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FASTEST way to factor a trinomial!
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- Опубликовано: 18 дек 2020
- Original video: 4 popular ways to factor a trinomial: • 4 popular ways to fact...
This is called the "Slide-and-Divide Method" that I learned from my colleague, Prof. Tchertchian!
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#shorts
bprp x
The title looks so familiar 😂
Eyes, I copied it from some guy named Dr. Payment!
And I am still waiting for him to comment on this video so I can pin his comment. BUT I guess I will just pin yours instead.
Hahahahaha
😂😂😂
@@ViratKohli-jj3wj Virat Kohli 🥺🥺 aap bhi math krte hain Kya,
This really changed my life , I’ve never been so confused 💀
Different.
Lol 😂
Lol same
Jajajajaja *laughs in spanish
yh ikr haha
This tutorial needs a tutorial 💀
After doing some research, I figured out why it works the way it does. Firstly, you should know that the reason you're able to multiply AC is because of the nature/pattern of Quadratics. No matter the method you use, the factors you generate will always be factors of AC.
Secondly, this method works by compensation. When you multiply AC, you're actually increasing the constant by whatever the leading term is. This also means that the factors you generate are going to be as many times as large (based on whatever the leading coefficient is. As such, you have to divide by whatever the leading coefficient is to compensate for this.
Now, I want to address by you swing up the leading coefficient when it can't be divided out. What you're actually doing is multiply the fraction by whatever the denominator is -- in order to put it in standard form. Technically, leaving it as a fraction is not wrong. The fraction and standard form are actually equivalent. I hope this helps.
Not work in everywhere
Actually quite simple once you get it. And it's much, much faster then the normal method's, at least for me.
@@rocketpencil5948 It is simple, once a person "gets it." However, one problem I continually see is that too many math will call things "tricks," don't explain the mathematical logic behind what they are doing, and just expect students to memorize steps. The longer approach isn't much longer than the shortcut, and you can actually see your work. Using area models are pretty quick as well.
ts so simple bruh ☠️
Literally been tryna figure this out for an hour and a half and this dude came in with a min video and randomly all makes sense💀
This is why you need to focus and learn the fundamentals.
Real💀 I was already thinking about life after failing my math exam
THANK GOD HE MADE THIS!!!!
I have some steps, label each trinomial as eithrr a b or c (going in order) multiply a and c factor the resulting expression, devide everything by a. If a term cannot divide by a slide a infront of the variable. Hope this helps
@@onikabuto5064 did you pass?
another method:
P = 4 (the coefficient in 1st term)
Q = -6 (the coefficient is 3rd term)
P × Q = -24
Try to find 2 numbers when added is equal to the coefficient of the middle term (which is -5) and if multiplied then equal to P × Q (-24)
You can do this by the process of elimination or do this in a calculator which is easier. (Scroll down to see how)
In this case it is -8 and 3 which adds to -5
And -8 x 3 = -24
So. 4x² - 8x + 3x - 6
and then 4x(x-2) 3(x-2)
Last step (x-2)(4x+3)
Too smart for me
yes, the middle term breaking
Yes ! It is how I learned it
THANK YOU SO MUCH
Yeah dont understand and im a junior in highschool with a math final tomorrow
It's a perfect method I have been using it since your 100 Trinomials video came out. Thank you for it!!
Amazing! : )
I've been using it from 6th, it's called splitting the middle term. Yes Im an asian🤣
@@zerksez9963 stop saying i am an asian on every comment only the most stupid people brag about their knowledge
@@aayushswami3022 Is splitting something to brag about. I was just lettin them know how much better asians specially indian people are then them in calculating and yet we are not recognised so much. I don't think splitting is something to brag about. Even simple Calculus isn't, I don't know, maybe u should fix ur Preassumptions bout someone.
@@zerksez9963 bruh thats literally bragging be a little bit humble brother yes its a stereotype that indians are bad at everything but we dont need to prove it to anyone
You just explained what took my teacher 2 weeks to explain to my class.
you have all of that background information from the two weeks of class and this is one small, simple method of an entire family of concepts
bruh what my teacher explained it in 1 class and everyone understood
@@maxpie16 my teacher spent a whole month on matrices. im serious.
@@_quixotewhy though you can learn it in a few hours.
@@maxpie16it depends on your teacher, mine barely teaches, she just did one class and assumes everyone understood, we always get 6-13 over 20 as our score because of that.
