if you are a high school student living in the NorCal area, then you should definitely consider participating in the Berkeley Math Tournament on Saturday, November 5th, 2022. Please visit www.ocf.berkeley.edu/~bmt/ for more info. I will be going to BMT as a guest. We (meaning all my subscribers and myself) will be sponsoring BMT so they can offer scholarships of $1000 to the first place, $500 to the second place, and $300 to the third place. Big thanks to all my subscribers and special thanks to all my patrons and channel members. I know this is not possible without you guys! Also, Shout out to the whole BMT team (all the organizers and problem writers)! I can’t wait to meet everyone there! If you are also interested in sponsoring BMT and making an impact, please contact communications@bmt.berkeley.edu Thank you!
@leonardobarrera2816
2 жыл бұрын
And for Southtowns América pearsons :(
@wowZhenek
2 жыл бұрын
For the second method it is arguably better to do this substitution: u = x^2 - 2x + 4, so you end up with (u-1)*(u+1) - 3 = u^2 - 4 = (u - 2)*(u + 2)
@tbg-brawlstars
2 жыл бұрын
Yes
@fiathla
2 жыл бұрын
I would have done that, too
@spikypichu
2 жыл бұрын
Exactly what I was thinking!
@adityaraajsingh4322
7 ай бұрын
Just use the quadratic formula 🤣
@rollingdigger19
Жыл бұрын
You really understand my needs as a calculus student. Factoring are one of most obstacles I get stuck at.
@MyOneFiftiethOfADollar
Жыл бұрын
Thanks for being honest about how the problems you create are "designed" to work out nicely if one applies the correct factoring technique. Lotta Love!
@o_s-24
Жыл бұрын
Math is really satisfying when you get what's going on
@fafa47t
2 жыл бұрын
Thanks It's really great and your solutions are really intelligent😯 I liked the "double cross method" it's so useful👍🏼
@timeonly1401
2 жыл бұрын
For #3: I also made the sum of cubes & factored, x^3+8 = (x+2)(x^2-2x+4) = (x+2)*u, where u = x^2 - 2x + 4. Then I rewrote the squared quadratic factor terms of u: (x^2-2x+5)^2 = [(x^2-2x+4) + 1]^2 = (u+1)^2 = u^2 + 2u + 1. The expression becomes: (x+2)u + (u^2 + 2u + 1) -1 = (x+2)u + u^2 + 2u = u ( u + x + 4) = (x^2 - 2x + 4)(x^2 - x + 8). Done!
@samueljehanno
10 ай бұрын
Wow
@jin_cotl
3 ай бұрын
How did (x+2)u + u^2 + 2u = u (u +x+ 4) ?? Shouldn’t it be u(u^2 + 2u + x + 2) ??
@jin_cotl
3 ай бұрын
I think I got it, yes the answer is correct, steps are right too but it skipped like 3 steps so I got lost.
@prefabrication
9 ай бұрын
I've always had a problem with factoring as if I can't visualize it well. The second substitution method (specifically U-substitution) is a brand-new favorite of mine. Thank you very much!
@tedkaczynski2727
2 жыл бұрын
in the second one i considered to use u =x²-2x+4 so we will have (u+1)(u-1)-3 and we will have u²-2² :)) and then we will have (x²-2x+2)(x²-2x+6) easily and fast
@johnnykrocker5604
2 жыл бұрын
why, we will have u^2-2^2?
@gdtargetvn2418
2 жыл бұрын
@@johnnykrocker5604 yes, that's the point
@aayushswami3022
2 жыл бұрын
Nice
@Arbion26
2 жыл бұрын
@@johnnykrocker5604 because after we factor we'll have u^2-1-3 Simplify u^2-4 Rewrite 4 as 2^2 u^2-2^2 Now you can factor (u-2)(u+2) Now you should resubtitute, but im to lazy to do that
@johnnykrocker5604
2 жыл бұрын
@@Arbion26 ohh, yes! I already see this. Thanks bro.
