#đă»general-2
1 messages · Page 143 of 1
here's an example of a trial error combination i could make (obviously this one won't result in the answer, but it's a combination to try)
2x * x = 2x^2 (left side of cross)
-2 * 2 = -4 (right hand side of cross)
isn't trial and error just plain worse than doing a strict method?
isn't it just inneficient?
depends
join 2x and 2, 2x(2) = 4x
join x and -2, x*(-2) = -2x
add them together. 4x+(-2x) = 2x (which is not 7x)
my brain tends to find the trial and error in my head instantly
especially when it's 2x^2, 3x^2, or 5x^2
you have the math virus
if it's 4x^2, the left hand cross could be 2x/2x or 4x/x
but that still is a pretty easy case
technially. with "ur method". if you have ax^2 + bx + c, you have to trial and error find two constants x,y such that b = x+y and x*(y) = a(c)
so my argument there is that they are both technically trial and error
one can be less brain worky tho than the other
https://www.youtube.com/watch?v=5QyeZ7KwFKg here are some other methods
and how would you factorize t^3+1 for example?
use the sum of cubes formula
i don't know it i think
but
difference of two cubes
a^3 - b^3 = (a-b)(a^2+ab+b^2)
t^3 + 1 = t^3 + 1^3
-1?
i mean -t
it's the formula. it's weird. but it works.
wheres the minus
you're meant to write minus that's why i was correcting u
difference of two cubes
a^3 - b^3 = (a-b)(a^2+ab+b^2)
@abstract scaffold but
no thats not the one
a^3 + b^3 = (a+b)(a^2-ab+b^2)
@abstract scaffold ohhh ok
t^3 + 1 is the sum of two cubes
yes, it's very easy to mix the two, since they are very similar to each other.
here's a proof
first line: cubic expansion
second line: isolate a^3+b^3 on one side, and that will be equal to something
third line: notice some group of terms can be factorised
fourth line: notice there's now a common factor of (a+b)
fifth line: "simplifying the right factor"
last line: simple algebra
and in similar fashion, we can prove the difference of two cubes
mi tu
@sweet cape yeah?
Anyway, i copied and pasted the exercise that you made on sheets, Iâm waiting if the professor just tell me the grate, but that was so hard!
like like 6 variables is stupid
I didnât copy that because this exercise didnât need to give him
Yeah
Because i didnât see decisionsmaking
it'd take me ages to do by hand.
Lol
bloody like 6x6 matrix
My teammates just told me, âwhy do you have some errors about results in those boxesâ
Hahahaha
That was hard, the reason of that is all of things to do any exercise, i didnât see that
hola
Maricon
@summer wolf We can use english here if you want haha
si es demasiado español en #đžă»ê±áŽâprincipiante
All this is shit
I canât even understand anything
And Iâve been doing non stop Duolingo for 64 days
And still nothing makes sense
âčïž
ahaha, take it easy :p this stuff takes time
what grammar rules have you been learning recently in duolingo?
CuĂĄl reglas de gramĂĄtica
đ I dunno what grammar stuff that includes, looks like general themes to me haha
like, you know all the present tense conjugations?
Yo bebo = I drink
Yo como = I eat
Tu bebes/Tu comes = you drink you eat
El Bebe/el come = he eats he drinks
Thatâs all so far
Thatâs like my only verbs I know
@chrome spear
ahhh
I wonder if there are lists of super common spanish verbs online đ€ there have to be lol
idk
it looks like right now, just expanding vocab would help
My friend tried to teach me verb endings
Itâs complicated though so I didnât really get it
because like, you can conjugate everything wrong but if you know the vocab, people will still get what you're saying
well, mostly :p not for the complicated sentences of course (no para las oraciĂłnes complicadas, por supesto)
Sometimes I make it sound like cave,an
Caveman
Like
I BUY CAR
I EAT APPLE
I DRINK WATER
I BUY APPLE AND EAT APPLE
like
Caveman talk
Also I heard itâs April fools
ah, but I suppose caveman talk can still transfer ideas ;)
Yo aprendo español - I learn spanish
Yo soy nuevo - I am new
đ
So I think I have the wrong role
Yeah but
Tmrw it will say Spanish native
And I, not
This is all so confusing and
you can change that pretty easily with bot commands, shouldn't be too bad
Itâs discouraging sometimes like
;iamnot spanish native
Iklaendia#6756 You don't have that role
My friend is taking Italian
;iamnot english native
Iklaendia#6756 You no longer have the English Native role.
