@@suvranilchoudhury6452 Yeah, I am refreshing my knowledge on Math on that time since I am lazy studying Maths when I was at elementary school to half of my Senior High lel.
I hate to be the one that makes a negative comment but this method does not work. All these examples just happen to be examples where the coefficient's add up to 0 (a+b+c+d=0) so x=1 will always be a solution. this method does NOT work because you could very easily change the leading coefficient to anything other than 1 and the 3rd term ( the 'x' term) to anything other than what is given in the examples and this method will fail outright. This method may look like it "works" for very specific Cubic's but it is not a tool I would recommend learning as it will fail for almost all other Cubic's. For example, try 3x^3 + 11x^2 - 25x + 13=0 . If you follow the same rules given in this video you will get the same factors but you get completely wrong solutions!!
So what method do you use to solve these type of equations ? Just asking because I want to find the universal method for solving all types of cubic equations rather than hit and trial.
@@AlexonHughes Thanks for replying to my comment. I posted this response such a long time ago I actually forgot about this. LOL.... for me solving any random cubic equation is a pretty specific process. I would start with synthetic division to see if there happens to be an even solution to the equation. so if x=1 or x=2 etc. happens to be a solution then you can factor (x-1) or (x-2) out of the cubic equation and the remaining quadratic can be solved with the quadratic formula. If there are NO even solutions to the equation then you can use the cubic formula (math.vanderbilt.edu/schectex/courses/cubic/) to get at least one solution. But ultimately cubics that don't have at least one even solution can be very cumbersome to deal with...... I don't even want to get into quartics like x^4 powers (major headache!!) LOL
@@andrewhealey7698 Thanks for this reply. I'm currently taking Linear Algebra (and have taken calc 1, 2 already) and had no idea how to solve these sorts of cubic equations (I don't recall ever learning this in math 😭). The method in the video just so happened to work with the polynomial I had obtained from trying to calculate the eigenvalues of a 3x3 matrix. I will look into this more reliable method now for future reference!
@@coolkaw4497 That's really cool. I graduated with my math degree back in 2007. To be honest, I was never really taught in any college class how to solve cubic equations either. I was more or less just self taught. and like said, there is really no great cookbook recipe for how to solve cubics. They are relatively straightforward if there happens to be at least 1 whole number solution (i.e x=1, x=2,x=3 etc...) Then you can factor that solution out and you are left with an easy quadratic to solve. Without at least one whole number solution they can be a bit of a headache. But I know one thing the method taught in this youtube video is complete non-sense and will only work for very VERY specific cubics and the creator of this video really does a disservice to anyone trying to learn how to properly solve a cubic equation.
Glad to hear that Sakthi! I'm sure you are an awesome student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃 Keep smiling😊
You are very welcome Jonathan! Thank you so much for taking the time to leave this comment. I'm glad you liked it! Your feedback is always appreciated. Please keep sharing my channel with your family and friends. Take care dear and all the best 😃
Really fantastic method of solving cubic polynomials. Very easy and time saving method too. Thanks a lot for sharing this type of shortcut methods sir.
Only works if the product of the two constant in two binomial roots + the sum of the two constant in two binomial roots multiplied to the constant of the third binomial root is equal to the coefficient with the linear variable x. (x + a) (x + b) (x + c) = [a + b] [c] + [a x b] Otherwise, a different answer would pop out, which is most probably irrational or imaginary.
You are awesome my friend! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃 Have a very happy and blessed New Year!
You are very welcome Ayush! I'm sure you are an awesome student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
before seeing this video , i was also thinking about the relation between constant and any other number , but i also couldn't find the relation. Finally i see your video and understand much better
Thank you bryan johnson for this crucial point. I'm in a processing of uploading more vid tutorials that would answer your questions. Please stay tuned. Kind regards
These cubics all have 1 has a root (since sum of all coefficients = 0). Thx to your channel, a couple others, I always check +/-1 (if sum[coeffs of odd-power terms] = sum[coeffs of even-power terms, then -1 is a root). Using synthetic div to get the quotient quadratic... which can be factored using inspection, grouping, or quadratic formula to get remaining roots.