Better way and makes more sense:
Do A times C because quadratics are in the form ax^2 + bx + c
So 4 x 6 = 24
Factors that make up 24 and sum to -5 is -8 and 3, So rewrite original equation as
4x^2 - 8x + 3x - 6
Then factorise
4x(x-2) + 3(x-2)
Factorise again
(4x+3)(x-2)
Bro this looks good but is applicable always, huh?
@@Einstein_rocks yes, but of course when the numbers are very large like a * c is very large then it will be hard to find the factors
It's (4x+3), not (4x-3)
@@omarskigiza5289 oh yes youre right i misread the sign
how this work in 6a^2+13a+6😢
I have an undergrad degree in mathematics. I feel cheated that I never learned this! All my future students that I tutor will be thanking you for sharing this!
this is the intuition for the AC method
But I thought most schools teach this
Really? They usually teach the AC method in Algebra 1
Can you explain how or why this works? Like, what's going on under the hood here, how would I ever know to do this?
@@astralbeatz9950 A self contradiction!
What. What is this monstrosity. Who figured this out. My brain has completely deflated due to utter confusion.
Hi
@@alvinbowley4230 Hi.
@@RJ-sm2cg Math teachers arent there to teach the easy way, they are supposed to teach you so that you understand *why* the math works, if all they ever did was give you the formula that you plug things into, it would be pointless and you would leave the class having learned nothing
@@RJ-sm2cg Teacher is supposed to teach you how and why math works. You can find formulas in textbooks or on the internet, so there is no reason to memorize them. Btw everything up to 12th grade level is easy enough that if you take your time to understand it deeply, you won't have to know any formulas
@@TabbyVee You say that but were taught so many methods when we’re younger that we have to forget bc it was just an easy way of teaching.
That is much faster! That was awesome!
I'm going have to use this next time I have to factor. Thank you!
U can use split middle term
1. Multiply first term coefficient and last constant
2. Find a factor which products to that and sums to the middle ( keep in mind the signs)
3. Split the middle term by making product factors different (eg- x2-5x+6 factors of 6 are 1.6,2.3 and - 2+( -3)=-5 ie the middle term. Then write it as x2-2x-3x+6. Take common factor i.e. x(x-2)- 3(x-2) .Then group the terms outside and the common thingie inside.
(x-3)(x-2)
Hope I might be clear
They cut the main thing plz try to understand 🙃
Any *legal* proof for this?
Yes. (Hint : Quadratic Formula. Do you see that the discriminants are equal ?)
@@donaldbiden7927 he doesn't want just this concrete equation.
He wants to prove that the method works for every factorable quadratic equation.
He does it pretty similar to the way I know how.
Quadratic equation:
ax^2 + bx + c
Let 2 factors of a×c be p and q such that:
p × q = ac
p + q = b
ax^2 + px + qx + c
px(qx/c + 1) + c (qx/c + 1)
(qx/c + 1)(px + c)
(qx + c )(px + c)/c [qx/c + 1 = (qx + c)/c]
Using blackpenredpen method:
ax^2 + bx + c
write as :
x^2 + bx + ac
Let two factors of a×c such that:
p × q = a×c
p+q = b
Then:
x^2 + px + qx + ac
(x + p/a) (x + q/a)
Put a = pq/c
(x + cp/pq) (x + cq/pq)
(x + c/q) (x + c/p)
(qx + c ) (px + c) / pq
(qx + c) (px + c) / ac
ignoring a in denominator
(As bprp does in last step)
(qx + c) (px + c) / c
sloppy ending but is this good enough?
@@LetsSink you are the mvp
I love how he builds up hype and all in the end by saying "THIS IS IT!" and then just normally says "done"
Find 4 x -6 = -24
And then what two number you would add and get -5 and multiply them and get -24
-8 and 3
And boom
😎
I think I know why this makes sense. So basically, you take the quadratic equation formula, where it's and you can move it from a to c, and it doesn't change value under the square root, only decreases the denominator, so you can solve that by decreasing the values by the correct amount
Asslam o alikum. We need to find 2 values that multiply to -24 and add -5.
4x^²-8x+3x-6. So we use the grouping method.
4x(x-2)+3(x-2).
So the answer is:
(4x+3)(x-2).
I think he did something similar to the grouping method.
Have a good day.
