Cubic Eqn Trick Faster Way to Solve Cubic Equation

how to solve cubic equation in faster way
• Cubic Eqn Trick Fast... Fast and Easy Cubic Eqn Trick

[@alvinchan403]

Awesome but it doesn't work for all cubic equations.

[@MatthewFearnley]

It's worth saying that this method only works when the roots are all whole numbers, and when the x^3 coefficient is 1.

[@biologyigcse]

Matthew Fearnley not whole nos, the correct word is integers and it works for every number it is just the factors and infinite

[@joserivera6776]

Amazing! I can now instantly solve for my eigenvalues

[@rpc-os8qg]

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.

[@MegaSwati007]

I don't know how to thank you enough :) you made our nightmare equation turn into a quick handy solution

[@YourMutualFunds]

why do u change the signs though

[@bonbonpony]

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.

[@nathandaniel5451]

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.

[@Shiv2626]

+hills nathan i m not clear with your question

[@aymanalgeria7302]

+hills Nathan I agree with you

[@alexandermathis2955]

+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.

[@baristurkmen2342]

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.

[@Ensign_Cthulhu]

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.

[@addisonslack2049]

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...

[@jaymgraf]

I think this just saved my sanity, and this will surely help me in my College Algebra class. Thank you!

[@DiamondSapphire93]

What would you do if the coefficient of x^3 was greater than 1???

[@rudzanifoster5170]

yes ur ryt it only work on the even number not when b=even n d=even

[@deependumandal6544]

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.

[@cricketfans766]

genius bro it is working a new trick has been added in my mind

[@adityamehta9729]

amazing way of solving complex equations !!! thanks a lot ! :)

[@ghost9816]

is there a similar way for 4 degree equation

[@biologyigcse]

Vishu Malik yes if you want it contact me at my email

[@sushmamahadev7735]

Thank you soo much sir this is the one i learnt quickly ...

[@smitpatel5389]

pls solve 4x^3+21x^2+28x+8

[@hirensolanki5522]

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

[@TheChaptstick1991]

This is a life saver! You don't happen to have a trick to factor quartic polynomial equations, do you? Or higher ...?

[@priyanka_dey]

Thank you so much sir. I always faced problems while solving this kind of equations. Your video really helped a lot.: )

[@vikramsinghmaran5089]

Just one word to say Amazing Amazing.......... Thank u so much. This made my day

[@AbhishekSingh-if1uw]

Really very helpful and easy thanks man

[@Shiv2626]

+Abhishek Kumar my pleasure

[@Ak-hc4rr]

man you are awesome..... respect.. thanks for helping us....

[@vasudevaraju6796]

What if constant term d is large like 100,200,...Did we write all the factors and calculate?

[@amanpatel9]

thankyou so much. you have no idea what shit other websites posted. this really helped. thanks a lot again

[@Yamatoikiski]

So how do you factor when all the solutions are irrational or fractions?

[@hrishikeshtelang5396]

the method was good, but if the x³ gets a coefficient, for eg: 3x³-10x²+9x-1=0 it will be difficult to find out the factors, because, in that case, we can divide each term by 3, but it won't be so easy to find factors of a fraction, which is ⅓. How to solve it then?

[@pratikchavan8057]

take the coefficient common if possible

[@hrishikeshtelang5396]

Thanks

[@deepakpatil6245]

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

[@m.wilkinson9559]

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.

[@rommuelevangelista377]

wow! thanks for this! this helps me. I didn't learn this at school.

[@balanivandana6429]

Thanku very much sir....god bless u

[@akenow.]

Good thumbnail 👍

[@vic4955]

woww!!!, i like this guy. he simplifies it a lot. now i know it more than ever

[@sdsnahranofficial3419]

Sir, what about the question: 2x cube - 5x square - 14x + 8 ???

[@aniksheikh902]

Build a formula of the solution of: ax3+bx2+cx+d=0

[@jenemi1934]

Im in class and i laughed when he said “it looks like an anaconda”

[@vatsal888]

Hey, first of all thanks for such an amazing video!! Can you please help me how to solve x^3 -2x +2=0 using this method?? Thanks Again!!

[@abubakarabdullahidauda]

thank you for the this contributions

[@gamereplayhq]

haha this is good for solving eigens equation in case of 3X3 Matrix :P ROFL thnx :D

[@GurpreetSingh-so1bu]

thanks ji...can you help with this equation y^5+3y^3+4y^2+12

[@JarlinJamesNDK13]

thx this video really helped me

[@yasmin-dw4zh]

how about x³ + 2x² - 13x + 10 ??? Please help, that's my homework to be done tomorrow :'(

[@strikestar007]

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

[@rexlin7903]

You are a genius, this is so much helpful. Thank you!