You are very welcome Isaac! I'm sure you are a brilliant and bright student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
I've tried this method on different cubic equation and re-multiply them to check whether it's right or wrong and it's not always right. This solving method is too dangerous to be used, it might solve some equation of the cubic factorisation but not all of them. I do not recommend this method. but this is just my opinion on solving cautiously, if you're reckless and a simple person i don't mind. Honestly no offense, rather i respect you "PreMath" for creating such a shortcut, i couldn't done such things.
@@abdeslamelhaddouchi9857 I think you may have mistyped this. It should be: s = {2, -1+i√2, -1-i√2} Whereas if you plug in say (-1+2√2) into f(x) = x³-x-6, you get f(-1+2√2) = -30 + 20√2 [or f(-1+2√2) = 10(-3 + 2√2)]
but sir, when picking the three numbers I can choose different signs and get the same answer but the sign will be different so how can I choose the correct sign
It’s actually simple. Generally, you find all the factors of “D” term, then plug those factors into the cubic. If one of those factors let’s the equation to =0, then you have already found a solution. If that doesn’t work, then take all the “D” factors and divide them by the “A” coefficient. Plug those “(Factor-ofD) / A” into the cubic and if any of those let the cubic equation =0, then you have found one of the solutions. From there, you need to convert it from a “x=_____” into “x-_____=0”. Then use polynomial long division or synthetic division to pull out a quadratic. Factor the quadratic if possible. The other method is to just graph the cubic and use graphing technology to solve it lmao.
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Glad to hear that JoeyLou! I'm sure you are a brilliant and bright student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
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You are very welcome Tangikina! I'm sure you are a brilliant and bright student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
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Greetings, In this video you are solving all cubic equations having coefficient of x^3 is 1 i.e., a=1. Can you please explain how to solve if a is not equal to 1 also. Thank you.
You are very welcome Rohan! I'm sure you are a brilliant and bright student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
This to be taken with a grain of salt, since it works only in the rare case that the solutions, or at least one solution, is an integer number. It does no help for the general cubic, although such videos do always suggest...
Dear Celine, in that case, you won't be able to solve it that quick as the problem is going to get complicated. We may have to take different route to solve for that case. Hope I answered your question. Take care dear and all the best😃 Have a very happy and blessed New Year!
3:00 Sorry, but this method only works if you are certain in advance that all three solutions are real and integer, which is not realistic. In particular, you did not check if the coefficient of the first power matches Vieta's rules, too. For the cubic equation x^3 - 7x + 6 = 0, you must check if the three divisors t1 = -1, t2 = -2, t3 = 3 fulfill the three conditions t1 + t2 + t3 = 0 t1*t2 + t1*t3 + t2*t3 = -7 t1 * t2 * t3 = 6 But you did not check if the second condition is fulfilled! In this case, you were lucky because the second condition is indeed fulfilled, but what if the original equation were x^3 - 8x + 6 = 0 or x^3 - 5x + 6 = 0, just to name a few examples? All these equations have coefficients of 0 for x^2 and 6 as the constant, but different (non-integer!) solutions.
This doesn't always work though. It works with x^3 + 11x^2 - 25x + 13 But what happens if you change the co-efficient of x? This doesn't work with x^3 + 11x^2 - 24x + 13. Expanding the expression (x+a)(x+b)(x+c) gives us x^3 + (x^2)c + (x^2)b + bcx + a(x^2) + acx + abx + abc Factoring this gives us: x^3 + (a+b+c)x^2 + (ab+ac+bc)x + abc So, what we _really_ need, is three numbers that satisfy three conditions. The sum of the three numbers must be the co-efficient of x^2 The product of the three numbers must be the constant. And the sum of all possible product pairs must be the co-efficient of x. Also, unrelated to the above, but: Sometimes, however, you may end up having a situation where the numbers that multiply to give you the constant may involve fractions. For example, 13 can be made from -1, -26 and 1/2 If we take -1, -26 and 1/2 and add these together, we get -53/2 If we then take product pairs, 26, -1/2 and -26/2 then add these together, we get 25/2. Therefore, (x-1)(x-26)(x+(1/2)) would be the solutions to the equation x^3 - (53/2)x^2 + (25/2)x + 13 = 0 To remove those fractions, we could then double this, to get 2x^3 - 53x^2 + 25x + 26 = 0
That is not going to work in many cubic equations. He was ignoring the number in front of the x(in the first problem that number is 7, in the 2nd that number is -25 and in the third one, the number is 5). He totally ignored those numbers. A change in any one of those numbers, and his solution does not work.
