You should not take calculus if you can't solve this. This video utilizes the most common strategies adopted in solving calculus problems. I used substitution, factoring, theorems, synthetic division and the quadratic formula
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@rimantasri45785 ай бұрын
It's great to see how such many mathematical theorems or manipulations appear in one simple-looking problem! You explained it flawlessly!
@jadapinkett16565 ай бұрын
One of the primary reasons why math is bullshit.
@ndotl5 ай бұрын
"simple-looking problem" You need to have your eyes checked. The algebra was easy, but I have been away from it too long to remember the rules/theorems/steps to apply.
@elio90085 ай бұрын
it is better to use "1 1" at the top of the Pascal triangle. This way. "1 2. 1" follow the rule
@mozvi14365 ай бұрын
Wanted to comment that as well. We're talking about (a+b)^n where n=1 so -> a+b with coefficients 1 and 1. If one really wants to have that leading row with only one 1, one should not look at n=1 but rather one row above 1, where n=0: (a+b)^0=1 except a=-b. One 1, but not one nice one.
@emanuelborja20095 ай бұрын
I substituted y just as you did. Then I subtracted 2 from each side, and grouped it as ((y-1)^2-1) + y^3 + ((y+1)^2-1). This allowed me to factor the two groups as a difference of squares. Y then factored out nicely as a common factor, revealing y=0 as a solution. A bit of further work also allowed (y+2) to factor out nicely, revealing y=-2 as a solution. The remaining quadratic was solved just as you did. This method avoided the need for polynomial division and distribution of the quartic term.
@TheRenaSystem3 ай бұрын
I've been binging your channel since discovering it, and just wanted to say I love your style of presentation and how you teach, 10/10. You have perfect handwriting, great explanatory skills, you speak clearly, and you have a really soothing voice to boot lol.
@nstarling885 ай бұрын
You have a forever subscriber from this video. As a math teacher my self this was flawless.
@laman89145 ай бұрын
Great. Never seen this solution before. Very interesting. We have watched a number of clips by this dude. His patience and systematic approach are excellent. Born to be a teacher.
@gregmackinnon36635 ай бұрын
Great teaching. So many techniques in one problem. Brilliant!
@robertlezama19585 ай бұрын
Excellent delivery ... love his manner of teaching, he makes math seem fun and not scary. More importantly, he applies techniques using the formal theorem names, so if you need to brush up you can go find the theorems and study them outside of solving actual problems. Really well done. Thanks!
@chao.m5 ай бұрын
This is way cool. Never seen the rational root theorem and remainder theorem explained and applied in such simple and easy to follow manner. Also that long division with an addition rather than subtraction is excellent too. Goes to show the difference between teaching and effective teaching: solve complex problems without breaking a sweat
@TheFrewah4 ай бұрын
That’s what I miss from my collection of math formulas. Maybe the rational root theorem is there, if it is, it’s not explained with an example but there’s a loooong proof. So it’s easy to miss such gems.
@seansmith14575 ай бұрын
I’m so happy I found your channel. Your explanation is amazing.
@petejackson79765 ай бұрын
I used y=x + 3 instead and it was a lot easier using a difference of 2 squares as part of the factorisation .
@sauronbadeye5 ай бұрын
That was the way to go, and NOT the one selected by the teacher This approach generates a polynomial which is quite easy to manipulate , and at first glance has less terms than the one on the blackboard, and leads to a very simple Factorization.....
@schlingel00175 ай бұрын
First I could not solve it like this but now I see what you did there. Yes, I agree that this is a better solution.
@TheFrewah4 ай бұрын
I think it’s important to realise that in math, there are several ways to get from problem to solution. It makes perfect sense to try more than one way as you learn something along the way.
@gandalfthegrey9116Ай бұрын
Worked this out and I agree. You basically cut the time in half by solving it this way.
@TheFrewahАй бұрын
He said y=x+3 didn’t turn out well.
