/sci/ - Science & Math

[Anonymous No.1956165]

I need to find the real zeroes of f.

f(x)=3x^3+4x^2+4x+1

I know the answer will be irrational.

Pic unrelated.
Do I use the quadratic formula?
How would I go about doing that?
I don't understand how to solve it.

[Anonymous No.1956192]

the cubic formula is a bitch..

all these can be written as (x-x1)(x-2)(x-3)
guess one root (usually +-1 or 0 for solvability).
here try -1

then do long division with (x- (-1) = (x+1)

and use cubic formula on the rest

[Anonymous No.1956197]

>>1956192

sorry, the short form should be (x-x1)(x-x2)(x-x3)

[Anonymous No.1956206]

competitive mathlete here.
this is a cubic function, quadratic formula is irrelevant here.

one thing i suggest is to remember that when you find the first root, it's conjugate is also a root, as is the case with all imaginary factors of polynomials.

do your own homework.

[μαθμαν !!vnLoxcfyRGs No.1956210]

-1/3, use quadratic formula after dividing out the root.

[Anonymous No.1956212]

>>1956192
imaginary roots, +/-1 and 0 are not roots, it has 2 pairs of conjugate imaginary roots

[Anonymous No.1956235]

>>1956212

http://www.wolframalpha.com/input/?i=0%3D3x^3%2B4x^2%2B4x%2B1

>real root

sorry, was a bit quick with the ones, but my point stands.

they dont usually give these to do by hand so you need to spot the complex roots

[Anonymous No.1956242]

>>1956212

>cubed
>has two pairs of imaginary roots
>fundamental theorem of algebra

mfw

[Anonymous No.1956969]

Are you fucking serious? I posted a problem similar to this a couple weeks ago and /sci chewed me up saying I should just quit school and calling me a dumbass. WTF

Anyway, use long division on an integer you think is a zero. If you get 0 for the remainder, it works! Then use quadratic formula.

[Anonymous No.1957013]

or if you have a TI graphing calculator you can enter the function into y= and go 2nd ---> calc, and one of the first few options are zeros. then just put in left/right bounds and you've got your zeros.