Thank you so much!! Cubic equations like "x^3-7x+6" is very rare. Luckily, I found it here so your video helped me a lot. Thanks again LOL!
maybe u just didn't focus on maths properly, it isn't at all rare.
@@suvranilchoudhury6452 Yeah, I am refreshing my knowledge on Math on that time since I am lazy studying Maths when I was at elementary school to half of my Senior High lel.
@@andrii21 oh that's great
Hi
@@andrii21 yes it is super rare ive been looking for a video explaining how to do it and i only found this one
I hate to be the one that makes a negative comment but this method does not work. All these examples just happen to be examples where the coefficient's add up to 0 (a+b+c+d=0) so x=1 will always be a solution. this method does NOT work because you could very easily change the leading coefficient to anything other than 1 and the 3rd term ( the 'x' term) to anything other than what is given in the examples and this method will fail outright. This method may look like it "works" for very specific Cubic's but it is not a tool I would recommend learning as it will fail for almost all other Cubic's. For example, try 3x^3 + 11x^2 - 25x + 13=0 . If you follow the same rules given in this video you will get the same factors but you get completely wrong solutions!!
So what method do you use to solve these type of equations ? Just asking because I want to find the universal method for solving all types of cubic equations rather than hit and trial.
@@AlexonHughes Thanks for replying to my comment. I posted this response such a long time ago I actually forgot about this. LOL.... for me solving any random cubic equation is a pretty specific process. I would start with synthetic division to see if there happens to be an even solution to the equation. so if x=1 or x=2 etc. happens to be a solution then you can factor (x-1) or (x-2) out of the cubic equation and the remaining quadratic can be solved with the quadratic formula. If there are NO even solutions to the equation then you can use the cubic formula (math.vanderbilt.edu/schectex/courses/cubic/) to get at least one solution. But ultimately cubics that don't have at least one even solution can be very cumbersome to deal with...... I don't even want to get into quartics like x^4 powers (major headache!!) LOL
@@andrewhealey7698 Thanks for this reply. I'm currently taking Linear Algebra (and have taken calc 1, 2 already) and had no idea how to solve these sorts of cubic equations (I don't recall ever learning this in math 😭). The method in the video just so happened to work with the polynomial I had obtained from trying to calculate the eigenvalues of a 3x3 matrix. I will look into this more reliable method now for future reference!
@@coolkaw4497 That's really cool. I graduated with my math degree back in 2007. To be honest, I was never really taught in any college class how to solve cubic equations either. I was more or less just self taught. and like said, there is really no great cookbook recipe for how to solve cubics. They are relatively straightforward if there happens to be at least 1 whole number solution (i.e x=1, x=2,x=3 etc...) Then you can factor that solution out and you are left with an easy quadratic to solve. Without at least one whole number solution they can be a bit of a headache. But I know one thing the method taught in this youtube video is complete non-sense and will only work for very VERY specific cubics and the creator of this video really does a disservice to anyone trying to learn how to properly solve a cubic equation.
I watched around six videos today on solving cubic equations, best one so far. thanks a million!
Glad to hear that Sakthi! I'm sure you are an awesome student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃 Keep smiling😊
simple, elegant...why isn't this being taught as the norm?? thank you
You are very welcome Jonathan! Thank you so much for taking the time to leave this comment. I'm glad you liked it! Your feedback is always appreciated. Please keep sharing my channel with your family and friends. Take care dear and all the best 😃
Really fantastic method of solving cubic polynomials. Very easy and time saving method too.
Thanks a lot for sharing this type of shortcut methods sir.
Only works if the product of the two constant in two binomial roots + the sum of the two constant in two binomial roots multiplied to the constant of the third binomial root is equal to the coefficient with the linear variable x.
