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!
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!
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 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
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.
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.
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!
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)
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!
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 🇨🇷
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?
*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.
-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
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 !
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.
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!
And for Southtowns América pearsons :(
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)
Yes
I would have done that, too
Exactly what I was thinking!
Just use the quadratic formula 🤣
You really understand my needs as a calculus student. Factoring are one of most obstacles I get stuck at.
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!
for the 1st question i personally wouldve just polynomial long divide after factor theorem i didnt even know this sht is possible thxs
Thanks
It's really great and your solutions are really intelligent😯 I liked the "double cross method" it's so useful👍🏼
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
why, we will have u^2-2^2?
@@johnnykrocker5604 yes, that's the point
Nice
@@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
@@Arbion26 ohh, yes! I already see this. Thanks bro.
Math is really satisfying when you get what's going on
This video is very powerful
For Q1, how do you know to choose 5 and 3 rather than 1 and 15 ?
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.
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?
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.
That was perfect
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!
Wow
How did (x+2)u + u^2 + 2u = u (u +x+ 4) ??
Shouldn’t it be u(u^2 + 2u + x + 2) ??
I think I got it, yes the answer is correct, steps are right too but it skipped like 3 steps so I got lost.
Props for the donation!
Thank you.
Thank you sirrr :)
I knew someday we'd be double crossed by blackpenredpen.
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)
That was great! I like the third problem best :)
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!
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 🇨🇷
Can I do the cross method for cubic formula?
Hello, I am new subscriber. I saw your graduation picture and was wondering at what age did you graduate from UC Barkley😊
The cube trick really blew my mind! Thanks
my patreon halted because my cc expired. I'm going to fix that today!
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
There’s a BMT online. U can check that out. 😃
These are some great techniques, thanks!
From Morocco 🇲🇦🇲🇦🇲🇦
I fellow you just to understand how est stupids were somes of my math's teachers🤣🤣🤣
Did u come to morocco?!
What happened to the "main dish" video?
Your expressions being all proud about Makinh up your own methods makes me so happy😭I truly wish to achieve that soecific achievement
GREAT!!! Saludos desde Chile
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)
Not as quickly for a math tournament
@@DaMoNarch91 it takes 2min
Why couldn't you "just express it" as the product of a linear and a cubic? That requires some explanation.
All the expo markers in the bottom right corner of the screen waiting for their turn to be used
Nice video as usual
Thanks for teaching me this amazing method
Regarding the BMT are the scores race adjusted? Or is it fair where ONLY the work counts?
I am not sure how they will score this time. You can find their contact info in the description
This is the youtube channel Steven He's dad wishes he ran.
Very nice. Finding on your own is the best.
Very useful video 😁
10:57 how did one (x^2-2x+4) go away? I’m so confused
Double cross method is greatest method ever
whats annoying is ive been subbed for a while but literally never see your work in my feed.. my brain has been missing this
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?
*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.
Excellent point which detracts from the viability of the technique, especially for large constant terms.
The cross method gives you a right result... This means that the double-cross method actually screws you up? (Haha)
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.🎉🎉
中文亂入,看無,看無。
前三名有錢?真是不好。賽馬?
鼓勵童學參加挑戰,因資金有限只能提供前三車馬費,會不會比較好。
另,比的是解題速度?解題邏輯與方法無法評斷?舉個不可能例子,如果有學生把簡單題寫成像愛因斯坦般申論題如何辦,一張紙不夠寫?
歡迎一起來贊助喔 這樣你要怎樣分配都可以
In the first method/problem how did you know how to split the 15?
Basically, never use 1 times 15 unless there is no other choice like 1 times 3 or something.
Double-cross method?
Treacherous.
Thank you sir!!!
Good luck with the tourney, may the best wins!!
Was a square sign missed in the 3rd (blue) expression in the thumbnail? (Sorry not sure if it is mentioned).
You are right! thank you.
Hey there. Can you explain to us what tetration, pentation, hexation, etc is? 😈
Equation zaddy is back 😶
Mr. Burp is an absolute whiz!
I will also say hi can cheer you up too
2:33 shoudnt that be -2 and -3 ? cuz X1+X2=-p ?
-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
We are proud of you for figuring that out bro
I run the math club at Acalanes high school in lafayette. We will be there!
Excellent!!! See you there!!!
Very useful content
Ty man. Ur top tier
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 !
Imagine facing bprp in a math tourney you'd get smoked 10 different ways (with an elegant proof for each)
try x^15+1, it's very tricky but fun.
Interesting, what's your final answer?
WFA gives this: www.wolframalpha.com/input?i=factor+x%5E15%2B1
@@blackpenredpen Just as WFA gives it.
The point is that extracting x²-x+1 requires finding the roots of x¹⁰-x⁵+1.
He's really happy because he made those equations 😭💗
Fantastic
These are olympiad question!!!!😂
no
Great👍
The chad answer: Ferrari's method every single time
Hi
Buen video.👏🏽
Sir find the factor of the equation X ^ 4 - 4 x cube + 8 x square minus 8 x + 4 equal to zero
Thank u sir but my brain popped😅
whoa i did this individully at home!!! 2:50
for the second you can use difference of squares
thanks teacher this was very nice i am impressed especially last one.
Nice tricks! Factorization is an art.
I agree
Why we factored 15 as 3 into 5 not 15 and 1
🙂
How about x^100+1? Lol.
Great video...i liked it 👍👍
really great sir 🙏🏻😊
thank you sir!
Splendid, Thanks.. Can these methods become standard for resolving quartic polynomial
Absolutly
How’d we know the first one wasn’t a product of cubic and linear instead of two quadratics
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.
@@ultimatedude5686 that’s true yeah thanks!
This is because all cubics have at least one real solution
I forgot about that
I am from India
I am not from India
Me too but no one asked bro
@@chessematics I did 🗿
Pucha kisi ne
@@user-et1up1nk9k Where are you from?
有名次有錢拿哦
對
"because i made this" LOL
I think basic factor theorem would be a lot more straight forward 😂
What’s basic factor theorem?
😂
I think this is just guess and check for zeroes right? Then you can reduce the degree of the polynomial.
@@owlsmath yeh I just couldn’t be bothered to explain it 😂
@@Dylan-oq8xb I googled factor theorem and then it all made sense 😃
ooorrrrr just try and guess one solution then long divide?
Missed chance to use fibonacci numbers for the prize money :)
🍉
Please do some lebesgue integrals
NO, PLEASE DONT
@@hydropage2855 haha just don't watch it 😅
Request: how are trigonometric values of angles beyond 90° found?