@coltonboxell1960
2 жыл бұрын
The cube trick really blew my mind! Thanks
@Swiftie-d2u
2 жыл бұрын
for the 1st question i personally wouldve just polynomial long divide after factor theorem i didnt even know this sht is possible thxs
@reddestglizzest
Жыл бұрын
Your expressions being all proud about Makinh up your own methods makes me so happy😭I truly wish to achieve that soecific achievement
@holyshit922
2 жыл бұрын
x^4-4x^3+12x^2-16x-15=0 Firstly group the terms with x^4 and x^3 and rest (x^4-4x^3) -(-12x^2+16x-15)=0 Now we complete the square the expression in leftmost bracket (x^4-4x^3+4x^2)-(-8x^2+16x-15)=0 (x^2-2x)^2-(-8x^2+16x-15)=0 Expression in the other bracket is quadratic and will be perfect square when its discriminant is equal to zero If we try to calculate discriminant now it may appear that discriminant is not equal to zero so we have to introduce parameter to make discriminant dependent on it We introduce parameter in such way that expression in leftmost bracket remains perfect square (x^2-2x+y/2)^2-((y-8)x^2+(--2y+16)x+y^2/4-15)=0 Calculate discriminant and force it to be equal to zero (-2y+16)^2-4(y^2/4-15)(y-8)=0 (y-8)(y^2-60)-4(y-8)(y-8)=0 Cubic resolvent is partially factored and is easy to see that y=8 is solution (x^2-2x+4)^2-1=0 (x^2-2x+3)(x^2-2x+5)=0 x_{1}=1-sqrt(2)i x_{2}=1+sqrt(2)i x_{3}=1-2i x_{4}=1+2i This method for quartic is quite easy and never fails but usually needs to solve cubic equation (so called cubic resolvent)
@KazACWizard
2 жыл бұрын
First one you can just express it as f(x) and set it equal to the product of two quadratics. Expand coefficients and equate them to f(x)
@DaMoNarch91
2 жыл бұрын
Not as quickly for a math tournament
@KazACWizard
2 жыл бұрын
@@DaMoNarch91 it takes 2min
@MyOneFiftiethOfADollar
Жыл бұрын
Why couldn't you "just express it" as the product of a linear and a cubic? That requires some explanation.
@EulersEye
4 ай бұрын
We are proud of you for figuring that out bro
@owlsmath
2 жыл бұрын
That was great! I like the third problem best :)
@spikypichu
2 жыл бұрын
These are some great techniques, thanks!
@Enormousguy
2 жыл бұрын
Thanks for teaching me this amazing method
@reidflemingworldstoughestm1394
2 жыл бұрын
Very nice. Finding on your own is the best.
@SuryaBudimansyah
2 жыл бұрын
Good luck with the tourney, may the best wins!!
@bryanjara477
2 жыл бұрын
Mr. Blackpenredpen I really enjoy your videos and it blows my mind how you solve things, it’s just awesome!! Do you have any resource for me to practice this types of factorization? Greetings for Costa Rica 🇨🇷
@dosedemaths
Жыл бұрын
@finrodf8442
Жыл бұрын
for the second you can use difference of squares
@SidneiMV
5 ай бұрын
Nice tricks! Factorization is an art.
@jin_cotl
3 ай бұрын
I agree
@hazelqawamid
10 ай бұрын
He's really happy because he made those equations 😭💗
@bananaman4705
2 жыл бұрын
This video is very powerful
@w__a__l__e
Жыл бұрын
whats annoying is ive been subbed for a while but literally never see your work in my feed.. my brain has been missing this
@SuperYoonHo
2 жыл бұрын
Thank you sir!!!
@BraddersVee
2 жыл бұрын
For Q1, how do you know to choose 5 and 3 rather than 1 and 15 ?
@robertpearce8394
2 жыл бұрын
As the video states, it is a guess but an educated guess as 5,3 looks mote likely than 15,1. You can try 15,1 but the answer will be incorrect and time is wasted as this is a competition. If you apply this method to problem 3 you have choices of 32,1 or 16,2 or 8,4. I tried 8,4 first as it seemed most likely.
@Wondering_human
2 жыл бұрын
Nice video as usual
@davidbrisbane7206
2 жыл бұрын
I knew someday we'd be double crossed by blackpenredpen.
@chadd9907
Жыл бұрын
thanks teacher this was very nice i am impressed especially last one.
