Question

f(x)=x^3+x+31

Slowly, posting bit by bit, please.

Answer 1

Well? :p

Answer 2

:'c

Answer 3

you want to derived?

Answer 4

\(\huge\color{blue}{Can~~~someone~~teach~~me?}\)

Answer 5

What do you need to do?

Answer 6

I don't need to do anything with it as in school, but I want to know how to find the zeros of this function.

Answer 7

Try to get it in this form.

\(x(x\pm\underline{~~~})(x\pm\underline{~~~})\)

Answer 8

How?

Answer 9

if you substitute 0 to x F(x)=31?

Answer 10

k, then....

Answer 11

or \[F \prime (x) =3x^{2}+x \] then \[3x^{2}+x=0 \[x(3x+1)=0\] \[x=0 \] and \[x=-\frac{ 1 }{ 3 }\]

Answer 12

what about 31, where did it go?

Answer 13

31 when we don't derived F(x)

Answer 14

That makes no sense. Sorry Solomon, I know you're upset with me. I missunderstood how this was going to work in the beginning. I apologize. But where did you get this problem?

Answer 15

batman, did you just say that the question makes no sense?
ganeshie8 gave it to me, as an example.

Answer 16

No, I was referring to her explanation. It threw me off. I was just wondering, because it didn't even start off in the correct form: ax^2+bx+c. It cannot be factored. I tried to do the opposite of b thing, and that doesn't work either, well, it might, but it's a really awkward example. You should have started off with something like x^2+6x+8 or something..

Answer 17

NO, I am trying to learn cubic functions, I know how to do things like ax^3+bx^2+cx+d, but this one is just ax^3+cx+d

Answer 18

Well then. When we did stuff like that, you had to plug in a false number to replace that empty spot, because it has a peice missing. But this one is still a really hard example. Like, for yours, we would do x^3+0x^2+x+31. But normally, you would factor by grouping.

Answer 19

but what do I do after plugging 0x^2 in?

Answer 20

Like I said, normally we would factor by grouping... But this problem is really awkward.

Answer 21

So how would be do this one, do you know?

Answer 22

I'm looking it up. There are a few things I've found. There's one where you factor c, and plug in the factor's for all the x's. There's another where you do: \[\frac{ -b \pm \sqrt{b^{2}-3ac} }{ 3a }\] But I wish you could have started off with an easier example.

Answer 23

I know how to do the easier once, hold on....

Answer 24

Which way do you do the simpler ones?

Answer 25

I find the zeros, as +- factors of the constant.

So I would say that they are +-1, +-31, but none of them work.

Answer 26

I am now fully determined to figure this out.

Answer 27

Can someone just show me, without giving me mysterious tips?

Answer 28

I just asked my goto, and he said, " I don't think it can be solved/factored.
The roots aren't "real"
There IS a quadratic formula equivalent for cubics but I think it's a lot longer and impractical by hand.

Answer 29

Like I said, bad example. Why doesn't the guy who gae the problem to you show you how to do it?

Answer 30

Whatever the roots are...

Answer 31

Imaginary, complex, can you show me how to find them?

Answer 32

*gave. Find what?

Answer 33

\[f(x)=x^3+x+31~~~~~find~~the~~roots.\]