[@sukhkaur5195]

Very easy method 😃really too much helpful 👌👌👌

[@rishikeshkumar4427]

wonderful tricks,never seen before... thanks a lot sir.

[@dicksonmanongi823]

wow an easy way to deal with whole numbers, similar process to solving quadratic equation...thanks a lot kindda help..

[@SADRACFIRMINO]

you deserve an oscar award!!!!!!!!!! thanks bhaai

[@keralacodingacademykca4597]

youre the best in the world

[@amishaparmar1368]

Thanks for this amazing video

[@craigjenquin3416]

this deserves a like

[@vivekvishwakarma6533]

Really helpful method. It is quick method. Thanks a lot..

[@ellamorgan2899]

This doesn't work... I've just tried using the terms and expanding out the first and second examples. It's close, but the signs come out wrong. It'd be lovely if this did work, but it doesn't.

[@maazkhattak2156]

AMazing ! I can now easily find eigen values . Thankyou

[@kkeconomicsgroup6159]

excellent to understand for class 9 students

[@MouseTrack]

Beautiful explanation dear. Very simple and useful method Expecting more 👍🏼

[@haribabu9671]

x^3-6x+8x-2=0 then the roots of the equation are

[@haribabu9671]

for this equation getting imaginary roots for this type of equations there is any short cut

[@bibhilusonpadi3655]

Thank you sir for this. I m fortunate to find out this channel☺

[@arsalaanuddin4501]

This was the easiest method i have ever seen. I was so much confused in solvinv these cubic equation!! Thanku soo much sir Thank u

[@willjohnston2959]

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.

[@SocratesAlexander]

But we knew that, if the sum of odd degree term's coefficients are equal to the sum of even degree term's coefficients, then you have a factor (x+1), hence a root of -1. The other factors can be found by dividing the polynomial by x+1. Similarly, if the sum of all coefficients are equal to zero, then you have a factor (x-1), hence a root of +1. The other factors can be found by dividing the polynomial by x-1. But your method is nicer and quicker. By the way, who invented this method?

[@sakshamgupta692]

superb sir

[@rickyvalentinej7465]

superb ..... it really helped me

[@vartikasingh664]

reallyy an awsm way of solving such eqns thank u so mchhhhh

[@AbidAli-wp2fs]

-x^3+6x^2-9x+4 please solve this and explain the steps thanks and your trick is very good

[@darkhouse397]

I should have known that this is too good to be true...

[@thatbrownchick96]

Really helpful thanks!!!!!

[@dhanshreelade2827]

Thank you so much. This is a very helpful video. Cheers! 😀

[@dropagemonem]

Best way oh hell. still cant believe. Thank you

[@ling6701]

Awesome tricks ! Thanks a lot.

[@jayuchawla1892]

thankyou so much for the first trick

[@dailyenglishepisodes]

Wow....great....

[@morningwood3150]

what about x^3-×^2-1=0? it doesn't work on that

[@katlehosenatsi6318]

awesome tips indeed

[@narayan3400]

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

[@sandeepkrverma8636]

y³-15y²+74y-120=0 1,2,3,4,5,6,8,10,15,20,24,30,60,120 Here the number of factors are very much So How to decide quickly? Find the factor

[@DYoung2112]

Good explanation.Ive learnt something.Thx.

[@mavulandlela]

Veunol, I found your solutions much much easier to work with , thank you for sharing.

[@MuhammadQasim-pm8vh]

It was very helpfull.. Thankyou so so very much!

[@JingweiZhong]

Help a lot!!!! Thank!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[@Hacker7763]

What if the coefficient of x^3 is greater than 1? How does the method change/does it work at all?

[@biologyigcse]

Alexander Barnes divide the whole equation by coeff of x^3

[@bonbonpony]

+Kartavya Kothari Sure, but then there could be some fractions as other coefficients and this method will no longer apply ;J

[@biologyigcse]

This method is too specific and won't work if all the roots of the equation are integers And If they are There won't be any fractions

[@bonbonpony]

