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