"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.
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.
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.
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.
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.
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.....
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.
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
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.
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.
Your channel is so good and informative. I remember when I would have looked at this problem and wouldn't have even been able to find a way to solve it. Now I solved it in a few minutes.
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.
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.
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!
@@architlal8594 True, but the quartic you are left with is n⁴+n³-2n²-n+1=0. This is easily factorable by writing -2n² as -n²-n². Now you can isolate n²+n-1 from the first and the second three terms, ending up with (n²-1)(n²+n-1), which saves you from the syntetic division.
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.
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.
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.
i just learned something new about factorial expansions today (a + b)^n with the triangle u showed, that really helped a lot, and solving cubic equations.. yeah that is something new too.. not to mention you can convert it to quadratic equation by using another theorem division and develop a new equation.. that is just crazy
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
I like your way of explaining or the way you speak. Calm speech, thoughtful explanation. I first heard the term synthetic division in RUclips videos, in high school we called it Horner's algorithm. Named after William George Horner.
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.
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.
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
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.
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
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.
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.
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.
By observation, x = -2 and x = -4 are solutions. If we expand the polynomial and divide by [x - (-2)] and [x - (-4)] we find the remaining quadratic factor x² + 7x + 11 = 0
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.
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}.
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).
I use y = x + 3 to avoid expansion of (x + 3)^4. I think it is an easier method. With that, the equation becomes y^2 + (y + 1)^3 + y^4 = 2 y^4 + y^3 - 2y^2 - y + 1 = 0. (y + 1) and (y - 1) are the factors. The equation becomes (y + 1) (y - 1) (y^2 + y - 1) = 0. So, there are 4 roots y = - 1, y = 1, y = (− 1+ √5)/2 and y = (− 1 − √5)/2 When y = - 1, x = - 4 When y = 1, x = - 2 When y = (− 1+ √5)/2 , x = (− 7 + √5)/2 When y = y = (− 1 − √5)/2, x = (− 7 − √5)/2
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.
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.
Solve: (x+1)² + (x+2)³ +(x+3)⁴=2 It's always worth testing integer values of x to see if you can discover one or more roots of the equation. X=-2 is good (it makes the middle term zero) moreover 7:03 Substituting x= -2 shows that x=-2 is indeed a root of the equation of the equation. The same strategy,(choose an x value to make other bracketed terms zero) determines that : -1 is not a root, but x=-4 is a root Now let P(x)=(x+1)² + (x+2)³ + (x+3)⁴ -2 Since we know that x =-2 and x=-4 are roots of the equation P(x)=0 then (x+2) ,(x+4) are factors of P(x) so we may write P(x) = (x+2)(x+4)(ax² + bx +c) where a,b,c are integers. Furthermore,since P(x) is monomic ( leading coefficient is 1) a=1 So we have the identity (x+2)(x+4)(x² + bx +c) ≡ (x+1)² + (x+2)³ + (x+3)⁴ -2 (Note this is true for all real values of x). Now we may choose values of x to calculate the values of b and c. x=0 gives 2x4 x(c) =1² + 2³ + 3⁴ -2 so c =11. x= -1 gives (1)(3)(1-b+11)=0²+(1)³+(2)⁴−2 i e -3b = 0+1+16-2 -36 So b=7 P(x) = (x+4)(x+2)(x²+7x+11). P(x) = 0 has four roots they are x=-2. x=-4 and the two roots of x²+7x+11=0 These are given by the formula x = 〈-7±√(7²-4.1.11)〉/2 = 〈-7±√(49-44)〉 /2.1 - 7/2 ± √5/2.
Actually the x + 3 = y was better in my case as even if u end up with a biquadratic, it is easily factorised using the rational root theorem (u get 2 factors) and then just do some simple synthetic division and you get a quadratic, use the quadratic formula and get the other two factors, now as y = x + 3, just minus 3 from the roots; i felt this was easier and it took less than 2 mins
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)?
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 }
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.
![Sqrt of 2 is irrational[ Proof by Rational Root Theorem]](http://i.ytimg.com/vi/IChzhKYmRnU/mqdefault.jpg)
![Sqrt of 2 is irrational[ Proof by Rational Root Theorem]](/img/tr.png)
It's great to see how such many mathematical theorems or manipulations appear in one simple-looking problem! You explained it flawlessly!
One of the primary reasons why math is bullshit.
"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.
Consider this approach..
