finding one root isn't solving the problem, you only solved the polynomial for one root, when there possibly are 2 more. Luckily, it was a triple root in this case, but in other cases you would have been wrong.
@@CeRzfinding one solution makes it much easier and trivial though, after polynomial division and solving for the quadratic it doesn’t really matter, that was completely different from OPs point
@@CeRz Lol maybe if you don’t force your weird ideology or ego onto everyone who wanted to originally make a simple suggestion you would get what you wanted
Instead of factoring like you did, I used Ruffini's factoring method twice and discovered that it was, in fact, the product of three equal binomials. I like your method too, but I couldn't help myself! Every time I see a cubic or higher polinomial with all coeficients higher than zero I simply have to use Ruffini's!
Well, this was long. You can just use the cube of difference formula instead, which is (a-b)^3 = a^3 - 3*(a^2)*b + 3*a*(b^2) - b^3 Obviously a is x, b is 2. By that you can already see that what we have here is exactly the (x-2)^3, cause 6x^2 is 3*2*x^2 which is our 3*(a^2)*b and 12x is 3*4*x which is our 3*a*(b^2)
I have always wondered. How do we solve a cubic equation that has no “easy” roots? Because till now the only method I am aware of is just guess and check one root then reduce to quadratic. But what if all 3 roots were irrational? How do we solve it then? Thank you in advance.
There are general formulas for third and fourth degree equations. They are rather complex (both in the everyday sense and in the mathematical meaning of the word). Mathologer made a great video on the subject.
But it depends because in school teachers wouldn’t give you insanely hard problems but if you want to learn it for fun then you can check out some of the other math channels
factoring is much simpler, no need for the difference of cubes: x^3 - 6x^2 + 12x - 8 = x^3 - 2x^2 - 4x^2 + 8x + 4x - 8 = x^2(x-2) - 4x(x-2) + 4(x-2) = (x-2)(x^2 - 4x + 4) and after that like in the video: (x-2)(x-2)^2 = (x-2)^3
But this method isn’t very intuitive and most people wouldn’t think of breaking up the terms like that. I like bprp’s method better because most people who have done enough practice can recognize that x cubed and -8 are both perfect cubes.
Bro again, if it is asked in competitive exam that means we have to think slightly out of box Usually for any cubic or higher order equation we 1st have to find factors by substituting Here we can try x=-2,-1,0,1,2 Since x=2 satisfies equation that means x-2 is a factor of the equation So above equation reduces to (x-2)(x^2-4x+4) i.e (x-2)(x-2)^2 i.e (x-2)^3 = 0 ie x=2 is the only root
your third sentence is not a complete description of all possible roots. All possible roots are all the factors of the constant term with a plus or minus sign. That is, -8 has factors +-1, +-2, +-4 or +-8. If +-1 or +-2 doesn't work for you, you didn't describe in your set, all the possible roots. How are you then going to find roots for the cubic polynomial? 0 is not possible when you have a constant term not equal to 0, you will have a contradiction in this case 0 is not equal to -8.
i dont like this solution. find a root, factorize it, then do long division x^3-6x^2+12x-8 / (x-2); you know one root will be +-1, +-2, +-4 or +-8 due to the constant term being -8, these are all factors of -8. It is clear that 2 is a root, thus (x-2) is a factor to the polynomial. The answer from long division results in a 0 rest term and x^2-4x+4 and this can in turn be factorized because -4x = -2x-2x and +4=(-2)*(-2) (this shows that, again, the polynomial has a double root and it is 2.) and thus x^2-4x+4=(x-2)(x-2). This is basic calculus.
Solving x^3=8 and it's NOT just x=2.
th-cam.com/video/7ac4fp7M4t0/w-d-xo.htmlsi=NiPc_fHQCUy90lMK
You could also recognize that the equation is the the form of a^3-3a^2b+3ab^2-b^3 for a=x, and b=2
Try roots using the rational root theorem. Then use long division to find the quadratic. Then use the quadratic equation.
I sometimes use Calculus to find it lol x- f(x)/f'(x). Bad iterate it a few times and get to the original solution.
no
That's what's taught in school but I like the solution in the video better
Instantly saw x=2. Programming has warped my brain.
finding one root isn't solving the problem, you only solved the polynomial for one root, when there possibly are 2 more. Luckily, it was a triple root in this case, but in other cases you would have been wrong.
@@CeRzfinding one solution makes it much easier and trivial though, after polynomial division and solving for the quadratic it doesn’t really matter, that was completely different from OPs point
@@CeRz Bro im not reading all this im not trying to act smarter than u
@@CeRz no you may not add, I didn’t really ask
@@CeRz Lol maybe if you don’t force your weird ideology or ego onto everyone who wanted to originally make a simple suggestion you would get what you wanted
x^3-6x^2+12x-8=0
x^3-2x^2-4x^2+8x+4x-8=0
(x^3-2x^2)-(4x^2+8x)+(4x-8)=0
x^2*(x-2)-4x*(x-2)+4(x-2)=0
(x-2)*(x^2-4x+4)=0
(x-2)*(x-2)^2=0
(x-2)^3=0
x-2=0
x=2
Instead of factoring like you did, I used Ruffini's factoring method twice and discovered that it was, in fact, the product of three equal binomials. I like your method too, but I couldn't help myself! Every time I see a cubic or higher polinomial with all coeficients higher than zero I simply have to use Ruffini's!