So glad I ran into my algebra teacher at an adult store, she passed me with a C- just to make sure I didn’t have to retake the class with her again.
At an adult store? 😭
I'd show her my thanks if i were you, if you know what i mean ;)
omg what
Lucky
Math really is hard 😂
We were taught this method in our school in 6th grade. Yes I'm Asian 😂
Yep what's so interesting in that
Same here 😂
I came here looking for indian comments 😂😂
@Wayne Bhai yep man😂
@@anuragsinha3670 ah I see,a man of culture
That's called Splitting the Middle Term/Middle Term Break,learned it in 8th Grade, Super useful
Omg same it's also called middle term break in my area I learnt it in 7 grade
@@rc-shadow9691 are u from asia?
@@abd_sh_321 ye
@@rc-shadow9691 which country
I am from pakistan,we learnt it in 8th grade
SAVED MY LIFE ON THIS QUADRATIC QUESTION MAY THIS CHANNEL GO FAR.
A straightforward way for those not getting it is to have the 4 be brought down when factoring already, then reduce by Greatest Common Factor.
(4x-8)(4x+3)- GCF of 4
(X-2)(4x+3)-solve for x
I LOVE THIS! Why was I not taught this? So many hours lost to needless trial-and-error factoring and quadratic formulating that I’ll never get back. This is BRILLIANT.
BRO SAME. This is like genius!
i learned it when i first started to factor, however it doesn’t always work i don’t think. But completing the square is my go to
The fact that this is so much easier
Same 🤫🧏
your feelings are irrational
I learned the ac method. You multiply a and c (a is the leading coefficient c is the constant) and find the numbers that add/subtract to the product. Then you put them into a factored group with the leading coefficient and divide by the leading coefficient and then simplify. For example: 3x^2 -4x -15; ac: -45 -9 + 5; 1/3(3x - 9)(3x + 5); (x-3)(3x + 5)
The sound affects are what really helped me
Tysm I understood this the 1st time I watched it, and it saved me a lot of time for my super rushed math final (my teacher literally gave us 7 new topics to understand the day before).
… this is what i have always needed in my life
😮
So nostalgic. I remember doing a demo during Grade 7 of this method when I first studied in a western school. Damn I suddenly felt old.
After seeing some of these people that have went through college not knowing this, I'm so glad I'm learning this now, in my first semester of college.
This confused and amazed me at the same time💀 i remember 2 years ago me and my friend were doing a similar thing, she obviously knew so much more than me and she was the one who gave the main idea of it. But our teacher said that it just doesn't make sense and even said that she would have to make us leave the class if we don't respect the rules💀 i don't know where my friend is right now, but that just reminded me of her, she was a very kind, smart and lovely person! I can now practice this method too and get reminded of her every single time lol! 💕
A good memories !
Oh yeah I also remember when I was small I was damn scared when I was instructed to sit between two girls.
But after sometime we started eating lunch together, talking together. and solving some sum together and now sometimes it reminds of those days
Those good days when we were just innocent kid
@@xninja2369 wow that was so cute and wholesome! 🥰
@@thebox7673 thanks bro
Your one was wholesome too,
Hope you would be doing great :)
thats a bad teacher. y'all shoulda not been threatened for doing things a different way just cuz teacher didn't understand it
@@OpLapDancePikachu69 agree you
I legit felt my life changing 🙆♂️🔥
Bruh as an Asian why this is in my text book that I'm just reading now.
This looks like a very unique interpretation of Vieta’s formulas.
Is it? Would be interesting to see how someone could prove this works
@@nikey2110 I probably wrote this while I was half asleep in the middle of the night, because this is definitely not Vieta's formula xD
I was taught this in school, the moment I saw the title I figured he was gonna talk about the slide and divide method
‘Here’s a super easy way…’ *quickly explains something super complicated* lmao
I've been using this my whole life 😂
@@Animesh_Chaturvedi LoL f
are you a teen?
Same
Story of every indian
Does that means you do not know how to use the scientific calculator ?
I learned this in Algebra I at my school. We called it the A-C method, because you multiply the A and C terms of the quadratic.