(x + a) (x + b) (x + c)
= [a + b] [c] + [a x b]
Otherwise, a different answer would pop out, which is most probably irrational or imaginary.
Right
Try solving this one: th-cam.com/video/eF9T3cQREPw/w-d-xo.html
It's like now every expert mathematics teacher is Indian🥰🥰🥰🥰
You are awesome my friend! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Have a very happy and blessed New Year!
This is the best method of solving cubic equation i found on youtube. Thanks!
You are very welcome Ayush! I'm sure you are an awesome student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
What a beauty man i love this method! Waiting to show off to my friends!
Sir pls tell in comment section that how to factorize this one through this method
Question) x³+3x²+3x-7
Pls tell
X=1
1+3+3-7=0
before seeing this video , i was also thinking about the relation between constant and any other number , but i also couldn't find the relation. Finally i see your video and understand much better
What about when the coefficient for the cubic expression is something other than +1?
Thank you bryan johnson for this crucial point. I'm in a processing of uploading more vid tutorials that would answer your questions. Please stay tuned. Kind regards
ive been looking for this as well. cant find that example anywhere
It's equal to 0, you can divide everything by the first coefficient.
3x³+9x²+12x-15=0x³+3x³+4x-5=0
Then use the "rational root theorem"
This is one of the most helpful methods I've ever learnt. Thank you!
X^3-7x+6 is very I found everywhere but I can not found this help me a lot
These cubics all have 1 has a root (since sum of all coefficients = 0). Thx to your channel, a couple others, I always check +/-1 (if sum[coeffs of odd-power terms] = sum[coeffs of even-power terms, then -1 is a root). Using synthetic div to get the quotient quadratic... which can be factored using inspection, grouping, or quadratic formula to get remaining roots.
Amazing thanks
the best math yt channel!!!
does this also apply to cubic polynmials that has leading coefficients greater than one (ax3+bx2.... wer a>1
I wish I had been taught this method in college!!! SUBSCRIBED!
What we will do for high no.like 250 or 70..
Thank you sir
Your video is very helpful for me in solving cubic equations
YAAAAAS!!!! thankyou so much this an immense help with exams around the corner especilly
You are very welcome Isaac! I'm sure you are a brilliant and bright student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
I've tried this method on different cubic equation and re-multiply them to check whether it's right or wrong and it's not always right. This solving method is too dangerous to be used, it might solve some equation of the cubic factorisation but not all of them. I do not recommend this method. but this is just my opinion on solving cautiously, if you're reckless and a simple person i don't mind. Honestly no offense, rather i respect you "PreMath" for creating such a shortcut, i couldn't done such things.
What will happen if equation is
x³-x-6=0
The solution of thé équation is : (x-2)(x^2+2x+3) = 0 s={2 ; -1+2√2 ; -1-2√2 }
@@abdeslamelhaddouchi9857
I think you may have mistyped this. It should be:
s = {2, -1+i√2, -1-i√2}
Whereas if you plug in say (-1+2√2) into f(x) = x³-x-6, you get f(-1+2√2) = -30 + 20√2 [or f(-1+2√2) = 10(-3 + 2√2)]
Naman - you can try my method here: th-cam.com/video/H5y9nQ0abEo/w-d-xo.html
X=2
8-2-6=0
6-6=0
will this work for all degree 3 polynomials? what about x^3-2x^2-x+2=0
Simply superb,
Thank you so much, sir!!
Very helpful. Well explained.
O God unbelievable thanks a lotttttt
Finally found exactly relatable vedio♥️♥️🙏🙏💐💐
Excellent dear Ritu.
You are awesome 👍
Take care dear and stay blessed😃 Love and prayers from the USA!
@@PreMath thank you Sir💐
but sir, when picking the three numbers I can choose different signs and get the same answer but the sign will be different so how can I choose the correct sign
Very good explanation. Bravo.
sir what shall we do if there is a coefficient with X^3 also is it also workable if X is a fraction:
the first equation
Just remove the fraction but multiplying all the terms in the equation by the denominator of the fraction 😁
Thank you🌹
Your the king of math thanks very much i really appreciate cause now i can work polynomials by ma self please continue with the good work
Really help full video
How do you use this method if you're roots are non integers?