If you are too lazy to watch the other video here
x=2,x=-1±i(squareroot of 3)
Square root of 3 is ≈1.732
The ± sign means two more solutions so 3
sqrt(x) is short for "the square root of x" if you don't have the √ (radical) symbol.
I noticed that all the coefficients were even and wondered if we could polynomial long division it by x-2. And... yup works.
Well, this was long.
You can just use the cube of difference formula instead, which is (a-b)^3 = a^3 - 3*(a^2)*b + 3*a*(b^2) - b^3
Obviously a is x, b is 2.
By that you can already see that what we have here is exactly the (x-2)^3, cause 6x^2 is 3*2*x^2 which is our 3*(a^2)*b and 12x is 3*4*x which is our 3*a*(b^2)
Your way is just as long lmao. It just seems long cuz he’s trying to explain it clearly to everyone, even the less advanced people
Just notice that -8 is the cube of -2 and that both 6 and 12 have factors of 2 and 3. Check (x-2)^3 and be done with it.
Since it has double root (triple root actually), you can take gcd of the polynomial and its derivative. Not the simplest method in this case.
I love this channel
Ok😢
What constant cubed is equal to -8, -2. So x=2
Is my logic wrong here ?
I have always wondered. How do we solve a cubic equation that has no “easy” roots?
Because till now the only method I am aware of is just guess and check one root then reduce to quadratic. But what if all 3 roots were irrational? How do we solve it then? Thank you in advance.
There are general formulas for third and fourth degree equations. They are rather complex (both in the everyday sense and in the mathematical meaning of the word).
Mathologer made a great video on the subject.
It’s possible but it isn’t easy and it would usually require a calculator (not necessarily a graphing calculator)
But it depends because in school teachers wouldn’t give you insanely hard problems but if you want to learn it for fun then you can check out some of the other math channels
Using the binomial theorem, the question can be easily seen as (x-2)³ = 0.
factoring is much simpler, no need for the difference of cubes:
x^3 - 6x^2 + 12x - 8 = x^3 - 2x^2 - 4x^2 + 8x + 4x - 8 = x^2(x-2) - 4x(x-2) + 4(x-2) = (x-2)(x^2 - 4x + 4)
and after that like in the video: (x-2)(x-2)^2 = (x-2)^3
But this method isn’t very intuitive and most people wouldn’t think of breaking up the terms like that. I like bprp’s method better because most people who have done enough practice can recognize that x cubed and -8 are both perfect cubes.
watching sum1 solving math is a lot more fun than doing math
But wouldn't using Ruffini have been better?
hello
hi
Hey guys! How are you doing?
@@damiencarlocop7700 lo
so all 3 answers just happen to be x=2
its has two complex answers which are x=-1±sqrt(3)i
Bro again, if it is asked in competitive exam that means we have to think slightly out of box
Usually for any cubic or higher order equation we 1st have to find factors by substituting
Here we can try x=-2,-1,0,1,2
Since x=2 satisfies equation that means x-2 is a factor of the equation
So above equation reduces to
(x-2)(x^2-4x+4) i.e (x-2)(x-2)^2 i.e (x-2)^3 = 0
ie x=2 is the only root
your third sentence is not a complete description of all possible roots. All possible roots are all the factors of the constant term with a plus or minus sign.
That is, -8 has factors +-1, +-2, +-4 or +-8. If +-1 or +-2 doesn't work for you, you didn't describe in your set, all the possible roots. How are you then going to find roots for the cubic polynomial? 0 is not possible when you have a constant term not equal to 0, you will have a contradiction in this case 0 is not equal to -8.
Trust me bro, if you really love mathematics and not just wanna pass a test, guessing is not a very good way to go with.
An ambitious junior
i dont like this solution. find a root, factorize it, then do long division x^3-6x^2+12x-8 / (x-2); you know one root will be +-1, +-2, +-4 or +-8 due to the constant term being -8, these are all factors of -8. It is clear that 2 is a root, thus (x-2) is a factor to the polynomial. The answer from long division results in a 0 rest term and x^2-4x+4 and this can in turn be factorized because -4x = -2x-2x and +4=(-2)*(-2) (this shows that, again, the polynomial has a double root and it is 2.) and thus x^2-4x+4=(x-2)(x-2). This is basic calculus.
Where is the link to the other video you mentioned?
th-cam.com/video/7ac4fp7M4t0/w-d-xo.html
pinned comment and description