@ronaldomeeeessi5 ай бұрын
used t=x+3 and it was easier
@juanrobles92325 ай бұрын
I am amaze on how you broke down by explaining key concepts and theorems to justify your answer. Great explanation sir.
@leonznidarsic5 ай бұрын
I like your way of explaining or the way you speak. Calm speech, thoughtful explanation. I first heard the term synthetic division in KZbin videos, in high school we called it Horner's algorithm. Named after William George Horner.
@ndotl5 ай бұрын
Subscribed because you explain why each step was taken, which frees the learner from the rote memorization form of education.
@nullplan015 ай бұрын
Tried this before watching: (x+1)² + (x+2)³ + (x+3)⁴ = 2 All coefficients will be integers (obviously). Thus we get to use the rational roots theorem. Lead coefficient will be 1. Constant part will be 1² + 2³ + 3⁴ - 2 = 88. So, only possible rational roots are the divisors of 88 (positive and negative, of course). 88 = 2³ * 11, so divisors of 88 are 1, 2, 4, 8, 11, 22, 44, 88 Attempt at x = 1: (x+1)² = 4, adding further positive numbers will not decrease the value. No, we need negative numbers. Attempt at x = -1: (0)² + (1)³ + (2)⁴ = 1 + 16 = 17 ≠ 2. Attempt at x = -2: (-1)² + (0)³ + (1)⁴ = 2. Winner! Attempt at x = -4: (-3)² + (-2)³ + (-1)⁴ = 9 - 8 + 1 = 2. Winner! Attempt at x = -8: (-7)² + (-6)³ + (-5)⁴ = 49 - 216 + 15625 ≠ 2. The quartic term has far outpaced the cubic one at this point. Going lower will not help. So, it is time to pay the piper and face the music: (x+1)² + (x+2)³ + (x+3)⁴ = 2 x² + 2x + 1 + x³ + 6x² + 12x + 8 + x⁴ + 12x³ + 54x² + 108x + 81 = 2 x⁴ + 13x³ + 61x² + 122x + 88 = 0 (x⁴ + 13x³ + 61x² + 122x + 88) : (x + 2) = x³ + 11x² + 39x + 44 (x³ + 11x² + 39x + 44) : (x + 4) = x² + 7x + 11 We can solve the x = -7/2 ± √(49/4 - 11) = -7/2 ± √(49/4 - 44/4) = -7/2 ± √5/2 Thus the solutions are: x₁ = -2 x₂ = -4 x₃ = (-7-√5)/2 x₄ = (-7+√5)/2
@punditgi5 ай бұрын
Prime Newtons is awesome! ❤🎉😊
@SunflowerSerenade15 ай бұрын
Honestly ❤
@egyptian200915 ай бұрын
That is true
@rick57hart5 ай бұрын
I tried x = - 2, and it seems to be right. Or am i in error?
@jamesharmon49944 ай бұрын
@rick57hart yes, x=-2 is one of the four solutions.
@elmerhuamanpedraza31215 ай бұрын
You are great Sir. Nice and useful video.
@GonzaloMiguelGS3 ай бұрын
Excellent, Professor! Thanks you very, very much.
@MrNibiru21125 ай бұрын
From Tanzania, much respect...please keep up
@munkhjinmunkhbayar59525 ай бұрын
Amazing, looked like a pain but you explained it perfectly so it seems simple!
@vikasseth95445 ай бұрын
You are the coolest maths teacher I have seen. Super
@sankararaopulla525611 күн бұрын
Nice selection of sums and a wonderful explonation
@arbenkellici38085 ай бұрын
This is awesome! I am looking forward to watching more vodeos by you! Keep going!
@ahmedfanan31464 ай бұрын
You are so good in teaching. Thank you
@Bonginhlanhla5 ай бұрын
You are my favorite math teacher!