Kartavya Kothari I have an idea: Since fractions are just integers in disguise, we can still use factor theorem when the coefficient of x³ is not 1 :> Here's how: Suppose we need to solve this: 15x³ - 38x² + 17x - 2 = 0 If we divide it by 15 to have a monic polynomial with just one cube, x³, then we would get fractions for other coefficients: x³ - (38/15)x² + (17/15)x - 2/15 = 0 and we would have to deal with fractions, which are harder to factor. But since fractions are just integers in disguise, we can also temporarily multiply 15·2=30, and find the factors of 30 instead of factors of 2. These are: 1, 2, 3, 5, 6, 10, 15, 30 Now we need to choose such factors the sum or difference of which will give 38, the original coefficient of x². And this is easy: 30+5+3=38 :> But we have -38, and the -30 is also negative, which means we need to take _all_ these factors with negative sign, like this: -30-5-3=-38. But remember that we work with the scaled versions of these numbers: we multiplied the original constant by 15, so to get the final answer, we need to get back to the original scale, by dividing every factor by 15. This gives: -30/15 = -2, -5/15 = -1/3, and -3/15 = -1/5. So the factored form is: (x-2)(x-1/3)(x-1/5) = 0 and the roots are: 2, 1/3, 1/5. And this is correct! Check it :) Multiply the parentheses together and you will get the original equation ;) You can also substitute them into the equation to see if they really zero it out: 15·(2)³ - 38·(2)² + 17·(2) - 2 = 15·(8) - 38·4 + 34 - 2 = 120 - 152 + 34 - 2 = 154 - 154 = 0 [OK] 15·(1/3)³ - 38·(1/3)² + 17·(1/3) - 2 = 15·(1/27) - 38·(1/9) + 17·(1/3) - 2 = 15/27 - 114/27 + 153/27 - 54/27 = 0 [OK] 15·(1/5)³ - 38·(1/5)² + 17·(1/5) - 2 = 15·(1/125) - 38·(1/25) + 17·(1/5) - 2 = 15/125 - 190/125 + 425/125 - 250/125 = 0 [OK] So we're done :) The answer is correct, and the technique still works for fractions ;) Why does it work? Well, because when you convert all these fractions to the common denominator, they're all in the same "scale", so to speak, or you can think of them as using the same "unit" of 1/15, which is multiplied integer number of times for each fraction, so it is as if you were working on just integers, and you can ignore the denominators. Only at the end you need to tell that it is not the "standard" unit of 1, but the "scaled down" unit of 1/15 (accounting for the coefficient of x³). And that's it :)

[@love042005]

Amazing ,thanks !!!

[@viratdewan959]

x^3-x^2-5x+5=0 How about this question? I tried using this awesome method but failed to get the correct values...

[@odexpert9314]

+Virat Dewan x = 1 is clearly a root; (x - 1)(x^2 - 5) = 0 and the two other roots are - 5^(1/2) and + 5^(1/2)

[@michaelempeigne3519]

+Virat Dewan add every alternate term coefficient. 1 + ( - 5 ) = - 4 and - 1 + 5 = 4. if the signs are opposite as in this case, then x = 1 is a root. so ( x - 1 )* ? = x^3 - x^2 - 5x + 5 divide the first by the first and the last by the last. so (x^3 / x = x^2 ) and 5 / ( - 1 ) = - 5 Therefore, the factored form is ( x - 1 ) ( x^2 - 5 )

[@ranjeetrawat4197]

nice trick....

[@kesinenisireesha7799]

thank u sir. good explanation .

[@husaintaghi688]

this method doesn't work with all cubic equations i.e 2x^3 - 5x^2 - 23x - 10=0.. using this method would give x= 1,2,2 whilst the the correct answers are x 5, -2 & -1/2 !!

[@satyajitsahu6614]

Very helpful

[@kratigupta7292]

Thanks a lot sir for your kind information

[@rajendragupta.01]

Tq_sir, Its really a fastest trick

[@snchirag889]

I feel like a math god math god all my people now call me a ............

[@InfoandVlog]

Thank You so Much Sir

[@nayanamangalassery2941]

GREAT,VERY HELPFUL.....THANK YOU SO MUCH.......

[@EsperanceBG]

thank you sooo much!

[@gsuto]

This is misleading. It works for the terms shown in the examples. But if you change the coefficients of x^2 or x you don't get the right solution. For example, change the x^3 + 7x^2 + 14x + 8 to x^3 + 6x^2 + 14x + 8 and you only get one solution at x = -4.510. This solution only works if your equation is the result of the following multiplication (x+x1)*(x+x2)*(x+x3) , in which case the solution is -x1, -x2, -x3. For the first example given, the equation is the result of (x+1)*(x+2)*(x+4) which gives the cubic equation of x^3 + 7x^2 + 14x + 8 , with the nice solutions of -1, -2 and -4. This only works for these simple, utopian examples. The reality is much uglier...

[@evarandie4378]

Wow! That was fast and really helpful. What about the quartic equation

[@ahmedahmadyar877]

solve it as 2 quadratics

[@skystormgamingyt]

Thanx

[@sarmiray24]

Thanks!

[@aaron3438]

This is so cool. Thanks!

[@HackingDutchman]

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!