One thing i think you missed which could be addressed with a different example is splitting up the A term when it comes to dividing the M and N terms (x+m)(x+n). If the A term doesn’t go into either of the terms, you can split up the A term into two factors that the M and N terms can divide by, so long as the other term is applied to the X terms. For example, the equation (6x^2 + 5x + 1). You can use this method, but that results in (x+3)(x+2) before the last step can be applied. However, because 6 doesn’t divide evenly into either of those, you need to split it up. Splitting the 6 into 3 and 2 and then dividing results in (x+1)(x+1), but that isn’t right, so we need to apply the other factor of 6 to the X terms. So, for the first factor, if I divided the 3 by a 3, I need to add a 2 to the X term because that is the other factor of 6 other than the one I divided the M term by. This results in (2x+1). Now I need to do the same for the other factor, so, because I divided the N term by 2, I apply the 3 (the other factor of 6) to the X term. This results in (3x+1). And now, (2X+1)(3X+1) is your correct factored form!
You are such a legend
Thank yoy
You: Make me confused
Khan Academy: Literally puts the information inside my brain💀
I love his enthusiasm, there are also so many other ways to factorize! :))
Asian kids attendance here!!!!
We legit are taught this in school😁
This was a 6th grade indian question 😂
We got this in 6th grade in India
Great. But can that make India a great nation? I think not 😅
I'm not Asian but I learned this in school I think everyone did except for Americans
What school did you go to, bruh 😂😂
I love this method with my heart. Once I learned this in Algebra 2 I never used anything else lol
i did a similar method where u find 2 numbers that multiply to 4*6 and add up to -5, then u write it as (4x - ?) (4x + ?), which would be (4x-8)(4x+3) and then u simplify by dividing 4x-8 by 4, so the answer is (x-2)(4x+3)
This dude literally just solved my decades of doubt in a minute
If I was taught this... it woulda been 30something years ago... but somehow I learned to factor differently. At this point, it would be good to be able to retain this newly learned method... but my brain is now cluttered with having to remember 400,000 online userids and passwords... so badly that I can't remember what I ate for breakfast.
Seems to me you need the help of a password manager.
@@Kettwiesel25 - Seems to me I need to embrace my inner Amishness 😆
now for a word from our sponsor
400,000 userids and passwords? what you do
400,000 userids and passwords? what you do
CRAAAAZY BROOO I’ve always hated factoring, and even my schoolmates, we’re so bad at it the teacher even mentioned it lmao.. but this is crazy, i finally got it!
I am almost never given trinomials with a non-1 coefficient for the first term, but if I ever am, I am prepared.
Haha I independently discovered this method more than two years before this video, when I was teaching my son. Too many "degrees of freedom" when dealing with two coefficients you have to factor and juggle, so I wanted to make things simple for my kid by transforming it to a monic (lead coefficient one). I was inspired by the "ac" term in the discriminant of the quadratic formula, so thought of multiplying. And the rest of it is the same as yours. :)
well I discovered this method 3 years before this video so I am a better person than you
@@MrZootSuitz Well, bully for you. Your medal's in the mail.
Bravo 👏🏾 I honestly would’ve needed this in college 😂 it would’ve saved me a bunch of headaches lol
I’ve been using the box method for as long as I can remember, I’ll be sure to keep this method in mind. It actually works pretty well too, even on some complex problems. Why didn’t this show up in my recommendations earlier?
I love this! That said, if both of the factors have a coefficient on x other than 1, it’s a little more complicated.
Usually you look for the factors in the first coefficient and the last, then look for the combinations to see which ones fit. Like:
4x^2-5x-6
Factor first coefficient: (1,4), (2,2), (4,1)
Factor last coefficient: (1,6), (2,3), (3,2),(6,1)
This one works since of the answer is in the form of (ax+b)(cx+d) the first coefficient would be ac and the last coefficient would be bd. The middle one would be (ad+bc). A cross multiplication.
And you try for the mixes that work(remembering the +/-) with example, if you try (1,4) and (1,6) you would get 6+4.
The right case is (1,4)(2,-3). Which means (x+2)(4x-3).
But why did that algorithm worked?
Well, first, we are multiplying the last coefficient to the first and making the first equals to one.
We preserve that the multiplication of both would be the same.
Now we have 3 cases, in the monic form, there may be one factor that divides by the first coefficient, or none, or both.
In the first scenario, if one factor can be divided by "a", we are dividing this factor with "a", but since the other one is not divided by "a", we are multiplying the monic by "a", which means the multiplication stays the same.
In the second example, we have 4x^2+4x-15.
