Question

How do you find the zeroes of a cubic function?

Answer 1

- for example ???

Answer 2

-(x^3 + 4x² + x + 26)

Answer 3

There's a formula for the cubic, but it is very convoluted.

Answer 4

What you want to do is use the rational roots theorem to find a first root.

Answer 5

You can then factor out the remaining quadratic.

Answer 6

*or solve, the remaining quadratic.

Answer 7

There is almost no case in which you will be required to solve a cubic function at school, but there are formulas for the curious.

Answer 8

Nothing simple like the quadratic formula, but for cubics, eh? That's unfortunate, but I'll try your methor inkyvoyd! Thank you.

Answer 9

The process involved with solving them is very long and complicated, so I won't be able to explain it here without insanity.

Answer 10

You can attempt to factor it, or you can try to find a root by looking at...

ax^3 + bx^2 + cx + d = 0.

Look at any factor of d over any factor of a.

Look at http://www.sosmath.com/algebra/factor/fac10/fac10.html
where it says "Another example." Or better yet, read the whole thing.

Answer 11

Haha, I attend just the type of school at which we are required to solve a cubic function.

Answer 12

Once you find a factor r, divide the polynomial by (x-r) to factor out an (x-r).

Answer 13

Well, try by factoring.

Answer 14

Is the example you gave the one you need to solve?

Answer 15

If you are really asked to solve the cubic in the form ax^3+bx^2+cx+d=0, I can give you a basic outline of things you need to do to get it in it's solvable reduced form

Answer 16

I would greatly appreciate that! I'm running out of time, but I will take any help you can offer.

Answer 17

In fact, I have another question for you as well. How are you supposed to tell if a function will have complex roots just by looking at the equation?

Answer 18

uhm, what math class are you enrolled in?

Answer 19

IB Precalculus

Answer 20

Ok, let me try it out.

Answer 21

Thank you so much.

Answer 22

I'm not sure about the complex roots thing, but it has to do with complex conjugates being in a cubic equation.

Answer 23

Ok, first divide both sides by negative one.

Answer 24

0=x^3 - 4x² - x - 26

Answer 25

Mhm. It does. The point of my project is to work with shadow functions and see how they relate with finding the vertex and zeroes. Would you like to see the picture of the graph?

Answer 26

Sure.

Answer 27

Okay, here. http://screencast.com/t/f6hi9K7Oi

Answer 28

I understand that you need an exact solution?

Answer 29

what's the function?

Answer 30

x^3-4x^2-7x+10?

Answer 31

@AgentNao ?

Answer 32

I'll assume it is, because we are both running outta time (quite late here in my time zone)

Answer 33

quick wolfram search shows that that cubic actually factors into a nice pretty form.

Answer 34

@AgentNao ?

Answer 35

Yes! Oh goodness, sorry about that.

Answer 36

I need to find the roots for both the functions, but I know how to find the one that is black. I just don't know how to find the purple one.

Answer 37

Ok, see the quadratic that remains?

Answer 38

We can actually factor that with the quadratic equation.

Answer 39

gimme a moment, lemme write this on paper.

Answer 40

kay

Answer 41

My only problem with that is I don't know how I got that equation. This --> (x + 2)[x – (3 + 2i)][x – (3x – 2i)] is the ORIGINAL equation that I was given and we were required to put it in standard form and then graph it and its shadow as well as find its zeroes. I did not get -(x + 2) (x² - 6x + 13) on my own though and we're supposed to show every step but I don't know how to get that D:

Answer 42

Oh, can't you just simplify the equation you were given?

Answer 43

-[x – (3 + 2i)][x – (3x – 2i)] should equal (x² - 6x + 13) according to what you've said

Answer 44

Just multiply them out, you should get the result.

Answer 45

*expand, not simplify, sorry.

Answer 46

I multiplied it out multiple times, but I never got that! I must be doing something wrong, but I've checked it over and over and can't seem to get that.

Answer 47

Let me try.

Answer 48

With mathematica.

Answer 49

Would you mind showing me what you're doing using this website? http://www.twiddla.com/822073 That way I can watch you work it out.

Answer 50

-[x – (3 + 2i)][x – (3x – 2i)]
I just realized, you are missing an x!
-[x – (3x + 2i)][x – (3x – 2i)]

Answer 51

They should be complex conjugates, but they are not.

Answer 52

Oh, whoops! I accidentally typed it in wrong, but when I worked it out on my paper, I did have the appropriate Xs

Answer 53

did you get -4-4x^2 ?

Answer 54

I'm pretty sure I got the same thing, with just the opposite sign. 4x^2 + 4

Answer 55

Uhm, one sec.

Answer 56

If you click the link I sent you, I think you'll find it easier! It's like a public whiteboard room.

Answer 57

Yes, that's strange. Where did you get -(x + 2) (x² - 6x + 13)?