Factoring a cubic polynomial in two ways

This video is about factoring a cubic polynomial
If you need to post a picture of your solution or idea:
#ChallengingMathProblems #Polynomials

EXPLORE:

[hhgygy]

Before watching the video I tried a method and it seems I did it a bit differently:
2x^3+x^2+x-1
Add and deduct x^2 at the same time:
2x^3+2x^2-x^2+x-1
Deduct and add x at the same time and rearrange it:
2x^3+2x^2-x^2-x+2x-1
Rearrange again:
2x^2(x+1)-x(x+1)+2x-1
Rearrange again:
(x+1)(2x^2-x)+2x-1
Rearrange again
x(x+1)(2x-1)+(2x-1)
Rearrange again:
(x(x+1)+1)(2x-1)
Finally:
(x^2+x+1)(2x-1)

(Sorry I don't know the right mathematical terms for these operations in English)

[damianbla]

I solved this in that way:
f(0) = -1 <0
f(1) = 3 >0

So there must be a solution between 0 and 1.

From the rational roots theorem,
I have to check if 1/2 is a solution.
f(1/2) = 0 so x=1/2 is a solution.

P.S.
From the Carthesius' rule of signs,
there is (for sure!) one positive solution.

[billprovince]

For me, I just noted that 1/2 had to be a root. I used rational root theorem, and since the coefficients were relatively simple, using rational root theorem was relatively painless. So, that finds one of the roots, and the other can be obtained by synthetic division.

[vishalmishra]

Split 2x^3 into x^3+x^3. Then, LHS = (1+x+x^2)x + (x^3-1) = (1+x+x^2)(x + (x-1)). 2x-1=0 => x=1/2 and 1+w+w^2 = 0 since (x^3-1)/(x-1)=0, so x = cube-roots of unity but not 1. Hence, the polynomial can be factorized into (2x-1)(x-w)(x-1/w) where w and 1/w (or w^2) are complex conjugate cube-roots of unity. *Simple* . Right ?

[johnnath]

Your 1st method was excellent. Your 2nd method felt like it was skirting around the rational roots theorem without explicitly using it. If you turn your polynomial into an equation and divide it by 2, the only possible roots of the resulting equation are ± ½, ±1. The ±1 can be quickly ruled out, and the ½ fairly quickly established. This gives you the 2x – 1 factor. The other factor is then a long division routine.

[PokemonGamerBR]

3rd method: realize 1/2 is a root, so 2x-1 is a factor of the expression. Divide by 2x-1 to find the other factor.

[chandrashekharmehta]

When you multiplied the original expression by 4 the last term should be
-4 instead of - 1 ooh when I was writing this you have corrected it 🤣🤣🤣😂🤗🤗😏👍😉

[srijanbhowmick]

I have another NT problem :----
Solve the following system of modulo equations in integers :----
a = b (mod c)
b = c (mod a)
c = a (mod b)
This one's very easy !

[drinkingsodacomicsincorpor]

I dont understand how in the first method you split x^3-1 + x^3+x^2+x into (x-1)(x^2+x+1) + x(x^2+x+1). I understand how you got this (x-1)(x^2+x+1), but not the other one. How did x^3-1 factor into x(x^2+x+1)?

[Hobbitangle]

"Can you factor this cubic polynomial?"
I'll try to, not watching the video first.
Let's multiply the polynomial by 4 and make the substitution y=2•x
Then the polynomial will be
y^3+y^2+2•y-4
Quite obviously that y=1 is the root and we can divide this qubic polynomial by y-1 resulting to quadratic form.
But we will use the following transformation
y^3+y^2+2•y-4=
y^3-y^2+2•y^2+2•y-4=
y^2•(y-1)+2•y^2-2•y+4•y-4=
y^2•(y-1)+2•y•(y-1)+4•(y-1)=
(y^2+2•y+4)•(y-1)
This polynomial has the only one real root because the discriminant of the quadratic part is -12 that is less than 0.
So we're getting back to the x variable, making back substitution y=2•y:
(4•x^2+4•x+4)•(2•x-1)
Dividing the polynomial back by 4 we'll get the final result: (x^2+x+1)•(2•x-1)

[heribertogutierrez]

Excelente demostración es una clase magistral gracias por la información

[scottleung]

I actually took it a step further by solving the polynomial!

[kfjfkeofitorhf]

6X+2X+X-1=1, 2
6X-2X+X-1=
5X=1/5
X=1/5=
2(0, 8)+(0, 4)+(0, 2)-1=
1, 6+0, 4+0, 2-1=
2+0, 2-1=
2, 2-1=1, 2

[Heenakhan-kcze]

I enjoyed second method 👍❤️ it's quite tricky but interesting...

[aahaanchawla]

The solutions are rational or at-least we have rational solution
Every 60s in Africa, a minute passes

[mrhatman]

I use this method when I am able to find at least one solution by chance let for example a polinomial x^3+x^2+x-3=0 right fom the start we can notice 1 is a solution so we can say this polinomial becomes in the form of since its a polinomials of a third deegre we know it has 3 total soltutions so we might as well set the original expression to this (x-1)(ax^2+bx+c) where a, b, c are the cooficients now we just multiply everything out and it gives from now on we just set the right cooficients so it matches with the other side of the equation in this case a=1 b=2 c=3 replacing these cooficients to your quadratic equation and then solve for the other x and you know the rest but i have to say this method only is confortable when you can find a solution right away and don t want to deal with a dificult factorization of your polinomial

[golddddus]

Synthetic division solution:
2 1 1 -1
1/2 1 1 1
2 2 2 0, so

[coolmangame]

Lol i was testing some values of x after setting that polynomial equal 0 and found x = 1/2 works. No polynomial can stop my luck

[shatishankaryadav]

If we have option then it 10 sec problem

[krishnannarayanan]

elementary student should know 4 muliplied by 1 is 1 and then you can correct it to 4, but you might be out of your test and fail: good education to students!!!