By the method, we have. // x^2+4-60//(x+10)(x-6). Since both are not divisible by 4, the answer by the algorithm is (4x+10)(4x-6).
Which is weird. Since the result of the multiplication is different, it results in 16x^2. So we realize that we can factor everything with 2.
So the answer is 2(2x+5)(2x-3). Which is not the original answer, but it is double. If you only want to know the zero of the function it works. But be aware of that!
So, you always have some factor in common, because of the Aritmethic Fundamental Theorem, but not all.
So, if we compare with the explanation from above, we already have the multiplication right with 10 times 6. So what happens if we multiply both by 4? You are making it 4 times larger! But what happens next? We are dividing everything by 2, which means we are dividing the result by 4 again, that is why it also works.
This one is easier to prove because 4 is a square number, but I guess we follow a similar knowledge with other compound and prime numbers.
Didn't thought about the case where both can be divided, maybe it is impossible and we can prove it (if we put the additional conditions blackened pen put on one of the first comments)
I'm sorry I'm not solving for all cases, but I hope those insight helps someone have a better understanding on the method.
It is kinda funny because it came a bit natural to me. I have had some experience with polynomial equations with fractions as a result.
Better than my actual teacher who I pay to teach me lol
I actually did something a lil different...
4x²-5x-6
Focus on the factor for C
Factor for 6 is 2 and 3
The 6 is negative, so one of it's factor must be negative too
Then, just put 4 on the (x )(x ) skeleton
(4x )(x )
Then put the factors in
(4x+3)(x-2)
I am going into 10th grade and when I took my Keystone last year, I didn't know anything about trinomials, WHERE WAS THIS WHEN I NEEDED IT?! (I am SO glad, I get to retake my Algebra Keystone this year because I have ADHD and this way is more helpful and easier for me to remember, thank you.)
Am I the only one who learned this in school but had to watch over 5 times to understand that's what he was explaining?
Yep, I’ve been doing this since 8th grade, but I also didn’t understand the method he was using at first. But I’ve never actually written a new polynomial for it so.
It’s splitting the middle but more complicated
Omg that’s actually genius
if anyone doesnt get this, what we did in school is just find the -8 and 3 like he did, but then write it like 4x²-8x+3x-6, then you can notice the pattern of x-2, and get 4x(x-2)+3(x-2)=(4x+3)(x-2)
Thank you so much! I just started Pre Calc and my professor tried to teach us on the A.C. method, which I just didn't quite understand
How you get (X_8)and (X+3)
From usual factorising method
x²-5x-24 = x²+3x-8x-24
The rest should be obvious
My brain:error error error error error
There is a lot easier way to learn this method
Here is the tutorial for the tutorial
4x²-5x-6=0
Since it is 5x-6,we have to factor 5x through minus and it should multiply to be the constant term multipled by the term of x²(here,6×4)
4x²-8x+3x-6=0(-8x+3x=5x and 8×3=4×6)
Now take common
4x(x-2)+3(x-2) (note that the term in bracket should be same(here x-2))
Now make 2 brackets adding the common terms and taking other common
(4x+3)(x-2)=0
Ok let me make it simple for you guys like he took statement 4x^2-5x-6 ok so first mulitiply the first term with the last term and keep it mind that variables are not multiplied here after multiplying you will get 24 now take 2 numbers that such numbers which when multiplied we get 24 and when added or subtracted we get -5x the two numbers are 3x and 8x so we need so if i subtract -8x from 3x we get -5x and if we multiply them we get -24 now the statement will become 4x^2-8x+3x-6 now take common factor so it will become 4x(x-2)+3(x-2) now take (x-2) as common and the rest statement 4x and +3 write it in a different brakcet so the answer is
(x-2)(4x+3) hope it helped you 😊
I literally accidentally found out about this method a few years prior while I was doing some questions in my math lesson. Thought I was a genius and discovered something only to be met by reality letting me know it's been around for a while.
I remember doing the same back in grade school when trying to calculate the square root of numbers. I got really close to the real answers. I don’t remember the method too well anymore, but I saw a video on YT that looked like what I did. No genius here :|
My asian brain:
ToO eAsY
what u mean bruh i am asian and i just got 2.7 in math😢
This was how I learned it at first. Completing the square made things so much more complex when I got to Pre-Calc, imo.