(to avoid expanding 4th power)
let x+3 = y
Equation becomes (y-2)^2 + (y-1)^3 + y^4 = 2
after expanding using identities equation becomes
y^4 + y^3 - 2y^2 - y + 1 = 0
Rearranging the terms makes it easy to factorize
(y^4 - 2y^2 + 1) + (y^3 - y) = 0
(y^2 - 1)^2 + y(y^2 - 1) = 0
(y^2 - 1) {y^2 - 1 + y} = 0
problem is 90% done 😃
Yes, y=x+3 to avoid multiplying out the quartic🎉
I did the same way, but instead of rearraging the terms, I made it the harder way 😅
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.
You have a forever subscriber from this video. As a math teacher my self this was flawless.
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.
it is better to use "1 1" at the top of the Pascal triangle. This way. "1 2. 1" follow the rule
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.
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.
I used y=x + 3 instead and it was a lot easier using a difference of 2 squares as part of the factorisation .
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.....
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.
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.
Worked this out and I agree. You basically cut the time in half by solving it this way.
He said y=x+3 didn’t turn out well.
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
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.
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.
I used to drive too, and math videos were my entertainment.
Your channel is so good and informative. I remember when I would have looked at this problem and wouldn't have even been able to find a way to solve it. Now I solved it in a few minutes.
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.
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.
11:45 you have a marvelous way in explanation, interesting and full of simplicity
Thank you very much
I am amaze on how you broke down by explaining key concepts and theorems to justify your answer. Great explanation sir.
You are the coolest maths teacher I have seen. Super
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!
used t=x+3 and it was easier
t doesn't cancel out, so ur still left with a quartic equation instead of a cubic one.
@@architlal8594 True, but the quartic you are left with is n⁴+n³-2n²-n+1=0. This is easily factorable by writing -2n² as -n²-n². Now you can isolate n²+n-1 from the first and the second three terms, ending up with (n²-1)(n²+n-1), which saves you from the syntetic division.
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.
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.
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.
The instructor's love for math comes through! It's great!
Subscribed because you explain why each step was taken, which frees the learner from the rote memorization form of education.
i just learned something new about factorial expansions today (a + b)^n with the triangle u showed, that really helped a lot, and solving cubic equations.. yeah that is something new too.. not to mention you can convert it to quadratic equation by using another theorem division and develop a new equation.. that is just crazy
I’m so happy I found your channel. Your explanation is amazing.
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
I like your way of explaining or the way you speak. Calm speech, thoughtful explanation. I first heard the term synthetic division in RUclips videos, in high school we called it Horner's algorithm. Named after William George Horner.
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.
So true!
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.
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
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.
You are my favorite math teacher!
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
Great review of substitution, distribution, factorization, rational root theorem, synthetic division, and the quadratic formula in one take.
Забавные лекции. Решил подтянуть english. А вообще парень молодец, благодарность объявлю в приказе :))
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.
Prime Newtons is awesome! ❤🎉😊
Honestly ❤
That is true
I tried x = - 2, and it seems to be right. Or am i in error?
@rick57hart yes, x=-2 is one of the four solutions.
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.
Amazing, looked like a pain but you explained it perfectly so it seems simple!
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.
By observation, x = -2 and x = -4 are solutions.
If we expand the polynomial and divide by [x - (-2)] and [x - (-4)] we find the remaining quadratic factor x² + 7x + 11 = 0
Nice selection of sums and a wonderful explonation
This is awesome! I am looking forward to watching more vodeos by you! Keep going!
You are great Sir. Nice and useful video.
Great teaching. So many techniques in one problem. Brilliant!
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)
You are so good in teaching. Thank you
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.
From Tanzania, much respect...please keep up
What an information packed video!
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}.
I loved watching you solve this!!!
I use another substitution ,a=x+3, in order to avoid the 4th power distribution. Get the same result.
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).
Amigo, impresionante paciencia a escribir tanto.
Wow....
VERY USEFUL I HAVE A NEW APRACH IN MY MEMORY THANKS AND KEEP GOING
Excellent, Professor! Thanks you very, very much.
You're a great teacher, Mister *bowing
I use y = x + 3 to avoid expansion of (x + 3)^4. I think it is an easier method.
With that, the equation becomes
y^2 + (y + 1)^3 + y^4 = 2
y^4 + y^3 - 2y^2 - y + 1 = 0. (y + 1) and (y - 1) are the factors. The equation becomes
(y + 1) (y - 1) (y^2 + y - 1) = 0.