@DaveyJonesLocka5 ай бұрын
I like this. You could solve it by brute force, but taking advantage of specific characteristics of a problem to unravel a more elegant solution is just prettier. I do a lot of driving, and often try to solve problems like this one mentally. This was a fun one to do.
@PrimeNewtons5 ай бұрын
I used to drive too, and math videos were my entertainment.
@tyronekim35065 ай бұрын
Thank you for showing the detail. That was cool!
@oscarmontanez63044 ай бұрын
Thanks teacher! I enjoy your lessons.
@daddykhalil9095 ай бұрын
11:45 you have a marvelous way in explanation, interesting and full of simplicity Thank you very much
@jamesharmon49944 ай бұрын
I loved watching you solve this!!!
@schlingel00175 ай бұрын
I am happy to see that I did it exactly how you did it. But I now have more insight about which numbers to try to find a root for a 3rd degree equation. Thank you.
@diablobenson91685 ай бұрын
VERY USEFUL I HAVE A NEW APRACH IN MY MEMORY THANKS AND KEEP GOING
@SanePerson15 ай бұрын
Nice, clear exposition (and extremely nice board work!). I have few personal comments from the perspective of someone who took algebra and precalculus 50+ years ago and used it (along with trigonometry, calculus, linear algebra, differential equations, and various other upper division math) in my career as a theoretical chemist. I disagree with the opening note, "You should not take calculus if you can't solve this.") This is nonsense, BUT it is correct to say, "If you've forgotten how to do this, be prepared to relearn it when you take calculus" Almost all high school math will be needed for more advanced math, but you learn what is most important by the necessity of USING it. Personally, I always made fewer errors doing long division into polynomials than I made trying to remember exactly how synthetic division works. Pre-calc teachers use synthetic division all the time, but people who apply math, only come across these kinds of problems occasionally - tried-and-true long division I always remembered - synthetic division got hazy. By the same token, students will absolutely need trigonometry in doing calculus, but all those identities? Few people remember them. Just be prepared to relearn the most important ones when you're learning calculus.
@PrimeNewtons5 ай бұрын
I agree with you. And 8 understand every point you made. To relearn implies the student had learned how to perform that task before. That was my point. The biggest problem in any calculus class today is not the new material being hard to learn, it is that many students never learned the required algebra. Forgetting a concept or not mastering a concept is better than having never heard of it before if it is required for higher levels.
@anuragpriy5 ай бұрын
Love your passion and smile.❤
@Shubham215992 ай бұрын
Very beautifully solved!
@d.m.70965 ай бұрын
You are awesome! Very useful explanation!❤
@imlassuom5 ай бұрын
Thanks for Synthetic division shortcut method !
@Mustapha.Math_at_KUSTWUDIL3 ай бұрын
I used your method of Synthetic Division (Reduced Long Division Method) to reduce the quartic equation to cubic, used the same to reduced the later to quadratic and finally use the formula to get the solutions. I guess this one is more economical. Thank you
@mauludirachman78815 ай бұрын
You're a great teacher, Mister *bowing
@jensberling23415 ай бұрын
❤So enlightning , always rooted in and supporten by proven theorens. H is presentation is an example of hos things should be Done in mathematics.
@xyz92505 ай бұрын
After the substitution, move 2 to the right. (Y-1)^2 -1 + y^3 + (y+1)^4 -1 = 0 then follow the rule a^2 - b^2=(a+b)(a-b)
@jwvdvuurst5 ай бұрын
This was a nice problem! Thanks from a fellow math teacher.
@keinKlarname2 ай бұрын
Wonderful handwriting!
