#long division polynomials
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So I have the problem
x^3+3x^2-2x+5 divided by x+1
now I got the answer x^2+2x but it doesn't seem right
Wait
Do you know how number division works?
Not the process
The idea of what is it
Yea
What??
What is the part that's not the remainder
the quotient I assume?
That's its name, what is it though
The answer to a division problem
It's the biggest number c
So that a=bc+remainder
Where the remainder is between 0 and c
Dividend = divisor * quotent + remainder
Yep
I just haven't done long division in years
a,b,c,r
Can be polynomials
Only difference is what do "biggest" and "between" mean now
Right?
So, the biggest polynomial is the one with the highest degree
If you have two with the same degree, the one with a bigger coefficient
And "between" means the degree is between 0 and the degree of b
Almost there
I'm explaining context for a reason
For why this works
Once you understand why it works, you can understand how to do it very easily
Yea you're giving me a hard time with this philosophy talk man
I just wanna know if I placed my answer in the right spots
It's mathematics, not philosophy
The goal isn't to simply verify your work
Because if we just do that, you'll forget polynomial division once you stop using it
I just have an evaluation in a few days and I am studying for it
The goal is to understand what are you doing, so that you can verify your own work easily
If you don't want me to help with that you should wait for someone else to check your numbers, I don't believe it will help with anything
@stiff echo are you willing to follow through with this?
Yea sorry I had to go take a #2
So I dont get this entire "between 0 and c"
Oh, well
We could pick any number
As the divisor
And just choose the fitting remainder so it holds
But we want the remainder to be small
How small can we guarantee? No bigger than c-1
wait no let's have number 91, divisor 9
Aight
Lets have 98 actually
You'll see why
So,
98=9c+r
Find a number c
So that r is the smallest positive number you can find
Wait c is the quotient right?
Yep
well in this case it's obviously 10 quotient and the smallest number will be 8 (r)
Right
It can't be anythign under 8 since it wouldn't make sense
I claim
anything*
That no matter what number
The remainder is always between 0 and 8
If it goes over 8, it's not the smallest
If it goes below 0, it's not ≥0
Right?
Wait how is it between 0 and 8 if the only valid remainder is 8?
It cant be 7 since that wouldnt equal 98
Well for 98 it's 8
But for 99 it's 0
For 97 it's 7
However there is no number so that r=11 for example
yea that goes past 8
So, same with polynomials
I tested my answer with the division statement and it didnt give the right answer, god damn it
The difference is instead of size, we use degree
Wdym size? like the size of the remainder?
p(x)=g(x)q(x)+r(x)
Yeah
So instead of the remainder we use degree alright
So in the problem I posted the highest degree is 3
Well the size of the remainder is the degree
OH WAIT so the highest remainder is the highest degree?
So if it's 3 the highest remainder it can be is 3??
Wdym
If the highest degree (x^3) in this case is 3 the highest remainder it can be is between 0 and 3?
Well
0 and 3-1
It always goes one below
Like with number division
It's not 9, it's 8
Same with degrees
It's between 0 and deg(g(x))-1
So if we divide $x^{400}+x$ by $x^2+1$, the remainder is at most of degree $2-1=1$
yoavmal
Do I make sense?
Yep
Now, lets try find an algorithmic approach
To finding the quotient
We can try again with number division first
What we can do is start by having
So the long division with polynomials is like an algorithm?
a=0c+a
Exactly!
Now, we take some part of a
And transfer it to c
For example with our system from earlier
98=0*9+98
If the divisior is 0 then doesn't it all just become 0?
or is it undefined
This isn't the final division form
Since r isn't between 0 and 8
So, lets ask ourselves something to make our life easy
What is the biggest power of 10
So that, that power * 9
Is less than 98?
Wait so R here is 98?
Wdym by biggest power of 10
Where did 10 come from
We work in base 10
So it makes our life easy to divide using it
98 is actually 9*10+8*1
What is base 10?
827 is actually 8*100+2*10+7*1
The system we use for representing our numbers
Never heard of it before
There are different bases you can use, which give different values to different sequences of digits
Does this make sense to you though
Not really
Hmm
Just kinda side tracked from the whole between 0 and r
Then you have a different issue
Nevermind, not entirely relevant
Just, take it that we use 10 for that
We have different powers of 10
1, 10, 100, 1000
Etc
Yea
For 98?