And he has like 200 days and alllllllllll the Duolingo stuff for Italian
And he knows better spanish than me
Even though he is learning Italian
oof, ok maybe don't do what I did
What did u do
I tried removing roles, it disabled chat for a moment haha
well if you click on your name, it still says english native
I just need to do more Duolingo
it's just green
Oh
so, tomorrow it'll all return to normal, probably
K
I dunno. What do you think is discouraging about it?
alguien me llamo?
discord me dĂo un ping pero no sĂ© en cuĂĄl canal, o sea que comente un error 
@real imp
It's Ping Eternal
Pingternal


Oh it's April fool's...
đ | Stephiroth leveled up!
no te molestaré ver anime :o
adĂos y gracias đ
I've had cake set aside to eat for hours and forgot it was here until now
y ahora se va a desaparecer 
alguien tenĂa cuenta en Taringa! desde el 2017 o antes y tiene una cuenta de Steam asociada al mismo correo?
tengo unos 30 intentos de acceso a una cuenta de Steam que no uso
no concretados por el 2FA
y a un amigo le pasĂł lo mismo 
la Ășnica lista en comĂșn en la que estamos segĂșn https://haveibeenpwned.com/ es de Taringa!

@gusty ermine https://www.youtube.com/watch?v=1aLingC__P0 this video can also help with the "absolute value of log with integration"
and that derivative of ln(x) is not quite 1/x
it's actually if f(x) = ln(x)
then f'(x) = 1/x {x is a real number | x > 0}
and if g(x) = ln(-x)
g'(x) = 1/x {x is a real number | x < 0}
so a trap question might be.
"what is the derivative of ln(x) at x = -1"
people may say. well, derivative is 1/x then thus at x=-1, derivative at the point is 1/(-1) = -1
However, that's wrong because ln(x) doesn't exist at x =-1, thus, the derivative at that point actually does not exist.
yes
red is original f(x) and blue is f'(x)
notice none exist on the left hand portion
so the derivative of ln(x) does not exist when x=o or x= negative number
that's great to know
yes, because how can you draw a tangent at those points
the derivative at a certain point is the value of the gradient of the tangent of a function at the certain point
so, the domain of the function is the domain of the derivative
legs eleven 
do you have to do a lot of manual factorization?
Wtf
I know this, don't know how to apply case 2 to this one
Thatâs so many problems haha
it's practice
I was never a big fan of practice
only way to learn
for me at least
im math dumb
practice is the fun part though
nobody likes having to memorize theory
hahahaha @shrewd pendant story of my life
I like grocking theory, not memorizing it.
Itâs a bit of a problem for me haha
as soon as I understand something I have a false sense of security that I know it already.
I never study or practice for things
I figure it out as I go
yeah its grok apparently
No I graduated
I was shooting from the hip, didn't know how to spell it lol
because in high school you don't really need to study
i didn't, at least
ah nice well, im not that smart
also maths is my weakest subject
ever
Well I donât think itâs that Iâm smart
I do work but not for studying
I prefer poring over the theories
guys i need help with that exerciseeee :(((
Lol sorry to derail the conversation
it ok
danke schön
Yeah I donât know how to solve it đŠ
where does that picture come from, the guide?
yeppers
@gusty ermine can show you the solution, which I don't want to look at until I'm finished
bc she has the answers sheet
no I mean, where does the guide come from?
my prof
ah ok
hmm, I have the solution, but I don't understand how to use the method
well I know that the constant factor is 3, so the multiplication of the two constants in the parentheses have to be 3, so the only option is 1 and 3.
exactly
and now im stuck on number 9
yeah that's basically what im trying rn bc its too hard
here are all the combinations
oh god give me a sec to understand that lmao
whats the result
there
how do i find it
(3x+1)(2x-6)?
https://discordapp.com/channels/243838819743432704/383377678385938432/695265186893922344
This method is awesome
isn't it the same as this one
ever since my teacher taught it one class at my algebra class, I use it
yes, it's the same method
cool
A core in all of these is to find the factors of the coefficient of x^2 and the constant. 