@asiob3n505 ай бұрын
The beginning was the same as you. I subtituate x+2 by y and I use pascal triangle to expand the binomials. I sum and factor to find 0 as a solution. The other product is y³+5y²+7y+2. To solve that I factorised this polynomial. The two factor should look like this (Ay²+By+C)(Dx+E). If you multiply these factors, it give you Ay²Dy+Ay²E+ByDy+ByE+CDy+CE, and you can notice 4 equation. AD = 1, AE+BD = 5, BE + CD = 7, CE = 2 ; Directly, you can see that A = 1 and D = 1 too. So B+E = 5. I assume E = 1 but this doesnt work and try with 2. B = 3. So BE = 6; 7 - 6 = C; So C = 1; So the factors are (y²+3y+1)(y+2). You find y = - 2 with the second factor. You juste need to solve the first factor. To solve the quadratic equation, I complete the square. 2ab = 3y, a= y, b = 3/2; So I sqaure b and I found 9/4. This give me this binomial (y+3/2)²-9/4+4/4; I solve like that, (y+3/2)²=5/4 => y+3/2 = +/- √5/2. And I have just to find x. x = {-2,-4,-7/2 +/- √5/2}.
@kassuskassus62634 ай бұрын
God bless you. From Algeria !
@Maths__phyics5 ай бұрын
That's very good! Thanks!
@Jam.shed95 ай бұрын
Sir, you're born to be teacher.
@joelgodonou45673 ай бұрын
You are amazing Sir
@pierreneau58692 ай бұрын
Thanks to share such equation. Other way: finding trivial solutions. It's necessary to check with low value of power 4. X=-2 and x=-4 can be easily found. After that, it's necessary to develop x^4+13x^3+61x^2+122x+88=0 and factorize by (x+2) and (x+4) to obtain (x+2)(x+4)(x^2+7x+11)=0 The last 2 solutions are (-7+-sqrt(5))/2.
@sundaramsadagopan77955 ай бұрын
This teacher is different and good.
@mudspud5 ай бұрын
Very interesting method
@medabedhamzaoui21473 ай бұрын
Excellent
@parthas.chatterjee84405 ай бұрын
Really sir you are magician
@Adiii1355 ай бұрын
Nice 👍👍👍
@svyatoslavn97065 ай бұрын
Забавные лекции. Решил подтянуть english. А вообще парень молодец, благодарность объявлю в приказе :))
@TheFrewah4 ай бұрын
A good way to improve your English is to read an English version of a book that you have already read that you liked. I did at age 18 and now I only read novels in English.
@tusharsharma50535 ай бұрын
I think we can also use y = -2 is a solution then y+2 must be a factor of given cubic equation.Then we can also use long division because everyone is familiar with long division though there are some chances of mistakes
@nicolascamargo83395 ай бұрын
Otra alternativa: (x+1)²+(x+2)³+(x+3)⁴=2 Suma de potencias de tres números consecutivos luego 2=1+1=(-1)²+0³+1⁴, así x+1 debe ser -1 en está solución de donde x=-2 así (x+2) es factor y se puede hacer lo siguiente: (x+1)²+(x+2)³+(x+3)⁴=2 (x+2-1)²+(x+2)³+(x+2+1)⁴=2 (x+2)²-2(x+2)+1²+(x+2)³+(x+2)⁴+4(x+2)³+6(x+2)²+4(x+2)+1⁴=2 (x+2)⁴+5(x+2)³+7(x+2)²+2(x+2)+2=2 (x+2)[(x+2)³+5(x+2)²+7(x+2)+2]=0 Los coeficientes del polinomio en x+2 son 1,5,7,2, posibles raíces racionales para x+2 son: ±1 y ±2. Como 1+7≠5+2 queda descartado x+2=-1 Como 1+5+7+2=15≠0 queda descartado x+2=1. Como 2³+5(2²)+7(2)+2=8+20+14+2=44≠0 queda descartado