So that 9 times that power of 10
The highest is 10
Is less than 98
Yep
So we can say
98=9*10+(98-9*10)
=9*10+8
By this point we know we can stop because 0≤8<9
So how does that relate to the initial problem I had
But if it were 827 it would be different
Well, instead of 10
I dont see any use of power of 10s stuff
We just use powers of x
the x from x+1?
Well not from x+1 in specific
Just x in general
So lets try do the division the same way
We want to get
because x^2 * x is x^3
x³+3x²-2x+5?
Yea
Yea
Right?
Lets do exactly the same, and you'll see why it's actually what you were doing the whole time
And also see how it helps you spot your mistake very fast
So we start with x³+3x²-2x+5=(x+1)*0+x³+3x²-2x+5
Uh what?
Here we use r=x³+3x²-2x+5=(x+1)*0+x³+3x²-2x+5
That isn't long division
It's not the final form
It doesn't have the little thing that looks like the square root symbol
The symbols may be different, the process is identical
You're confusing me with the whole r = the entire polynomial
What do we subtract
When it was numbers
We had the highest power of 10, so that that times 9 was less than 98
Here
We have the highest power of x
So that that times (x+1) is less than x³+3x²-2x+5
Do you see it
Or is it a bit confusing still?
The highest power is x^3
Well, lets try
(x+1)x³=x⁴+x³
That has a degree of 4
So it's more than x³+3x²-2x+5
That isn't long division bruh
It is
long division is when you move it down and stuff
It's just a different way to write it
We didn't do the moving down
Yet
The highest power would be x²
So (x+1)x²
Where did x^2 come from
Well, x³ was too big
It would be the same orange one
Yes x only fix x^2 into x^3
yes I get that now
We then multiply that same one by (x+1)
x^3 and then x^2
Ok, so now we write the whole thing again
x³+3x²-2x+5=(x+1)*0+x³+3x²-2x+5
3x^2 minus the x^2
Becomes
We get 2x^2
We are now stuck with that what I showed above
Am i doing it correctly?
x³+3x²-2x+5=(x+1)*x²+(x³+3x²-2x+5-(x+1)*x²)
Dude like I cant read it like that, I can't form that in my head like I did above
I see hmm
You'd need to write it how I do it in order for me to understand
Like you get what I am saying?
am I doing it correctly above?
Yes I do
I am doing everything you said before
What you are doing always corresponds to an equation like we have
With p(x)=g(x)q(x)+r(x)
q(x) is the thing above the line
the quotient is on top
Wait wait so let me re-read this again
So the first part
is the dividend
Yes
equals the divisor
Which is x+1
and you did the multiply by x^2 I get that
Now the remainder is what's under
that last paren
with the 2 plynomials
Correct
Now I get it
Was it worth the effort?
So, basically every step
After the subtraction
the 3x^2 minus 2x^2 is x^2 and we bring down the 2x
You get an equation you can verify
So first equation is $x^3+3x^2-2x+5=(x+1)\cdot0+(x^3+3x^2-2x+5)$
Second equation is
yoavmal
Second equation is
Where did the 0 come from
x+1 just becomes 0 what
Before you wrote anything above the line
It was 0
$x^3+3x^2-2x+5=(x+1)\cdot x^2+(2x^2-2x+5)$
yoavmal
Oh yeah
That is the second equation
Now what you're doing is just secretly dividing 2x²-2x+5 by x+1
this is where I put the first thing on top of the line yea?
The same way we just did
Yeah the x²
Quick question when doing the long division, do you recommend bringing down all the terms down?
or only one of them
Or just all of them to form one polynomial
Whatever you find more convenient
Like I brought -2x here but left 5
Yea I'll just bring them all down then
So now we divide that new polynomial by x+1
and the same thing repeats
It's really irrelevant, all that matters is you don't make mistakes or work too hard
Exactly, what do we get then?
Wait what if the term goes outside the line
$2x^2-2x+5=(x+1)\cdot0+(2x^2-2x+5)$
yoavmal
Wdym
and just to make sure when dividing we only focus on the X and ignore the 1?
