This method is very efficient when the roots are integers or fractions. But in most situations, the roots of a cubic are either irrational or complex, so it's misleading to claim mastery of cubic equations after such a short video. You've only scratched the surface.
1. What if there is no quadratic term? :P (that is, the equation is in canonic form). Is my guess correct that then we need to choose the factors in a way so that their sum is zero? (e.g. 2+1-3=0) 2. What about the linear term? I can find a series of cubic equations all with the same constant term and quadratic term, but with different linear term. According to your method, they all should have the same solutions, because your method only looks at the constant term (its factors) and the quadratic term (how to add up these factors to get this coefficient), but ignores the linear term, which can be different. And it definitely plays some role in the equation, right? If this coefficient is "wrong", the entire method falls apart. I guess that when this happens, the solutions are not rational, perhaps even complex. (I'll check this hypothesis later and tell you if this is the case.) 3. What if there is some coefficient of the cubic term? Should it be divided out or left alone? But if we divide by it, fractions may appear as other coefficients. What then? How to find rational factors of the constant term? And what's more important... 4. How to solve cubic equations which have irrational or complex roots? This method doesn't seem to work for anything else than integer solutions. But where have you seen integer solutions besides "educational" examples like these you use? In real-world equations, it is a miracle if you encounter an equation with integer solutions. Most often the roots will be irrational, and quite often only one of the roots is real - the other two are complex. Well, I guess that it's enough to find the real one, because then we can factor it out by division and reduce the degree of equation to quadratic, which then can be solved for the other two complex solutions. But finding that first real factor can still be hard if it is irrational :P What then? At 11:28 you said that every cubic equation has 1 or -1 as one of its roots, because 1 is a factor of every number. But this is not true even when all the solutions are integer. Here's one counter-example (out of many): x³ - 4x² - 11x + 30 = 0.
Thank you so much. You are a life saver. Now I can solve the eigen values very easily for 3X3 matrix otherwise earlier I used to go mad solving the cubic equation. Again thank you very much.
Put the roots that you get back into the equation to check it and if they do not make the equation homogeneous, then there are no real roots... pretty simple really and would still be quicker than using other methods like factorising and algebraic division.
This method only works with cubic equations which have three real roots. One root of the cubic must always be real; the remainder might be complex. The trick then is to find the real root (e.g. by graphing or Newton's method), divide it out, and go hunting for the other two with the quadratic formula.
if the last constant like 1,3,6,8,9,5 is smaller then the sum difference no. like -----x^3+2x^2-x+1 . so the factor of 1 are 1,1,1 so 1 is less then 2 then how we will solve sir
This trick will ONLY work for cubic equations that have three INTEGER roots. Maybe you will be asked to do this kind of "cooked" up problem in math class. But you need to look up "Cardano's formula" if you are interested in solving cubics where the answers might be rational or irrational real numbers or complex numbers.
Firstly all the factors should be positive not negative. A good check is that multiplying 3 negatives gives a negative result. Multiplying two negatives and one positive would give a positive result. In this case though all factors are positive.
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Awesome but it doesn't work for all cubic equations.
It's worth saying that this method only works when the roots are all whole numbers, and when the x^3 coefficient is 1.
Matthew Fearnley not whole nos, the correct word is integers
and it works for every number it is just the factors and infinite
Amazing! I can now instantly solve for my eigenvalues
why do u change the signs though
You are a genius, this is so much helpful. Thank you!
wow! thanks for this! this helps me. I didn't learn this at school.
Just one word to say Amazing Amazing.......... Thank u so much. This made my day
So how do you factor when all the solutions are irrational or fractions?
Awesome tricks ! Thanks a lot.
Really great,i didn't know cubic eqns are so easy or may be u made it.thanks a lot.looking forward to see many of ur videos
Thank you so much. This is a very helpful video. Cheers! 😀
woww!!!, i like this guy. he simplifies it a lot. now i know it more than ever
This is a life saver! You don't happen to have a trick to factor quartic polynomial equations, do you? Or higher ...?
What would you do if the coefficient of x^3 was greater than 1???
I don't know how to thank you enough :) you made our nightmare equation turn into a quick handy solution
Best way oh hell. still cant believe. Thank you
thankyou so much. you have no idea what shit other websites posted. this really helped. thanks a lot again
Really helpful thanks!!!!!
amazing way of solving complex equations !!! thanks a lot ! :)
Awesome dude!
I think this just saved my sanity, and this will surely help me in my College Algebra class. Thank you!
this goes crazy man
This method is very efficient when the roots are integers or fractions. But in most situations, the roots of a cubic are either irrational or complex, so it's misleading to claim mastery of cubic equations after such a short video. You've only scratched the surface.
Oh my god thank you man! You saved me!
you deserve an oscar award!!!!!!!!!! thanks bhaai
Thanku very much sir....god bless u
Thank you soo much sir this is the one i learnt quickly ...
pls solve 4x^3+21x^2+28x+8
thx this video really helped me
Really helpful method. It is quick method. Thanks a lot..
This is so cool. Thanks!
genius bro it is working
a new trick has been added in my mind
Thank you sir for this. I m fortunate to find out this channel☺
1. What if there is no quadratic term? :P (that is, the equation is in canonic form).
Is my guess correct that then we need to choose the factors in a way so that their sum is zero? (e.g. 2+1-3=0)
2. What about the linear term? I can find a series of cubic equations all with the same constant term and quadratic term, but with different linear term. According to your method, they all should have the same solutions, because your method only looks at the constant term (its factors) and the quadratic term (how to add up these factors to get this coefficient), but ignores the linear term, which can be different. And it definitely plays some role in the equation, right? If this coefficient is "wrong", the entire method falls apart. I guess that when this happens, the solutions are not rational, perhaps even complex. (I'll check this hypothesis later and tell you if this is the case.)