x+2=2. Como (-2)³+5(-2)²+7(-2)+2=-8+20-14+2=22-22=0, x+2=-2 funciona así x=-4 es solución y por lo tanto (x+4) es factor así se puede hacer lo siguiente: (x+1)²+(x+2)³+(x+3)⁴=2 (x+2)[(x+2)³+5(x+2)²+7(x+2)+2]=0 (x+2)[(x+4-2)³+5(x+4-2)²+7(x+4-2)+2]=0 (x+2)[(x+4)³-3(2)(x+4)²+3(2²)(x+4)-2³+5(x+4)²-5(2)(2)(x+4)+5(2²)+7(x+4)-7(2)+2]=0 (x+2)[(x+4)³-6(x+4)²+12(x+4)-8+5(x+4)²-20(x+4)+20+7(x+4)-14+2]=0 (x+2)[(x+4)³-(x+4)²-(x+4)]=0 (x+2)(x+4)[(x+4)²-(x+4)-1]=0 (x+2)(x+4)[(x+4-(1/2))²-(4/4)-(1/4)]=0 (x+2)(x+4)[(x+(8/2)-(1/2))²-((√5)/2)²]=0 (x+2)(x+4)[x+(7/2)+((√5)/2)][x+(7/2)-((√5)/2)]=0 Así: (x+1)²+(x+2)³+(x+3)⁴=2 Tiene como soluciones: x_1=-2, x_2=-4, x_3=-(7/2)-((√5)/2) y x_4=-(7/2)+((√5)/2).
@jim23765 ай бұрын
Amigo, impresionante paciencia a escribir tanto.
@IlIlIllIlIl-xj2zx5 ай бұрын
Wow....
@AzmiTabish5 ай бұрын
Thank you, Sir for this video. Indeed if we learn algebra properly, calculus should be much easier. Though instead of synthetic division I would have normally taken y+2 as a factor by breaking the cubic equation so that y+2 comes out as a factor, i.e. y cube + 2* (y squared)+3*(y squared)+6*y+y+2=0 and then take y+2 as common and we get the quadratic equation multiplied by the factor y+2, and that expression being zero and then solve the quadratic.
@schlingel00175 ай бұрын
This is great, but unfortunately not everyone is that much gifted to immediately recognize such a factorization. I would never thought of expressing those terms like that to find a common factor.
@AzmiTabish5 ай бұрын
Actually, I am not good in mathematics. Just a coincidence perhaps. Just noticed, for example that if we break the term containing one degree less y in a manner that coefficient of 2nd term of the one degree less y is 2 times the coefficient of first term of one degree greater y, and add what remains and so on, etc. to make the expression same.
@nulakiustha3 ай бұрын
That was awesome 👍🏻😎
@joelgodonou45673 ай бұрын
Your teaching is too sweet
@InverseTachyonPulse3 ай бұрын
Loved it 😊
@jim23765 ай бұрын
By inspection, x = -2 is an obvious solution. Mr. Gauss tells us there will 3 other solutions. Grinding out the expansions and adding like terms would be a major pain in the ass. Doable, but tedious as hell. Excellent lesson. Dredges up a lot of algebra.
@jamiujabaru74683 ай бұрын
👍 Good Job.
@tpsb055 ай бұрын
cela me rappelle ma jeunesse merci !
@naturalsustainable61165 ай бұрын
I use another substitution ,a=x+3, in order to avoid the 4th power distribution. Get the same result.
@rcnayak_585 ай бұрын
You are always my adorable. I love you seeing over here again and again. At the same time, I also think how to simplify in a more better way of the problems! Here is a suggestion in this problem. Instead of assuming y = x + 2, if we we write y = x + 3 and solve it , we will get rid of expanding yours (y+1)^4 term as it will be only y^4. Thus we will only expand (y-2)^2 + (y-1)^3 + y^4. This can perhaps be an easier way of solving the problem.
@PrimeNewtons5 ай бұрын
So true!
