 3 intercepts, so if you put it in your calculator, youll get a number like i+1.234 or something like that, meaning that it is an imaginary number (i is basically the square root of a negative 2) and therefore indivisible, as you cant root a negative number.
Woo who?
Any 7th graders here that have no clue what he is saying
I understand him though because I'm take higher classes (Geometry and Algebra 2).