@comingshoon2717
2 жыл бұрын
GREAT!!! Saludos desde Chile
@dandark5554
2 жыл бұрын
I'm sorry to be stupid but i don't get one thing: Why didn't you consider 15 as 1x15 in the first method and went straight on 3x5?
@Sampreet.
2 жыл бұрын
I guess one clue would be to look at the coefficients of the terms. They all have factors of 2 or 3 and 2 + 3 = 5 which is a factor of the constant term. These patterns suggest that it might be easier if we were to proceed with 3 * 5 instead of 1 * 15.
@MathwithMarker
2 жыл бұрын
That was perfect
@numericalcode
Жыл бұрын
Props for the donation!
@blackpenredpen
Жыл бұрын
Thank you.
@hardeepsinghg13
2 жыл бұрын
Will you conduct any online competition in future for students not in your country? Like India Pls try doing it some day, for Grade 12 students
@blackpenredpen
2 жыл бұрын
There’s a BMT online. U can check that out. 😃
@brendanconnelly1075
2 жыл бұрын
I run the math club at Acalanes high school in lafayette. We will be there!
@blackpenredpen
2 жыл бұрын
Excellent!!! See you there!!!
@davidemasi__
2 жыл бұрын
Very useful video 😁
@اشکانمحمدی-ز1ث
3 ай бұрын
Very useful content
@Madtrack
2 жыл бұрын
Ty man. Ur top tier
@NarenderKumar-ll6xg
2 жыл бұрын
Sir find the factor of the equation X ^ 4 - 4 x cube + 8 x square minus 8 x + 4 equal to zero
@@blackpenredpen Just as WFA gives it. The point is that extracting x²-x+1 requires finding the roots of x¹⁰-x⁵+1.
@christopertianzon3232
2 жыл бұрын
thank you sir!
@jaii5955
Жыл бұрын
Double cross method is greatest method ever
@giselleformon
2 жыл бұрын
Thank you sirrr :)
@pragyapathak3918
2 жыл бұрын
really great sir 🙏🏻😊
@greensalad_1205
8 ай бұрын
Can I do the cross method for cubic formula?
@Bplayer1113
2 жыл бұрын
Hello, I am new subscriber. I saw your graduation picture and was wondering at what age did you graduate from UC Barkley😊
@niom9446
Жыл бұрын
Fantastic
@FreemonSandlewould
2 жыл бұрын
Regarding the BMT are the scores race adjusted? Or is it fair where ONLY the work counts?
@blackpenredpen
2 жыл бұрын
I am not sure how they will score this time. You can find their contact info in the description
@jin_cotl
3 ай бұрын
10:57 how did one (x^2-2x+4) go away? I’m so confused
@charlesgoodson5774
4 ай бұрын
Sir, I was wondering, if you have a quartic polynomial with integer coefficients, is it always factorable into the product of two quadratic polynomials (even if all roots are complex OR irrational real)? If so, will the coefficients in those quadratics be integers necessarily or might they be non-integer rational numbers?
@alejrandom6592
2 жыл бұрын
What happened to the "main dish" video?
@subodhmeena9918
2 жыл бұрын
Great video...i liked it 👍👍
@fabiangn8022
2 жыл бұрын
Buen video.👏🏽
@sayanjasu
2 күн бұрын
This is the youtube channel Steven He's dad wishes he ran.
@garyhuntress6871
2 жыл бұрын
my patreon halted because my cc expired. I'm going to fix that today!
@vishalmishra3046
Жыл бұрын
*First Technique is not Generalizable* The first technique ONLY works when you predict the correct factors of the constant term. What if it is not an integer in the equation. What if its factors are not integers (e.g. 15 = 9 x 5/3). If the constant term has many factors (large N), then you will discover these issues after a very large number of attempts (Number of pairs of factors = N x (N-1) / 2 attempts). And then the same number of attempts with negative factors (e.g. 15 = -3 x -5). That's a lot of factor-pairs attempted just to eventually find that this technique will not work.
@MyOneFiftiethOfADollar
Жыл бұрын
Excellent point which detracts from the viability of the technique, especially for large constant terms.
@charlesbromberick4247
Жыл бұрын
Mr. Burp is an absolute whiz!
@enjoywatchingyoutube3346
2 жыл бұрын
2:33 shoudnt that be -2 and -3 ? cuz X1+X2=-p ?