@mohammedel-gamal34555 ай бұрын
good solving but Y = X + 3 is easier and I have the same four answers
@mahmoudboutaglay54785 ай бұрын
good keep it up
@hamzaemad83385 ай бұрын
Thanks brother
@danny896203 ай бұрын
That's really cool nice vid
@user-ep6hr8tg4c5 ай бұрын
Beautiful
@bdb-music16085 ай бұрын
Yes very good indeed, but I'm Italian, I call this stuff 1 - Triangle of Tartaglia, 2 - Th. of Ruffini 3 - Practical rule of Ruffini 🙂
@PrimeNewtons5 ай бұрын
I just learned that synthetic division is Rule of Ruffini. Thank you 😊
@christiaan33155 ай бұрын
This synthetic division resembles at Horners rule.
@Hobbitangle5 ай бұрын
Hint: make the equation symmetric by substitution y = x+2 (y-1)²+y³+(y+1)⁴=2 after opening all the parentheses the right hand side coefficient drops out.
@italixgaming9155 ай бұрын
Of course I used the same method, because having y-1, y and y+1 creates some symetry and makes our life easier. However, you don't really need to develop each term individually. You can do everything at once very easily. We can see that we're going to obtain a quartic equation so just do this: - where does the coefficient for x^4 come from? Only from (y+1)^4 so we have y^4. - where does the coefficient for x^3 come from? From the y^3 and from (y+1)^4 so we have y^3+4.y^3=5y^3. - for x², it comes from (y-1)² and (y+1)^4 so we have y²+6y²=7y². - for x we have -2x+4x=2x - and finally for the constant we get 1+1=2. So we can directly write: y^4+5y^3+7y²+2x+2=0 For the cubic equation, you can obtain your solution a bit quicker. You rewrite the equation like this: y(y²+5y+7)=-2. If y is a relative integer, then y divdies -2, which means that y can be equal to -2, -1, 1 or 2. Now let's look at the function y ---> y²+5y+7. Its derivative is 2y+5, which is always positive if y>-5/2. Therefore, if y equals -2, -1, 1 or 2, the minimum value of the expression is 1, obtained for y=-2. But wait, if the value is 1, this means that y=-2 is a solution of our equation. Now, if y=-2 is a solution, you can factorise by y+2. You have y^3+5y²+7y+2=0, so let's start the factorisation: we can turn y^3+5y² into y^3+2y²+3y² and 7y+2 into 6y+y+2 and rewrite the equation: (y^3+2y²)+(3y²+6y)+(y+2)=0 then (y+2)(y²+3y+1)=0 And we can conclude like you did.
@burlino5 ай бұрын
ty so much❤
@lukaskamin7555 ай бұрын
We used to call that simplified division method the Horner scheme(or method), I wonder if this name is used wherever you're teaching (USA or UK or elsewhere)?
@math_qz_25 ай бұрын
Good video
@tungyeeso36375 ай бұрын
I like your smile more than anything. Nonetheless, the demonstration is awesome. Thanks for the effort.
@PrimeNewtons5 ай бұрын
Thank you! 😃
@325826575 ай бұрын
Is it OK to just look at the first equation and see that -2 is a solution, -3 is not, but -4 is? This makes it easier to see what substitution and factoring to try, but clearly would not work on something more complicated.
@user-dn5dk8mw4v5 ай бұрын
One more method to solve the cubic equation is absorption method which gives the result in a single line
@stephenlesliebrown59595 ай бұрын
I solved it the same way as you except did not introduce the intermediate variable y. Happy to report that not letting y=x+2 made for much MORE work 😅
@VanNguyen-kx6gx5 ай бұрын
Good.
@rick57hart5 ай бұрын
I tried x = - 2, and it seems to be right. Or am i in error?
@alipourzand64995 ай бұрын
Neat! Btw the golden ratio f is hidden in this equation since the two irrational roots are: f - 4 and 1/f - 3
@user-xw6ky8ob4l5 ай бұрын
Mona Lisa will still be elusive even if we offer her Golden Ratio for a smile.