@median trail the one i use?
well i would've sworn i had already tried that combination
seems like i missed it
thanks fcatorce
breathe
respirĂĄ che
3*3 is not 6
hahhaa
omg little errors like that
yeah i hate this
boy they can kill
Iâm pretty sure we spent like a week on factoring
Iâm surprised you guys have to do it so much
im just slow
No itâs not that
this is a precalc class im taking at university bc im dumb
We just used the quadratic equation mostly
I have no idea how to solve these either and Iâve taken most of the math classes
you probably just forgot
I donât think they emphasized it in my curriculums
Yes but that only works for certain ones iirc
Lmao
we tried doing that one with panda yesterday but neither of us knew how to
You have to factor?
hi vers
yes
yes berban
(a + b) (a^2 - ab + b^2)
"the more safe way"

it's less trial and error.
Always practice safe factoring
you have to equal the outer product with the center one?
yes, that's the method they taught at my school.
you have to equal the outer product with the center one?
@real imp ?
do a*c first
then find the factors of a*c
find which pair of factors when added match with b
assuming quadratic in form of ax^2+bx+c
because in the 6x^2 -5x - 6
there were a total of 16 cases to trial and error
but when doing the "safer one". there's only 5 cases
so in this case a*c = -36
ignoring the negative sign for a bit
1*36
2x13
3x18
4x9
6x6
then slap a negative sign somewhere since it's -36... u see the 4x9 combo works
since -9 + 4 = -5
yeah idk what i said
thus can split -5x into -9x + 4x
6x^2 + 4x - 9x - 6
wow thats big brain
group in pairs
(6x^2 + 4x) +(- 9x - 6)
fiind highest common factor in each pair
2x(3x+ 2) -3(3x+2)
then further factorise
(3x+2)(2x-3)
technically if you "include negative", then there's a total of 10 cases. since you can slap a negative on either 4 or 9... but only one works
but i like to say there's 5 cases... because it's not that hard to do addition and subtraction in one go
@real imp i told you about t^3+1 a bit before on another day
@real imp question 14 is also similar
to t^3 + 1
x^3 - 27
made up example:
x^3 - 8 = ?
a^3 - b^3 = (a-b)(a^2+ab+b^2)
x^3 - 8 = x^3 - 2^3 = (x-2)(x^2+2(1)x+2^2)
x^3 - 8 = (x-2)(x^2+2x+4)
ohhh so i gotta raise it to the same power
are you learning how to factorize stuff ticko
yes epic
this all seems familiar and unfamiliar to me lol
general cubics are harder to factorise, if not impossible.
wdym by general cubics?
ax^3 + bx^2 + cx + d
well, there is a cubic formula... but it has like imaginary numbers. so it's never taught
ew
Actually the fact that multiplying (1-i)(1+i) gives you ((1)^2+ (1)^2) if I'm not mistaken blew my mind
yes.
i*i = -1
yeah that
hahaha
xd
yeah but when working with real numbers you can't get a sum of numbers that are both raised to the second power
unless you allow "non-integer numbers to be factorised in"
i am so sorry
hahahaha
like you can't get this when working with real numbers
x^2 + 1
because that's equal to (x-i)(x+i)
like factorise it
I mean
ah ye
Oh I see
if one of them is negative, does it still work
I guess so
since they are both squared
is this already factorised
then why are we taught that we can not factorise that? don't tell me it's because engineeringâą
because "it's not a real number" it doesn't exsit, it's not real
@real imp short answer: no
i only needed to factorise with complex numbers when doing complex analysis
@real imp no, when something is factorised is because it's equal to factors that you can not reduce anymore and they are multiplying each toher
contour integration e.g.
contour integration sounds familiar to mw
is it related to green's theorem?
or something like that
@real imp no, when something is factorised is because it's equal to factors that you can not reduce more and they are multiplying each toher
@median trail so like
how do i get the factor of those two parts
@real imp recall the "special case" in ur picture
ok 1s
it's the difference of two squares
expand that out to see if you got it correct
9*9=???

81
im
confused
oh
uh
yeah
(2t + 3s) (2t - 3s)
whoops
but like then this formula is wrong
it should say sqrt(a) and sqrt(b)
no.
