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Homework Help: Factorising Cubic Functions

  1. Apr 6, 2009 #1
    Hello again. :smile:

    I'm having more trouble with my homewrork, but this one really isn't fair. We haven't done the work in class, and as far as I can tell it isn't on the syllabus, but we're still expected to do it.

    1. The problem statement, all variables and given/known data
    http://img17.imageshack.us/img17/4379/questions7.jpg [Broken]

    2. Relevant equations
    http://img23.imageshack.us/img23/122/equation7.jpg [Broken]

    3. The attempt at a solution
    http://img219.imageshack.us/img219/9325/attempt7a.jpg [Broken]http://img23.imageshack.us/img23/4422/attempt7b.jpg [Broken]http://img23.imageshack.us/img23/7175/attempt7c.jpg [Broken]

    For part c, I have no idea where to go. From the googling I've done, I'm supposed to find a value of x where f(x)=0. I would have thought that the point of part b was to get a value of of x to make f(x)=0, so the trial-and-error bit would be removed.

    I used some software called Mathematica to plot the graph of this function between x=-4 and 4, and the only point where it crosses the x-axis is somewhere between x=2.3 and 2.4.

    I plotted between those points because later on I have to sketch the graph and plot the points when x=0 to x=3, and only those points.

    I originally made a mistake where I have -250 - 16. I added 16 and got -234, then came up with this :

    http://img23.imageshack.us/img23/5506/wrongh.jpg [Broken]

    This is my third day on this question, and it's driving me nuts.

    Does anyone have any hints/tips or know what I'm supposed to do?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 6, 2009 #2
  4. Apr 7, 2009 #3
    Thanks for the reply. :smile:

    I really appreciate it, but I haven't seen anything like that before. I don't even know how the maths of it works. I can't see us being expected to do that.

    I'm going to email my teacher and see if it's meant to be +16. There have been typing errors before.

    I'm going to keep a note of your reply though, just in case it does pop up on the course at some point. :smile:

    Thanks again. :smile:
  5. Apr 7, 2009 #4


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    Hi Matty R! :wink:

    I take it you're confused by the "three possible answers"?

    If you haven't done complex numbers, then just ignore that …

    follow elect_eng's :smile: hint, and get to 8/27=(x-5/16)3

    then x - 5/16 must be the cube root of 8/27, which is … ? :smile:
  6. Apr 10, 2009 #5
    Hello again Tim. :smile: I'm sorry I didn't reply sooner. There's been a family emergency. :frown:

    I've got the cube root of 8/27 as 2/3, so does that mean x=47/48?

    I don't understand how to get (x-5/16), and according to my graph, the line doesn't cross the x-axis at 47/48.

    I'm so confused.

    Complex Numbers is on the syllabus, but we haven't done it yet.

    I haven't had a reply from my teacher either.
  7. Apr 10, 2009 #6


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    Hello again Matty R! :smile:

    on loooking at the original equation, I think elect_eng copied wrong, and meant 5/3, not 5/16 (which you should have spotted :rolleyes: … you should check everything you use here) …

    does that put it right? :wink:
    Forget complex numbers. :rolleyes:
  8. Apr 10, 2009 #7
    Yes, I'm very sorry about that. I'd like to say it was just a "slip of the pen", but since we don't use pens anymore, I've got no excuse. :smile:
  9. Apr 10, 2009 #8


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    we goldfish have to use pens that write underwater …

    we can't get underwater computers …

    i can only access the internet with the help of of my mouse :wink:
  10. Apr 11, 2009 #9
    That looks great, and fits with my graph. x=7/3. So I only have one value for x because the line crosses the x-axis only once. I'm getting there. It's slow, but I'm getting there. :blushing:

    elect and tim, thank you so much. I was stuck on that question for over a week, and the answer you've helped me work out has also cleared up the next question. :smile:

    :rofl: :rofl: :rofl:
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