It’s actually simple. Generally, you find all the factors of “D” term, then plug those factors into the cubic. If one of those factors let’s the equation to =0, then you have already found a solution. If that doesn’t work, then take all the “D” factors and divide them by the “A” coefficient. Plug those “(Factor-ofD) / A” into the cubic and if any of those let the cubic equation =0, then you have found one of the solutions. From there, you need to convert it from a “x=_____” into “x-_____=0”. Then use polynomial long division or synthetic division to pull out a quadratic. Factor the quadratic if possible.
The other method is to just graph the cubic and use graphing technology to solve it lmao.
Thank u given nice trick..
You are very welcome Bijoy Sharma! I'm sure you are a brilliant and bright student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
Very nice sir. Sir what if the x^3 term has a coefficient ??
Did you find your answer? because i have that question too
Thank you very much you 've helped me.i didn't know that cubic function is this simple
can you write the final factor in binomial form for the first question? thx
Yes, since the binomial form has two terms.
Thanks for asking😀
Take care dear and all the best😃
Have a very happy and blessed New Year!
Really nice
Thankyou soo much ❣️
Thank you very much for this helpful video
Thank you, finally I got it.
Thanku sir ji
Thanks sir i find this formula very much helpful
You are very welcome Raj! Thank you so much for taking the time to leave this comment. I'm glad you liked it! Your feedback is always appreciated. Please keep sharing my channel with your family and friends. Take care dear and all the best 😃
I was looking for the Equation 1 kind of questions like how to solve them and yours video helped me..... thanks alot😊😊
Can we use this method in board exam sir
OMG TYSM NOW IT'S A LOT EASIER FOR ME TO ANSWER NOW
Glad to hear that JoeyLou! I'm sure you are a brilliant and bright student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
Thanku soo much Sir!
You are very welcome Richa Parekh! Thank you so much for taking the time to leave this comment. I'm glad you liked it! Your feedback is always appreciated. Please keep sharing my channel with your family and friends. Take care dear and all the best😃
@@PreMath Sure sir, i will😃
great method!
This is really best method to quickly and easily solve cubic equations .. thanq vry much sir ❤️
Thank You :)
What if the constant is 0
Thanks alot sir.. sir plz clear my doubt that when there would be a no multiplied with x^3 then what changes do we need to make.
Thank you so much❤. This video helps me a lot.....
You are very welcome Tangikina! I'm sure you are a brilliant and bright student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
Thank you for this Sir, big help. I'm now able to give my son a clear explanation. 👍🙏👍
Super helped me tnks
You are very welcome Dhanvanth! I'm sure you are a brilliant and bright student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
Little confused on the first example, my question is x^3-x-6 and it doesn’t split cleanly
How to solve if A is not 1 in Ax^3
thxxxxxxxxxxxxx. ... so much sir
You are very welcome Lakhvir ji! I'm sure you are a brilliant and bright student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
What if we have a number before X cubic?
Thank u thank u thank million thanks to you sir💚🧡💜❤️💚
You are very welcome Atul! I'm sure you are an awesome student 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and all the best😃
Greetings,
In this video you are solving all cubic equations having coefficient of x^3 is 1 i.e., a=1. Can you please explain how to solve if a is not equal to 1 also.
Thank you.
My question too
Thank u sir...
You are very welcome Rohan! I'm sure you are a brilliant and bright student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
Best trick sir
Thanks dear😀
Thank you
Thank you again for your amazing videos!
Try solving this one: th-cam.com/video/eF9T3cQREPw/w-d-xo.html
Shocks, thank you so much❤️
You are very welcome! I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
@PreMath how do we factorise 2y3 + y2 - 2y -1
I tried this,,,,it was hard for me to find factors
Any idea on this one
An awesome simple method to factor cubic equations❤
This to be taken with a grain of salt, since it works only in the rare case that the solutions, or at least one solution, is an integer number. It does no help for the general cubic, although such videos do always suggest...
Best one i saw on youtube thank u sir❣️
What b - 3ac > 0
@@W81Researcher nob
Omg sir u r a life saviour 😍🥰.
what do i do if the last number is a fraction?