Oh alright
Lets try do that
Yea one sec
How do I know which term the quotient starts to go on top of
If I do 2x^2 divided by the x do I put the answer on top of the 2x^2 or the -2x?
(It doesn't matter)
Yeah
After it's just remainder 5
Did you do this?
Yea I showed a screenshot
Oh I just realized

So quotient times divisor plus remainder
=2x²+2x+5
Wait no it's not 2x it's x^2+2x
Is that 2x²-2x+5
Oh it works!
It doesn't lol
Wait what
Is that
Bro the fucking minus bruh
And that
How do I make it fucking minus
You don't
So leave it like that?
The difference is just not 5
There's nothing we can do 
the bottom one?
Yeah
Oh yeah I see now
So that is the divisor * quotient
move to the other side with the dividend
remainder stays on right side of equals sign
Yeah that's how we calculate the remainder each time
Wait so what even is this then
So
Remainder
Not =5
what why? it said = 5
The remainder is just -4x+5
Oh
You said that
wait wait so x^3+3x^2-2x+5 divided by x+1 is x^2+2x remainder -4x+5?
That is a way to say it but
The degree of x+1 is 1
The degree of -4x+5 is 1
So the division is not yet done
Bruh are you shitting me there's more division?
This is the 1st question to my homework bruh
Yeah
It feels longer because we're making a lot of explanations and checks for each step
In practice it's pretty short
It is but
We want the quotient
if we have the dividend, divisor and remainder can't we find the quotient with all three without needing to even do the long division stuff?
Galaxy brain moment?
You're mostly correct but
This only works for division of the form x-r
Dividing by x²+bx+x no longer works
Yea if the leading coef is greater than 1 then we're cooked
Or wait no so i thas to be a binomial?
it has*
(x-p)
Idk I haven't slept in nearly 24 hours
This discord thread has to be one of the biggest learning curves I've had in my 3+ years of highschool.
What grade are you in
Bro how tf do you know this shit so well
99% of the time everyone in my school learns this, does the test and forgets right after
How do you make sure to remember this stuff for longer periods of time
Wait btw bro I need some help with another problem
It's in the same question I just dont know what it's asking
What do you think are we doing right now
You won't remember the technique
But the concept of
"Fitting the biggest polynomial"
That does stick in the bsain
I see
What does it mean by restrictions
Oh yea
Like, can't divide by 0
Wait how would it look for the quotient anyway
x^2+2x
move everything to the other side but x
x = √(-2x)
≠*
Is that correct?
Well you're not done remember
So we're not even done the division?
No, just one step more
Whats the last step
Dividing -4x+5 by x+1
Since we want the degree of the remainder to be less than the degree of the divisor
Less than 1
Let's go dude
yoavmal
?
.
Oh shit yea
Yeah just gotta be attentive
Dude this is a crazy learning curve for me
I'm glad it is
It's just our teacher told us to do this and get remainder
When we finish i'm dividing sobbed though
Wdym
I haven't slept in like
24 hours
I'm really tired so when we finish i'll go sleep
So want to retry doing the division, but this time without side steps?
So you see it's not that slow
Sure
Let me try to write the equation
Aight
3x^4-4x^3-6x^2+17x-8 = (3x-4) Q(x) + R(x)
So like this
The first one will have the quotient 0 obviously
and remainder is also 0?
No, equation has to be true
Remainder must be the whole polynomial
At the first step
So (3x-4) Q(x) + the whole polynomial (r)?
Oh wait I remember why
because thats all we have under the line thingy
and it changes throughout the solving process
The next one is lemme type it out
3x^4-4x^3-6x^2+17x-8 = (3x-4) (x^3) + 3x^4-4x^3
that one should be good right?
Seems off
Which part specifically?
Left and right of this equation
Ohh alr
That seems right though
How does 6x^4 appear?
Here?
3x*x³
+3x⁴
On right
Sorry i'm falling asleep
Yea all good man
am
Can other people help though?
If they see this, yes
Yea
Yea so it should be 4 anyway
And u know that u divide remainder by your divisor
is help still needed?
@stiff echo
Hi
