# Finding x intercept

1. Dec 1, 2005

### gr3g1

Hey, Is there any simple way of finding the x intercept for

x^3-6x^2-15x+4

2. Dec 1, 2005

### Livingod

Basically find the roots of the equation?
I bet you already know that there are three roots.
to solve for x in a cubic of form ax^3+bx^2+cx+d, use a program like mathematica, or check out this cubic forumla:

http://www.math.vanderbilt.edu/~schectex/courses/cubic/cubic.gif

plug in 1 for a, 6 for b, 15 for c, and 4 for d.
pretty long, but it always works.

Last edited: Dec 1, 2005
3. Dec 2, 2005

### HallsofIvy

Staff Emeritus
As above, the only way to find the x-intercepts is to set y= 0 and solve the equation. This equation has no rational roots so there is no "simple" way to do it.

4. Dec 3, 2005

5. Dec 3, 2005

### andrewchang

what about it? $$x_{n+1} = x_n - \frac {f(x_n)}/{f'(x_n)}$$
find the derivative, and start solving...

edit:

there's supposed to be a "x_n - "before the fraction, but it isnt showing up...

6. Dec 3, 2005

### gr3g1

Thats another way of finding the x intercepts? Right?
(Our teacher never mentioned it, i think im gonna have to learn it myself)

7. Dec 3, 2005

### HallsofIvy

Staff Emeritus
Finding the x-intercepts for
x is exactly the same as finding the solutions to
x^3-6x^2-15x+4= 0. There is no "trivial?" way of doing that.

8. Dec 19, 2005