4t^2 - 9s^2 = (2t)^2 - (3s)^2
the minus is outside

(a+b)(a-b)=a^2-b^2 a=4t, b=9s (4t)^2-(9s)^2=16t^2-81s^2
2*2 is 4
yeah
(2t)^2=2^2*t^2
(ab)^2 = a^2*b^2
yeah
What does ^ mean
is there a difference between (a * b)^2 and a * b^2?
yes, huge difference
yes
if inside brackets, the WHOLE thing needs to be squared
idk
you distribute the exponent power to both variables
if inside brackets
only the b gets powered
yeah
hm ok
a*b^2 = (a)*(b^2)
ok ok got it
thanks
so in
x^3-27
the factorization would be
(x-3)(x^2-3x+9)
right?
cube difference
https://www.youtube.com/watch?v=LmbqB_pfcDU here is a 4hour 37min video of factorising 100 quadratics
blackpenredpen?
yeah
hahaha
ive seen that guy do integrals for
cant remember how much
(dunno how to do integrals)
oh i get it
he uses a black pen and a red pen
xd
oh wow this is a cool way to present it
contour integration sounds familiar to mw
@median trail sorry. got buried. greens theorem tiene alguna relacion
ok so
i got that x^3+2x^2+x = (x^2+x)(x+1)
just by looking at it which is cool and all
but i dont know how i got it
is it any different from grade 2 ones?
x^3 + 2x^2 + x has a common factor of x
x(x^2+2x+1), then now you have to factorise that quadradic
x(x+1)^2
with your answer.
x^2 + x can be factorised further
so x(x+1)(x+1) = x(x+1)^2
how u arrived at (x^2+x)(x+1), i have no idea.
oh so
its not factorized yet
how u arrived at (x^2+x)(x+1), i have no idea.
@abstract scaffold idk but it equals it
yes
alrighty
I see how you arrived at x^2 + x and x+1
not the normal way to think about cross method, but ahhaha
xd
A short story of a talented player in SA:MP.
more episodes soon, maybe
How can I find a number value for x when the equation is 2x^2-8x
I think the question is wrong
at a guess I would set it equal to zero
wait no
I'm retarded

did you figure it out?
minimum is at derivative = 0
you have to check the sign before the point that makes the derivative zero and after
if the signs are + and then -
you have a minimum
wait
the other wait around
if the signs are - and then + you have a minimum
Yep
x = 1 works
?
we call it algebraic divison
(x-1)(___) = x^3 - 4x^2 + 5x - 2
to work out what it is in the blank bracket we do something called algebraic division
should i use ruffini
tickoo
@real imp yes, absolutely
tickoo
@fathom flame ?
hi ticko
hi
@real imp ok. for cubics. in the form of
ax^3 + bx^2 + cx + d
test values of the form h/k
where h is factors of a
and k are factors of d
I was too tired to even see there were cubics 15 and onwards haha
so in the case of 16.
it's d = -2, a = 1
factors of d are {-2,-1,1,2}
factors of a are {1,-1}
thus, values to test are {-2,-1,1,2}
test x = -2
(-2)^3 -4(-2)^2 + 5(-2) - 2 = -8-16-10-2 which is not zero
someone in the engineering server had me do the full form of the cubic thing
thus (x+2) can't be a factor
forgot the name
and i got to the result
but i can not do that on an exam
its just too long
this is what i did apparently
nice nick
if it's polynomial division, then that just takes time to get used to.
yeah ok so now im trying to do it by polynomial division
sent me this website but i haven't been able to understand it yet
like i know how to do ruffini
i just dont understand how to apply it here
what i do know is that there is an evident root
which is 1
ok. so when testing out all the points. you see x = 1 works
because a + b + c + d equal 0 or 1
thus (x-1) is a factor
ok.
the rest of the possible rational factors {-2,-1,2} didn't work
yes, because we set P(x) = 0
yeah i just did it because i was told to before
so that if P(a) = 0, then (x-a)(x+....).... = 0
the rest of the possible rational factors {-2,-1,2} didn't work
@abstract scaffold how and when did we prove that
ok yes
so possible factors of d {-2,-1,1,2}
possible factors of a {-1,1}
d/a = {2,1,-1,-2,-2,-1,1,2}
getting rid of repeat cases d/a = {-2,-1,1,2}
a = 1 because its the only way that the first factor is 1, right?
like, the x from ax^3 is x1?
so thus (x+2) not a factor
(x+1) not a factor
(x-2) not a factor
sorry, ax^3 = x1
we figured out (x-1) is a factor just by setting the cubic equal 0 and tested x=1 as a possible point
@real imp the method will make more sense on question 19
yep. and the others ones didn't so we NOW HAVE TO use polynomial division
why can't we keep going on guessing?
do we always stop at +-2?
or do we always stop at d?
