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.
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.
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.
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?
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
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.
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 !!
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.
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...
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.
x^-6x^+11x-6=0 not only this eq'n..for many eq'ns I didn't get the correct answers by using your tricks!! just tell me what would be the answer for this..
Hey ! Getting tricked by an equation : -x^3+5x^2-48x+13 ... I can't find anything and it's turning me mad, I tried everything : factoring, grouping, your trick... Nothing worked... I asked WolframAlpha to factor it for me and it gave me a mad factorization with complex numbers... Does your trick work with that equation ?
+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 )
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 :)
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?
Not working if your last term is 1, in this case, you only have 1 as your factor number, therefore this trick is dead at that point. But it works on every other term.
24z^3-26z^2+9z+1=0 can't solving this question as there is coefficient at first term. After taking common 24 so there is changes in equation but there is problem in factorization. HELP ME!
Great trick for some cubic equation but it did not work for: x^3-6x^2+3x+10 The roots are: 5,2, -1. the sum is not -6.. neither will 5*2*(-1) give me +10 Or did I misunderstood something?
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
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.
So how do you factor when all the solutions are irrational or fractions?
why do u change the signs though
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
is there a similar way for 4 degree equation
Vishu Malik yes
if you want it contact me at my email
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.
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.
What if constant term d is large like 100,200,...Did we write all the factors and calculate?
how about x³ + 2x² - 13x + 10 ???
Please help, that's my homework to be done tomorrow :'(
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...
pls solve 4x^3+21x^2+28x+8
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.
thanks ji...can you help with this equation y^5+3y^3+4y^2+12
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
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?
take the coefficient common if possible
Thanks
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
yes ur ryt it only work on the even number not when b=even n d=even
what about x^3-×^2-1=0? it doesn't work on that
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
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.
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 !!
what if the first term is not 1...plz answer this.....
I think this just saved my sanity, and this will surely help me in my College Algebra class. Thank you!
then how to factorise x^3-7x+5
This is a life saver! You don't happen to have a trick to factor quartic polynomial equations, do you? Or higher ...?
Sir, what about the question:
2x cube - 5x square - 14x + 8 ???
genius bro it is working
a new trick has been added in my mind
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.
Thank you so much sir. I always faced problems while solving this kind of equations. Your video really helped a lot.: )
Im in class and i laughed when he said “it looks like an anaconda”
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...
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!!
Thank you soo much sir this is the one i learnt quickly ...
how to solve the cubic equation when coefficient of x^3 is not 1 (without using trial and error method)?
how to solve cubic equation which is cubic term have some coefficient
Build a formula of the solution of: ax3+bx2+cx+d=0
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.
x^-6x^+11x-6=0 not only this eq'n..for many eq'ns I didn't get the correct answers by using your tricks!! just tell me what would be the answer for this..
amazing way of solving complex equations !!! thanks a lot ! :)
Is it applicable in every case
man you are awesome..... respect.. thanks for helping us....
Hey ! Getting tricked by an equation : -x^3+5x^2-48x+13 ... I can't find anything and it's turning me mad, I tried everything : factoring, grouping, your trick... Nothing worked... I asked WolframAlpha to factor it for me and it gave me a mad factorization with complex numbers... Does your trick work with that equation ?
haha this is good for solving eigens equation in case of 3X3 Matrix :P ROFL thnx :D
Question: What happens when the leading term's coefficient is not 1?
It only works if every term is positive, the coefficient of the cubed term has to be 1 and it only works with a few coefficients of the third term :)
Divide throughout by it.
you deserve an oscar award!!!!!!!!!! thanks bhaai
Good thumbnail 👍
Thanku very much sir....god bless u
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?
What if the roots are in decimal such as 1.45 , 0.22 , 6.23 then how to solve??????
thankyou so much. you have no idea what shit other websites posted. this really helped. thanks a lot again
thank you for the this contributions
x^3-x^2-5x+5=0 How about this question? I tried using this awesome method but failed to get the correct values...
+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)
+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 )
What if the coefficient of x^3 is greater than 1? How does the method change/does it work at all?
Alexander Barnes divide the whole equation by coeff of x^3
+Kartavya Kothari Sure, but then there could be some fractions as other coefficients and this method will no longer apply ;J
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
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 :)
Just one word to say Amazing Amazing.......... Thank u so much. This made my day
2s^3+6s^2 +19s+12=0 can u plzzz tell me how to solve this
Really very helpful and easy thanks man
+Abhishek Kumar my pleasure
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?
What when equation is x^3-18x^2+45x
Xcube-10Xsq+41X-10=0
What is the cubic values of this equation.??
Not working if your last term is 1, in this case, you only have 1 as your factor number, therefore this trick is dead at that point. But it works on every other term.
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
x^3-6x+8x-2=0 then the roots of the equation are
for this equation getting imaginary roots for this type of equations there is any short cut
wow! thanks for this! this helps me. I didn't learn this at school.
24z^3-26z^2+9z+1=0
can't solving this question as there is coefficient at first term.
After taking common 24 so there is changes in equation but there is problem in factorization. HELP ME!
I should have known that this is too good to be true...
this deserves a like
how will you use the first method if the value of a is a different value and not 1?
thx this video really helped me
Very easy method 😃really too much helpful 👌👌👌
wonderful tricks,never seen before...
thanks a lot sir.
-x^3+6x^2-9x+4 please solve this and explain the steps thanks and your trick is very good
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
49.9B^3 + 330.67B^2 -800=0 how get value of B?
please reply how to calculate manually.....
It doesn't work for all such equations!
what happens x^3-7x-6
0x^2
Great trick for some cubic equation but it did not work for: x^3-6x^2+3x+10
The roots are: 5,2, -1. the sum is not -6.. neither will 5*2*(-1) give me +10
Or did I misunderstood something?
Daniel Gall yes, the coefficeint should be negetive of sum of the roots, this was the reason that he changed the signs of the final answer
Thanks for this amazing video
how do you solve 8x^3-3x^2+1=0 ?? Please !
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
woww!!!, i like this guy. he simplifies it a lot. now i know it more than ever
What is this ?I can't understand.
x³+12x²+36x-32=0
what are the factors of these?
Either the equation should have 36x or +32.
+Abdeslam El Haddouchi (x+2)^2 (x+8)
+Abdeslam El Haddouchi (x+2)^2 (x+8)
didnt work for me. x^3 + -2x^2 - 5x - 6
It only works for integers. I used to calculate examples which has real non integer roots. How to solve them? Anyone.??
if equation has complex roots then how to find
wow an easy way to deal with whole numbers, similar process to solving quadratic equation...thanks a lot kindda help..
What if we have 1 as a constant term ??
excellent to understand for class 9 students
Would u plz explain some difficult questions
If you change the coefficient of x in these examples you'll realize this method doesn't solve for any of those values.
I feel like a math god math god all my people now call me a ............
What happens if the cubic coefficient is greater than 1?
What do I do if the question is f(x)=x^3+4x^2-3x+1?
what will be the factors if the expression has minus in between the terms?
pls reply fast
what about x^3+7x^2+10x+10=0
Beautiful explanation dear. Very simple and useful method
Expecting more 👍🏼