So, there are 4 roots y = - 1, y = 1, y = (− 1+ √5)/2 and y = (− 1 − √5)/2
When y = - 1, x = - 4
When y = 1, x = - 2
When y = (− 1+ √5)/2 , x = (− 7 + √5)/2
When y = y = (− 1 − √5)/2, x = (− 7 − √5)/2
❤So enlightning , always rooted in and supporten by proven theorens. H is presentation is an example of hos things should be Done in mathematics.
Thank you for showing the detail. That was cool!
I like your smile more than anything. Nonetheless, the demonstration is awesome. Thanks for the effort.
Thank you! 😃
You are awesome! Very useful explanation!❤
This was a nice problem! Thanks from a fellow math teacher.
Love your passion and smile.❤
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.
Sir, you're born to be teacher.
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 🙂
I just learned that synthetic division is Rule of Ruffini. Thank you 😊
Thanks teacher! I enjoy your lessons.
Thanks for Synthetic division shortcut method !
Wonderful handwriting!
Nice 👍👍👍
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 😅
This teacher is different and good.
One more method to solve the cubic equation is absorption method which gives the result in a single line
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.
Umm como sería
The nodding at 7:21 is glorious haha
I need a gif of this on repeat...
Neat! Btw the golden ratio f is hidden in this equation since the two irrational roots are:
f - 4 and 1/f - 3
Mona Lisa will still be elusive even if we offer her Golden Ratio for a smile.
Solve:
(x+1)² + (x+2)³ +(x+3)⁴=2
It's always worth testing integer values of x to see if you can discover one or
more roots of the equation.
X=-2 is good (it makes the middle term zero) moreover 7:03
Substituting x= -2 shows that x=-2 is indeed a root of the equation of the equation.
The same strategy,(choose an x value to make other bracketed terms zero) determines that :
-1 is not a root, but x=-4 is a root
Now let P(x)=(x+1)² + (x+2)³ + (x+3)⁴ -2
Since we know that x =-2 and x=-4 are roots of the
equation P(x)=0 then (x+2) ,(x+4) are factors of P(x)
so we may write
P(x) = (x+2)(x+4)(ax² + bx +c)
where a,b,c are integers.
Furthermore,since P(x) is
monomic ( leading coefficient is 1)
a=1
So we have the identity
(x+2)(x+4)(x² + bx +c)
≡ (x+1)² + (x+2)³ + (x+3)⁴ -2
(Note this is true for all real values of x).
Now we may choose values of x to calculate the values of b and c.
x=0 gives 2x4 x(c) =1² + 2³ + 3⁴ -2
so c =11.
x= -1 gives
(1)(3)(1-b+11)=0²+(1)³+(2)⁴−2
i e -3b = 0+1+16-2 -36
So b=7
P(x) = (x+4)(x+2)(x²+7x+11).
P(x) = 0 has four roots they are
x=-2. x=-4 and the two roots of
x²+7x+11=0
These are given by the formula
x = 〈-7±√(7²-4.1.11)〉/2
= 〈-7±√(49-44)〉 /2.1
- 7/2 ± √5/2.
1:37 let x+2 = y
please explain why you chose x+2 rather than x+1 or x+3 to set equal to y?
Loved it 😊
I tried x = - 2, and it seems to be right.
Or am i in error?
Actually the x + 3 = y was better in my case as even if u end up with a biquadratic, it is easily factorised using the rational root theorem (u get 2 factors) and then just do some simple synthetic division and you get a quadratic, use the quadratic formula and get the other two factors, now as y = x + 3, just minus 3 from the roots; i felt this was easier and it took less than 2 mins
Very beautifully solved!
8:26 what if we can’t factor the polynomial?
Love your content! The top of that pascal triangle is wrong though!!
You're correct. I knew something was off as soon as I wrote it but I was impatient. It's 1 1 not 1
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)?
Your teaching is too sweet
God bless you. From Algeria !
This synthetic division resembles at Horners rule.
You are amazing Sir
Really sir you are magician
That was awesome 👍🏻😎
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 }
👍 Good Job.
I would have substitute y = x +3 because that is the 4th power.... and make it simple.
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.
يمكن إيجاد الحل بنقل 2 إلى الطرف الآخر وجعله 1+1 والتفكير بالفرق بين مربعات والعامل المشترك.
There is another way to solve that which is much easier. I can send it for you if you are interested!
Very interesting method
Excellent
I expanded it without the substitution and divided out (x + 2) and then (x + 4), leaving only a quadratic.
good keep it up
It's very simple, I evaluated it in 15 seconds and found it.
17:04 words to live by