@KarlFredrik5 ай бұрын
Another way to solve it. Use x+3 = y. Rearrange such that we get after some algebra: y^4 + (x+1)y^2 - x -2 = 0. Solve for y and a conspiracy of numbers get us: (x + 3)^2 = - x - 2 or 1. Solve for x and get the solutions.
@nicolascamargo83395 ай бұрын
Umm como sería
@mohsenfarrokhrouz54535 ай бұрын
There is another way to solve that which is much easier. I can send it for you if you are interested!
@Olga75475 ай бұрын
Классный мужик ведёт математику! 👍😊
@invisiblelemur5 ай бұрын
Love your content! The top of that pascal triangle is wrong though!!
@PrimeNewtons5 ай бұрын
You're correct. I knew something was off as soon as I wrote it but I was impatient. It's 1 1 not 1
@MLinoAlfredoN5 ай бұрын
Finally a good add
@ceknoloslavanakyacumuruhye10445 ай бұрын
Which grade are you guys learning that level of questions in your country?
@user-jj8kg5ef2t5 ай бұрын
I would have substitute y = x +3 because that is the 4th power.... and make it simple.
@idkman6405 ай бұрын
17:04 words to live by
@realasianrizz3 ай бұрын
why didn't you make y=x+3? that way you don't have to expand a quartic [(y-2)^2] + [(y-1)^3] + y^4 much cleaner
@E.h.a.b3 ай бұрын
I used another approach, I know it may not be clear like yours. (x+1)^2 + (x+2)^3 + (x+3)^4 - 2 = 0 ---------> [1] Calculate the coefficient (C) of x^0 in equation [1] C = 1^2 + 2^3 + 3^4 -2 = 1+8+81 -2 = 90-2 = 88 C = 88 = 11*8, 22*4, 44*2, 88*1 It is clear that I must use x less than zero e.g. ( -1, -2, -4, -8) in order to get solution for equation [1] when testing x values I got x = -2 is solution. x = -4 is solution. C=88 = (-4) * (-2) * 11 Rewrite equation [1] using the solution values of x (x+2)(x+4)(x^2 + b x + 11)=0 ---------> [2] where b is some real constant (x^2 + 6 x + 8)( x^2 + b x + 11) =0 ---------> [3] Calculate coefficient of x^3 in equation [1] from (x+2)^3 + (x+3)^4 : (x+3)^4 = (x^2 + 9 + 6 x)^2 = (x^2+9)^2 + (6 x)^2 + 2(6 x)(x^2+9) coefficient of x^3 in (x+3)^4 = 2*6 coefficient of x^3 in (x+2)^3 = 1 Total coefficient of x^3 in equation [1] = 13 ---------> [4] Calculate coefficient of x^3 in equation [3] from (x^2* b x + 6 x * x^2) : Total coefficient of x^3 in equation [3] = (b+6) ---------> [5] from [4] and [5] we get b+6=13 b=7 Rewrite equation [2] and substitute b=7 we get (x+2)(x+4)(x^2+ 7 x + 11)=0 x^2 + 7 x + 11 = 0 ---------> [6] To get x we solve equation [6] using quadratic formula x = (-7 +√(49-4*11))/2 and (-7 -√(49-4*11))/2 x =(-7 + √5)/2 and (-7 - √5)/2 Solutions are x = { -2, -4, (-7+√5)/2, (-7-√5)/2 }
@jeffersonluizbento205 ай бұрын
Top!
@randvar29525 ай бұрын
Great! Just one note: the statement that y^2+3y+1 ‘cannot be factored’ (over real numbers or integers?) is misleading. It can, over the reals: y^2+3y+1 =(y-y_1)(y-y_2), where y_1,y_2 are the real roots provided by the quadratic formula.
@madonnacesso405 ай бұрын
Never stop learning ❤🇮🇹
@kirahen04375 ай бұрын
agreed madonnacesso40
@Jianlong-xp5li5 ай бұрын
Can you please make a video for log and natural log