@mcguides9176
2 жыл бұрын
-2 and -3 are the roots but this is dealing with factorizing the polynomial is (x-(-2))(x-(-3)) so when the x coefficient is positive the roots are negative but when you factorize it the negative negative becomes a positive
@cathaywongkt
2 жыл бұрын
Was a square sign missed in the 3rd (blue) expression in the thumbnail? (Sorry not sure if it is mentioned).
@blackpenredpen
2 жыл бұрын
You are right! thank you.
@EyeSooGuy
Жыл бұрын
Hey there. Can you explain to us what tetration, pentation, hexation, etc is? 😈
@ahmadsalehin8329
Жыл бұрын
whoa i did this individully at home!!! 2:50
@noelyvalisoarakotoarison7240
Жыл бұрын
Splendid, Thanks.. Can these methods become standard for resolving quartic polynomial
@dosedemaths
Жыл бұрын
@jin_cotl
3 ай бұрын
Absolutly
@alham9656
2 жыл бұрын
For the first method, does the answer need to end up being (u +- C1)(u +- C2) where c1,c2 are just constants? Ive tried workijg backwords with the u's not being equal and it doesnt really work out well, but for every one of tried in the form of yours it seems to work !
@tsaqifrizky5276
2 жыл бұрын
Imagine facing bprp in a math tourney you'd get smoked 10 different ways (with an elegant proof for each)
@crazybanana0272
Жыл бұрын
All the expo markers in the bottom right corner of the screen waiting for their turn to be used
@user-rx3iv7bl6b
Жыл бұрын
Why we factored 15 as 3 into 5 not 15 and 1
@sajjadali-taeminsteppedony8434
2 жыл бұрын
Equation zaddy is back 😶
@mantasr
9 күн бұрын
Double-cross method? Treacherous.
@michaelbaum6796
Жыл бұрын
Great👍
@VardiIntegral
Жыл бұрын
Is it true: The method which is applicable to factor a 4 degree polynomial is depend upon the way i represent it to the world.🎉🎉
@mahhar31
2 жыл бұрын
I am from India
@user-et1up1nk9k
2 жыл бұрын
I am not from India
@chessematics
2 жыл бұрын
Me too but no one asked bro
@guywhoasked6046
2 жыл бұрын
@@chessematics I did 🗿
@Aditya-zs8hz
2 жыл бұрын
Pucha kisi ne
@mahhar31
2 жыл бұрын
@@user-et1up1nk9k Where are you from?
@tomshimon
Жыл бұрын
In the first method/problem how did you know how to split the 15?
@jin_cotl
3 ай бұрын
Basically, never use 1 times 15 unless there is no other choice like 1 times 3 or something.
@afuyeas9914
2 жыл бұрын
The chad answer: Ferrari's method every single time
@CAFoundationExam
2 жыл бұрын
Hi
@user-et1up1nk9k
2 жыл бұрын
🙂
@killing_gaming0973
Жыл бұрын
I will also say hi can cheer you up too
@muhammedrehaan5750
Жыл бұрын
Thank u sir but my brain popped😅
@dosedemaths
Жыл бұрын
@lucasmontec
Жыл бұрын
The cross method gives you a right result... This means that the double-cross method actually screws you up? (Haha)
@olli3686
Жыл бұрын
"because i made this" LOL
@Happy_Abe
2 жыл бұрын
How’d we know the first one wasn’t a product of cubic and linear instead of two quadratics
@ultimatedude5686
2 жыл бұрын
Any cubic can be factored as a linear term and a quadratic term. If you have a linear term times a cubic, you can convert that to two linear terms times a quadratic. If you multiply the linear terms you get the original polynomial factored in terms of two quadratics. Therefore, all quartics can be factored as two quadratics.
@Happy_Abe
2 жыл бұрын
@@ultimatedude5686 that’s true yeah thanks! This is because all cubics have at least one real solution I forgot about that
@noobplayer5461
Жыл бұрын
These are olympiad question!!!!😂
@dosedemaths
Жыл бұрын
no
@youssefbenmMorocco
2 жыл бұрын
From Morocco 🇲🇦🇲🇦🇲🇦 I fellow you just to understand how est stupids were somes of my math's teachers🤣🤣🤣 Did u come to morocco?!
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