Dear Celine, in that case, you won't be able to solve it that quick as the problem is going to get complicated. We may have to take different route to solve for that case. Hope I answered your question.
Take care dear and all the best😃
Have a very happy and blessed New Year!
thank you, i figured it out it wasn’t easy. but this video was very helpful!
And what about x^3-4x-1=0
But is the name of that rule or trick
Thankyou sir
Finally i understood it!!
Sir, could you please solve this equation x^3-3x^2-4x-2=0
Thank you very much Sir
And please help me to solve it
x^3-7x^2+36
That means in this format
ax^3+bx^2+c=0
X=3
27-63+36=
-36+36=9
Thanks sir
This trick works only in particular questions not in every question.
Very helpful
3:00 Sorry, but this method only works if you are certain in advance that all three solutions are real and integer, which is not realistic. In particular, you did not check if the coefficient of the first power matches Vieta's rules, too. For the cubic equation x^3 - 7x + 6 = 0, you must check if the three divisors t1 = -1, t2 = -2, t3 = 3 fulfill the three conditions
t1 + t2 + t3 = 0
t1*t2 + t1*t3 + t2*t3 = -7
t1 * t2 * t3 = 6
But you did not check if the second condition is fulfilled!
In this case, you were lucky because the second condition is indeed fulfilled, but what if the original equation were
x^3 - 8x + 6 = 0 or x^3 - 5x + 6 = 0, just to name a few examples? All these equations have coefficients of 0 for x^2 and 6 as the constant, but different (non-integer!) solutions.
good video
Thanks Vishvanathan! Love and prayers from USA!
I need answer for this equation: y^3+3root3y^2-3y-root3
Thanku sir
You are very welcome dear. Take care dear and stay blessed😃
what about x3_15x+2=0
Kaleem - for your example, try my method: th-cam.com/video/H5y9nQ0abEo/w-d-xo.html
How to solve x³-2x²+2x+1=0 by this method??
2x³+x²-5x+2=0.. can this equation solve by this method.
X=1
2+1-5+2=
3-5+2=
-2+2=0
How to do x∆3-3x-2=0
This doesn't always work though.
It works with x^3 + 11x^2 - 25x + 13
But what happens if you change the co-efficient of x? This doesn't work with x^3 + 11x^2 - 24x + 13.
Expanding the expression (x+a)(x+b)(x+c) gives us x^3 + (x^2)c + (x^2)b + bcx + a(x^2) + acx + abx + abc
Factoring this gives us:
x^3 + (a+b+c)x^2 + (ab+ac+bc)x + abc
So, what we _really_ need, is three numbers that satisfy three conditions.
The sum of the three numbers must be the co-efficient of x^2
The product of the three numbers must be the constant.
And the sum of all possible product pairs must be the co-efficient of x.
Also, unrelated to the above, but:
Sometimes, however, you may end up having a situation where the numbers that multiply to give you the constant may involve fractions.
For example, 13 can be made from -1, -26 and 1/2
If we take -1, -26 and 1/2 and add these together, we get -53/2
If we then take product pairs, 26, -1/2 and -26/2 then add these together, we get 25/2.
Therefore, (x-1)(x-26)(x+(1/2)) would be the solutions to the equation x^3 - (53/2)x^2 + (25/2)x + 13 = 0
To remove those fractions, we could then double this, to get 2x^3 - 53x^2 + 25x + 26 = 0
What if d is 1
if the sum of coefficients of all terms of a polynomial is zero, then x-1 is a factor of this polynomial.
Just wowwww😃
x^3-4x^2+4x+x-2=9
So, we get after transformation, equal solution!
Easiest way to have factor
What about expressions like 6X^3+11X^2-4X-4😭
Thanks dear for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard
Nice but doesnt work if the solutions are complex numbers.
Doesn't work for 4x³-6x²+1=0
That is not going to work in many cubic equations. He was ignoring the number in front of the x(in the first problem that number is 7, in the 2nd that number is -25 and in the third one, the number is 5). He totally ignored those numbers. A change in any one of those numbers, and his solution does not work.