Help Me Factorise~!~@#@! (1 Viewer)

[YannY]

haro, anyone have mathsmatica?? if you do can you plz factorise this for me
P(x)=x^3+3x+2. :burn: its the cubic function that proved my thought ideas wrong... my beautiful ideas.

Anyways, thanks

PS: for those who are trying to factorise with a pen and paper, all i can say is good luck.

 

[DownInFlames]

are you allowed to do long division, and do you want it over reals or imaginary?

edit: I forgot how to do maths. Damn. I'm working on it.

 

[undalay]

adsfasdfsdf
Use cubic formula : D
http://www.math.vanderbilt.edu/~schectex/courses/cubic/

 

[YannY]

are you allowed to do long division, and do you want it over reals or imaginary?

edit: I forgot how to do maths. Damn. I'm working on it.

i suggest you to stop.. theres no rational roots in this.

 

[YannY]

adsfasdfsdf
Use cubic formula : D
http://www.math.vanderbilt.edu/~schectex/courses/cubic/

Wtf bro lol, that formula stuned me for a minute. i need it to be factorised and that formula only will only give me one answer. i need all 3 to factorise and yeah i could just divide but that be way too messy lols.

But thanks for the site its quite interesting haha

 

[DownInFlames]

i suggest you to stop.. theres no rational roots in this.

Irrational roots I can handle... at least... I could handle...

but the formula brings it down to some nice shapely numbers since the b=0.

did the question specify how you were meant to do it, or did it just say: here's an equation, go?

 

[YannY]

Irrational roots I can handle... at least... I could handle...

but the formula brings it down to some nice shapely numbers since the b=0.

did the question specify how you were meant to do it, or did it just say: here's an equation, go?

Nah the question asked for a completely different thing with the equation but the equation itself kinda bugged me so i want to find the roots of it. And no knowing what the question wanted does not help even a tiny bit with my question.

Thanks for trying, all i can say is the irrational real root is a rational+irrational

 

[Affinity]

yeah the cubic equation would do the trick..

or if you understand how the equation was derived, you would notice that the cubic does not have the sqaure term (which reduces the formula)

so just let x = u-v and proceed:

u^3 - v^3 + 3uv(u-v) + 3(u-v) + 2 = 0

so if we can let
u^3 - v^3 = -2
uv = -1 then we are done.

so one has u^6 + 2u^3 + 1 = 0
so u^3 = 1 u = one of the cube roots of unity

etc.

 

[YannY]

yeah the cubic equation would do the trick..

or if you understand how the equation was derived, you would notice that the cubic does not have the sqaure term (which reduces the formula)

so just let x = u-v and proceed:

u^3 - v^3 + 3uv(u-v) + 3(u-v) + 2 = 0

so if we can let
u^3 - v^3 = -2
uv = -1 then we are done.

so one has u^6 + 2u^3 + 1 = 0
so u^3 = 1 u = one of the cube roots of unity

etc.

This is a wonderful formula but u^3=-1 and one of the solutions of u is -1 and that means x=0 is a solution of the polynomial which dosnt make sense?

the other solutions of the cube roots of unity does not work either

 

[Affinity]

This is a wonderful formula but u^3=-1 and one of the solutions of u is -1 and that means x=0 is a solution of the polynomial which dosnt make sense?

the other solutions of the cube roots of unity does not work either

made a hiccup,

got the expansion wrong

not u^3 - v^3 + 3uv(u-v) + 3(u-v) + 2 = 0
should have been

u^3 - v^3 - 3uv(u-v) + 3(u-v) + 2 = 0

so infact want u^3 - v^3 = -2
uv = 1

u^6 + 2u^3 - 1 = 0
which is a little nasty

you get u^3 = 1+/- sqrt(2)

proceed from there.

 

[YannY]

made a hiccup,

got the expansion wrong

not u^3 - v^3 + 3uv(u-v) + 3(u-v) + 2 = 0
should have been

u^3 - v^3 - 3uv(u-v) + 3(u-v) + 2 = 0

so infact want u^3 - v^3 = -2
uv = 1

u^6 + 2u^3 - 1 = 0
which is a little nasty

you get u^3 = 1+/- sqrt(2)

proceed from there.

yeah i got the answer, thanks for the method its really great. is this method only usuable if one term is missing like x^2 here?

 

[Slidey]

No. Use a substitution to 'fold' any 4-term cubic into a 3-term cubic.

I.e.:

ax^3+bx^2+cx+d=0
use substitution: x=y-b/3a
a(y-b/3a)^3+b(y-b/3a)^2+c(y-b/3a)+d=0
a(y^3-by^2/a+b^2y/3a^2-b^3/27a^3)+b(y^2-2by/3a+b^2/9a^2)+cy-cb/3a+d=0
ay^3+(c-b^2/3a)y+(2b^3/27a^2-cb/3a+d)=0

As you can see, this is of the form ly^3+my+n=0

 

[Affinity]

yeah... and that's basically how the cubic formula was derived.

 

[buchanan]

I factorised it over R for you:

It can be factorised further over C but I prefer it over R.

You don't need a computer. Pen and paper will do fine.

Use the formula http://users.tpg.com.au/nanahcub/cubic.gif

There's another one for quartics at http://users.tpg.com.au/nanahcub/quartic.gif

 

[YannY]

I factorised it over R for you:

It can be factorised further over C but I prefer it over R.

You don't need a computer. Pen and paper will do fine.

Use the formula http://users.tpg.com.au/nanahcub/cubic.gif

There's another one for quartics at http://users.tpg.com.au/nanahcub/quartic.gif

Oh thanks haha thats messy, thanks alot