3. What if there is some coefficient of the cubic term? Should it be divided out or left alone? But if we divide by it, fractions may appear as other coefficients. What then? How to find rational factors of the constant term? And what's more important...
4. How to solve cubic equations which have irrational or complex roots? This method doesn't seem to work for anything else than integer solutions. But where have you seen integer solutions besides "educational" examples like these you use? In real-world equations, it is a miracle if you encounter an equation with integer solutions. Most often the roots will be irrational, and quite often only one of the roots is real - the other two are complex.
Well, I guess that it's enough to find the real one, because then we can factor it out by division and reduce the degree of equation to quadratic, which then can be solved for the other two complex solutions. But finding that first real factor can still be hard if it is irrational :P What then?
At 11:28 you said that every cubic equation has 1 or -1 as one of its roots, because 1 is a factor of every number. But this is not true even when all the solutions are integer. Here's one counter-example (out of many): x³ - 4x² - 11x + 30 = 0.
AMazing ! I can now easily find eigen values . Thankyou
Amazing ,thanks !!!
wow an easy way to deal with whole numbers, similar process to solving quadratic equation...thanks a lot kindda help..
It was very helpfull.. Thankyou so so very much!
Thank you so much sir. I always faced problems while solving this kind of equations. Your video really helped a lot.: )
Thanks for this amazing video
thank you sooo much!
Terrific job!
It is just the use of the equation of eigen values of 3×3 matrix
Use trac(A) ,sum of co factors and determinant wisely
Good explanation.Ive learnt something.Thx.
This is brilliant!!! Thank you so much. This is so helpful.
Thank you so much. You are a life saver. Now I can solve the eigen values very easily for 3X3 matrix otherwise earlier I used to go mad solving the cubic equation. Again thank you very much.
reallyy an awsm way of solving such eqns thank u so mchhhhh
excellent!! whatever it is it made my day!!!!!
Beautiful explanation dear. Very simple and useful method
Expecting more 👍🏼
man you are awesome..... respect.. thanks for helping us....
Very helpful , thank you♥
What about the third terms? I'm sure when f(x) =x^3+10x^2+31400000x+18 You don't end with -1, -3, -6.
+hills nathan i m not clear with your question
+hills Nathan I agree with you
+Vuenol He wants to say, if you change the third terms, your solution with your method stays the same and the real solution obviously changes.
Put the roots that you get back into the equation to check it and if they do not make the equation homogeneous, then there are no real roots... pretty simple really and would still be quicker than using other methods like factorising and algebraic division.
This method only works with cubic equations which have three real roots. One root of the cubic must always be real; the remainder might be complex. The trick then is to find the real root (e.g. by graphing or Newton's method), divide it out, and go hunting for the other two with the quadratic formula.
Really very helpful and easy thanks man
+Abhishek Kumar my pleasure
God bless you.. great video
thankyou so much for the first trick
Genius, this is just a good way to look at the sum in a different way I first thought of it. Now I can solve them a lot faster. Thanks!
awesome tips indeed
thank you for the this contributions
Nice trick if you know the answers are whole numbers, sadly it fails when the answers are not whole numbers. The 1st and 3rd term still do matter...
Very interesting, and leads toward meaningful insights. Makes solving some of these an easy ROOTine!
thank u sir. good explanation .
Too much appreciated
Thanks a lot sir for your kind information
Very easy method 😃really too much helpful 👌👌👌
Loved this video
Amazing !!!
thanks a lot it helped me a lotthank you
I should have known that this is too good to be true...
Thank You so Much Sir
very helpful ,, thank you
yes ur ryt it only work on the even number not when b=even n d=even
ax3 plus bx2 plus cx plus d. in here if roots are A,B,C prove that A plus B plus C = b/a and also prove that ABC = d/a
is there a similar way for 4 degree equation
Vishu Malik yes
if you want it contact me at my email
awsome very very helpful and fast way 2 solve these equstions 5 stars
if the last constant like 1,3,6,8,9,5 is smaller then the sum difference no. like -----x^3+2x^2-x+1 . so the factor of 1 are 1,1,1 so 1 is less then 2 then how we will solve sir
sir , thanks .. for this ..
wonderful tricks,never seen before...
thanks a lot sir.
What if constant term d is large like 100,200,...Did we write all the factors and calculate?
superb ..... it really helped me
how about x³ + 2x² - 13x + 10 ???
Please help, that's my homework to be done tomorrow :'(
VERY VERY THNKS......
Thanks!
Tq_sir,
Its really a fastest trick
haha this is good for solving eigens equation in case of 3X3 Matrix :P ROFL thnx :D
can u find it for x3-11x2+38x-40
i appreciate this trick
i can even take set of 5,4,2 or 1,8,2 but only one satisfies
this deserves a like
thanks....it was really useful....
Awesome sir....
Help a lot!!!! Thank!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
how would you do x^3 -4x^2-9x+36
Thanks a lot bro
This trick will ONLY work for cubic equations that have three INTEGER roots. Maybe you will be asked to do this kind of "cooked" up problem in math class. But you need to look up "Cardano's formula" if you are interested in solving cubics where the answers might be rational or irrational real numbers or complex numbers.
Firstly all the factors should be positive not negative. A good check is that multiplying 3 negatives gives a negative result. Multiplying two negatives and one positive would give a positive result. In this case though all factors are positive.
brilliant man
This only works if the equation is a result of expanding factors. How do you solve if you change the coefficients of the middle terms?
Thank you boss
Build a formula of the solution of: ax3+bx2+cx+d=0
what will be the factors if the expression has minus in between the terms?
pls reply fast
what about x^3-×^2-1=0? it doesn't work on that
how will you use the first method if the value of a is a different value and not 1?