LMFAOOO
we always stop when we exhausted the list of rational factors
but -2/3 for example is still rational
or or
or
uh
nkjgdjkjkdjkbnfkskdsldjslkjflsfjs
I DONT GET IT AAAAA
ljasdlkas
why did i have to be born so slow
ok i get that formula
what i dont get is
what limits x
to {2,1,-1,-2,-2,-1,1,2}
ohhhhhhhhh
so its
ok so
x = d
right?
its xd-1
and since its the denominator
it has to be diff from 0
wait
x not equals d
wahjkfhsdkfhsdkj
in this case. factors of d and factors of a
d = -2
ignore negative for a sec
ok d' = 2
factors of 2 are {-2,1,1,2}
yes
now factors of a
a = 1
{-1,1}
ok
take each term from {-2,-1,1,2} and divide it by the first term from {-1,1}
that is {-2/-1, -1/-1, 1/-1, 2/-1}
complete the list by taking each term and dividing by the second
-2/-1 = 2
-1/-1 = 1
1/-1 = -1
2/-2 = =-2
-2
-1
1
2
and get rid of redundant terms.... when it's a = 1.... we see this is fully redundant
Yeah
so u see in the end, list of x values to try are {-2,-1,1,2}
and when trying them, only x = 1 worked
only x=1 results the cubic to be zero
thus (x-1) is a zero
ohhhhh
so factors
are roots
well that would've been a nice thing to know two days ago when i started lol
then u have to do this shit, where i totally didn't make any mistakes along the way.....
ah ok
yeah
ok, so polynomial division isn't needed
1 is actually a double root btw
(x-2)(x-1)(x-1) are the factors
yep.
@real imp when u find two factors, the last one you can find
it was a looooong ride
so, x = 1, x = 2 worked
so (x-1)(x-2)(x-a) = 0
but (-1)(-2)(-a) = d
so -2a = 2
so a = -1
x(-1-2-a) = 0 right
if you expand out that cubic. you see those last three constant terms will be multiplied together to give "d"
try expanding (x-1)(x-2)(x-a)
(x-1)(x-2)(x-a) == (x^2 - 3x + 2)(x-a) =
= (x^3 - 3x^2 + 2x -ax^2 + 3ax + 2a) =
= (x^3 - x^2(3 + a) + x(2+4a))
yep. try expanding it in the form of ax^3 + bx^2 + cx + d
and you'll hopefully see what I mean
you made a slight mistake on the second line
(x-1)(x-2) expands to (x^2 - 3x + 2) {just like i found in my polynomial division}
so, the -3 should be +2
well, this is how it'd partially look like if done using division
but if 2 roots can be found, division can be avoided
ok so
(x-1)(x-2)(x-a) == (x^2 - 3x + 2)(x-a) =
= (x^3 - 3x^2 + 2x -ax^2 + 3ax + 2a) =
= (x^3 - x^2(3 + a) + x(2+4a))
right
i will go make dinner
don't forgeting the +2a at the end
cool, go do that
sorry
literally the only thing wrong now is the +2a term.
(x^2 - 3x + 2)(x-a) expands to
(x^3 - 3x^2 + 2x -ax^2 + 3ax - 2a)
a more general form can be found for the product of the roots {and the sum of the roots}
so another way to appraoch it is to use the product of roots formula or sum of roots formula
roots are x_1 =1 and x_2 =2
thus 1 + 2 + x_3 = -b/a = -(-4)/1 = 4
1 + 2 + x_3 = 4
=> 3 + x_3 = 4
x_3 = 1
so roots are x_1 = 1, x_2 = 2, x_3 = 1
thus factored form is:
(x-2)(x-1)^2
alternatively (product of roots)
x_1 = 1, x_2 = 2
(1)(2)*x_3 = -d/a = -(-2)/1 = 2
(1)(2)(x_3) = 2
2(x_3) = 2
x_3 = 1
thus roots are x_1 = 1, x_2 = 2, x_3 = 1
(x-1)^2(x-2)
@real imp the method we're using now for cubics works in general for quartics or even any polynomial of order n.
are we allowed to share server links? there is a math server im apart of that specializes with helping people
@wheat ravine You can if you ask an admin first, and since I am one, then yeah go ahead
đ I want to see
i almost feel like this server in the math one go together
i really love it how with discord i can find and talk to people who know certain subjects crazy well
it feels amazing to be able to just find people who know what they're doing with things and get them to explain stuff
Ah I already joined that server
Hi can someone help me with algebraic fractions(?