@nonchalant if you’re ever confused in calc, take the derivative of it. If that doesn’t work, integrate it instead. If that doesn’t work either, do the second derivative. If none of that worked, retrace your steps, you might’ve missed something.
Precalc… that was like 3-4 years ago, my memory isn’t that good, so I don’t have any good advice.
I love your energy!!
Always know he's Chinese from how he boxes stuff
He does the left side (shu), then the upper and right side (heng zhe), and then the lower side (heng)
That's some Sherlock level detection sh*t right there.
Wait, is everyone not taught this? 😂
Middle term split??
Indians Rock 🤟🤟
Yeah 😄
@@chetanya_sharma yeah BOI it's taught in 8th class nowadays in India
@@aayushswami3022nope in 8th grade Shri dharcharya method is used to teach
I m indian too and the fact ppl here are considering it a different method shows that u guys rote memorize maths. This is not a new method it is basically saying that the constant is (a.b) and the middle term is -(a+b) in a general quadratic equation of the form (x-a)(x-b)=0 which is taught everywhere in the very first class of Quadratic equations. It is the simplest property of Quadratic equations. If u r calling it a new method and giving it an entirely different name like "middle term split method" that shows that u don't actually understand it in depth. For u it is a new trick for ur competitive exams. So come down off ur high horse.
Thank you very much you literally changed my life
My teacher called me stupid for not using his method and marked my problem solving wrong 😂😂
This helped me sooo much! I literally went from "I can't factor consistently" to "I can factor in seconds"
Thank you very much, now I will be able to power through the 100 or so textbook questions asking to factor non monic trinomials.
COOL! However, be sure to factor out any common factor first otherwise the common factor will dispear.
Finally someone is using an easy method
Thank you, sir
4x^2 - 5x - 6
= 4x^2 - 8x + 3x - 6
= 4x(x-2) + 3(x-2)
= (4x+3)(x-2)
This is also known as middle term factorisation.
We do this by factorising the first and third terms' numbers. I.e- 4= 2,2 ; 6 = 2,3.
Now, we need to find two factors that add to form the middle factor, here 5. We can multiply the factors or add them. Here, we are taking the 2,2 from 4 and 2 from 6 and multiplying it, making it 8. And remaining is 3. So, we are getting -8 + 3 = -5.
I don't remember the full explanation, but this was how it was done in my class haha. Hope it helps someone.
This is exactly how we do it here
@@fardiah- Good to know! Where are you from?
@@LuciferAmoyai Bangladesh. Gonna sit for my Mathemathics Alevels ( As ) this month. :)
@@fardiah- We are neighbors! Good luck for your exam bhai 💯
@@LuciferAmoyai dhonnobat ☺😇
Wow I needed this like a year ago. Such a good method.
Saved my life
for the people who dont get it
this is called middle term factorization
it is done in trinomials in the form of ax^2 + bx + c
for e.g here
4x2 + (-5x) + (-6)
multiply the A and C terms (not in the equation but just to break down the middle term)
you get 24
now factorize AC in a way that you get two such numbers who’s sum is equal to BX but also equal to AC
so, 8x3 = 24 -8 + 3 = -5
we get now 4x2 - 8x + 3x - 6 Now we just simply take common so 4x(x-2) + 3(x-2) = (x-2)(4x+3)
With this knowledge I'll be able to take over the world!
The world of Trinomials...
Find 4 x -6 = -24
And then what two number you would add and get -5 and multiply them and get -24
-8 and 3
And boom
😎
What? 💀 wth is that method? i’ve never been confused before tho-
This is called factorisation method.
This is not a trick it's a proper method.
Let's just take an example of a quadratic equation 3x^2+4x-4
In this equation the product of the coefficients of x^2 and constant is -12 so you have to make the factors of -12 suck that the sum of the factors will become the coefficient of x
Here it is 6 and -2
Now we can also write the equation as 3x^2+(6-2)x-4
=3x^2+6x-2x-4
Now you have to take 3x common in first two terms and -2 in next two terms
=3x(x+2)-2(x+2)
Now just take x+2 as common
You'll get (x+2)(3x-2)
And these are the factors of the polynomial
FYI this doesn't always work. If there is a coefficient on the variable of both factors, you end up with an extra step where you have to find the gcf of both factors. For example,
6x^2+5x+1
X^2+5x+6
(x+2)(x+3) you can't pull the 6 out of either factor here. If this happens, you keep going to get
(6x+2)(6x+3) but foiling this doesn't give the original problem. The gcf of the first factor is 2 and the gcf of the second is 3, so you divide each factor by their gcfs to get
(3x+1)(2x+1)
Bro thank you so much. Half of the problems I did has this issue and I couldn’t figure it out 🙏
It does always work- you just need to make sure you simplify your fractions before you divide:
(x+2)(x+3)
@@Mameiadude ur a life saver
Thank you sir!!
so if you could divide in the second factor you would just divide right?
Yes.
I am Indian and going to explain in easy way :
We have our equation : 4x²-5x-6
We will first take the first integer and last integer
4 × 6 = 24 (keep this in mind)
We have 5x in middle
We have to split it like that if we multiply it we get 24x² and if we add it we get 5x
So ,
4x²-8x+3x-6=0
8×3=24
-8+3=5
We take 4x common in first and then 3
4x(x-2) 3(x-2)
(4x+3) (x-2)
Hope it helps you all
Thank you so much❤😊 I don't understood 😉 the 😂method of the video🎥 but you made it😝 easier thank you❤🌹🙏 duo
@@simerpreetkaur7502 your welcome 🗿🚬
Thank you, we appreciate that you took the time to explain this to us all and it was very understandable.
@@Focusquick welcome
thx bro
The fastest way to get a+ in math is just sit next to a nerd and watch his every moves undetected while he was focusing
Thank you so much 💓 ❤
How u can transfer 4 to k i.e. -6 its wrong i think bcoz its different term...Please guide me if im wrong
That's just a way to think about it. It's just multiplication; then you divide by that original 4 in the end
The method is wrong. Just take the ex: 4x^2 -28x -32
This was the only reliable polynomial factoring strategy ive ever learnt and I will always use it.
Assuming that the quadratic equation eq1: ax^2 + bx + c has a solution within the complex number range, and the factorization equation is a(x-p) (x-q) → (ax-ap) (x-q), the discrimination equation D_1 = (-b+-sqrt (b^2-4ac)/2a of the original quadratic equation.
For another quadratic equation eq2: x^2 + bx + ac, assuming that this quadratic equation is (x-m) (x-n), the discriminant in the original equation of eq2 is
D_2= (-b+-sqrt(b^2-4ac))/2.
| • • • • In aD_1 = D_2, we can see that the two roots of eq1 in (ax-ap)(x-q), the two roots of eq1 are p, q, and in | the two roots of eq2 are m/a = p, n/a = q. Substituting this into the factorization equation of eq1 yields (ax-m)(x-n/a) = (ax-ap) (x-q). (Since p,q multiplied by a is m,n, we can see that a is a non-complex real number, and even if p,q,m, and n are complex numbers, the values substituted in the original equation are the same.) maybe?
Please let me know if there are any mathematical discrepancies. It's the best to prove this because I'm a high school student
Now this is quality content.
This would have changed my life in school when I took maths instead of language.
Best way to do it hands down. Thanks so much. It’s helping me determine limits in calculus.
I'm a chemistry teacher and I applaud you for this! Taught one of my students this trick 👏
We called that the “bottoms up” method when I was in 6th grade.
Huh we called it middle term break....with some variations
Its really a bad Idea ...
Knowing the actual calculation is best.....😊
or
multiply the numerical coefficients of the first and the last term.
6*4=24
Prime factorise 24
and split the middle term accordingly
so u get:
4x²-8x+3x-6
=4x(x-2) +3(x-2)
=(x-2) (4x+3)
Thank you. I actually somehow understand it. Helpful but is going to take lots of practice to actually use it correctly and no mistakes.
This is one of the few things that I feel proud of knowing since it’s something my American school actually taught for some reason, along with a few other methods too of course as this one doesn’t always work.
can you make a video on factoring and solving cubic functions??
i love slide and divide its so easy 🫶
Isn't this what middle term breaking is called?
You multiply the coefficient of the first term with the constant and the term you get should be the product of the 2 terms you break the middle one into but it also has to sum back to the original equation
So basically its, 4x²-8x+3x-6
You factor out, 4x(x-8)-3(x+3)
And the answer is, (x-8)(4x-3)
i cant believe i was breaking my ass the past 2 weeks watching through hours of khan academy-style videos searching for a fast way to factor trinomials, now it's just showing up in my youtube shorts page