I can tomorrow after 12 pm
Ok
u can join that server 
algebraic fractions 
what du need help with?
hello
Hi F14
Theres an exercise titled "fracciones algebraicas" and the first exercise is just
2x+5/5Ă+3
And idk what im supposed to do. Simplify them?? Divide??
The teacher isnt online so :(
erm... i cant really think of anything if there's no instruction
I did polynomial division, but that it ends up not too pleasant
thanks for trying
Guess I'll just wait for him to get online tomorrow
other things i could possibly do to it is integrate or differentiate it. but there's no instruction, asi que no se que deberia hacer
@eager wagon qué significa la bufanda?
Es del autismo.. es el mes de concientizaciĂłn del autismo
bien. No lo sabĂa
What is broadband
it's a very wide elastic band
broadband = internet, Epic *
so uh
my gf did this without even blinking twice
she solved in 5 minutes what it took me 3 days what it took me to understand
What is broadband
@rain ermine the maximum internet speed capacity.
for example, if you have idk 1GB/s broadband
two people can use about 500MB /s
four people can use about 250MB/s
Oh
@real imp if you think when you find 1 root by hand and can do synthetic division just fine afterwards, stick with that.
I tend to make errors with polynomial division, so I try to exhuast my list first.
idk seems like she first used the constant
and then the evident root
which makes a lot of sense
since
(thing)+constant=0
(thing)=-constant
therefore 0=-constant
yeah. she used division each way through
thats nice
no, it works for all. it just mindfucking for me, because i'm not used to it.
No you're right
It works to check whether or not something is a zero for any factor
But to actually divide and get the dividend, it has to be a linear factor of the form x-a
i never learnt it in high school. wish i did. it's so much quicker.
plus it's technically less error prone since your adding rather than subtracting
i'll probably still stick to normal if it's not a monic
Yes but it's limited in application due to only be usable with a linear factor of the form x-a
mhmm, so i still have to have the traditional way in my toolbelt still
... course
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#1079276 đŁ xD
Puzzle
@copper fox
#1069850 đŁ @copper fox
What?
... xd
#1068482 đŁ @alpine cradle
.. xd bleh
Puzzle#3740 Quote Added
... xd
#1079437 đŁ bleh
Eso?
I have no idea what you want me to do
And i dont think i have powers to delete things in this channel anyway
you've created it so you can delete it haha
No quotes found which you can remove.
... xd
#1068482 đŁ @alpine cradle
Oh
Quote #1069850 deleted.
. qdel 1068482
sin espacio
.qdel 1068482
Quote #1068482 deleted.
Lo siento
đ”đ±
Tarkin * ?
(x-1)(x+4)(2x+1)
can I go further?
i tried solving the last equation
2x+1 = 0
2x = -1
x = -1/2
but when i calculate
(x-1)(x+4)(x+1/2)
which doesn't give me the right result
Notting Hill intesifies
the original polynome is
2x^3 + 7x^2 - 5x - 4
(x+1/2) or (2x+1)?
yeah exactly
i dont know if i can just leave it at
2x+1
or if i have to go further
if i leave it at 2x+1 the result is correct
@real imp you're missing the 2: (x-1)(x+4)2(x+1/2) if ur tryna to do that but its already factored in
i mean you're trying to factor the 2, right?
mm you can use ruffini
that's what i did
... thats what i did bruh
that's
what i did
i mean, i used ruffini but its the same
entendiste lo que pregunte?
estas hallando una de las raices (ceros)
@fathom flame y bueno
Los factores son las raĂces
not exactly, each factor helps you get the zeros
ok but in (x + a), -a is the root
yes
Because
x+a = 0
x = -a
this is okay
@median trail that was my question, thank uuuu
@median trail if you have time, can you explain this to me? I don't even understand the question
This must be calculus then?





