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Finding the dimensions

  1. Sep 25, 2006 #1
    A box has dimensions that are linear factors of (x^3-7x^2+14x-8) cubic centimetres. What dimensions give a volume of 12 cm cubed?

    I started it off by moving 12 to the left side giving the equation:


    then from then I don't know....If I substitute 5, the equation is satisfied and is a factor but from there it doesn't work. Plus 5 won't really make it 12 so...i'm lost
  2. jcsd
  3. Sep 25, 2006 #2
    Okay, so [tex] (x-5) [/tex] is a factor of your cubic. Now factorise it, and see what the other factors are.

    Edit: No wait - checked the other factors, are you sure your equations are correct?
    Last edited: Sep 25, 2006
  4. Sep 25, 2006 #3
    Well, remember that the dimensions are the factors of [itex]x^3 - 7x^2 + 14x - 8[/itex], not of [itex]x^3 - 7x^2 +14x - 20[/itex].

    You need to solve [itex]x^3 - 7x^2 +14x -20 = 0[/itex] and plug the solution for [itex]x[/itex] back into the factors of [itex]x^3 - 7x^2 +14x - 8[/itex] to get the dimensions.

    By the way, these are very odd questions you're being given!
  5. Sep 25, 2006 #4
    Ahh! Didnt notice he had changed his cubics between lines - that should sort out the problem :)
  6. Sep 25, 2006 #5
    dont know what u mean by find x because when I plug in 5 into the 2nd equation with the -20, I get = 0

    From there I can't factor any further
    And you say plug back into the first one? what will that do? It will only give me =12
  7. Sep 25, 2006 #6
    Good! So you've found one solution, x=5. That means x-5 is a factor of the cubic. Just use long division to reduce it to a quadratic and find the other roots (I'm not promising there will be any other real roots!).

    Then all you have to do is to factor [itex]x^3-7x^2+14x-8[/itex] and plug x=5 into each of the factors to find the dimensions.
  8. Sep 25, 2006 #7
    I did that but I used short division and got:


    I can't get [tex]x^2-2x+4[/tex] to equal 0

    and you can't factor the first equation with synthetic division
  9. Sep 25, 2006 #8
    use [itex] x^2 - 2x + 4 = (x^2 - 2x + 1) -1 +4 = (x-1)^2 + 3 [/itex].
  10. Sep 25, 2006 #9
    but that doesn't do me any good hargoth
  11. Sep 25, 2006 #10
    Whoops, I didn't remember you were looking for real solutions.
    I don't see why you set [itex] x^3-7x^2+14x-8=12 [/itex], since it isn't said that this polynomial is the volume. I'd suggest factorising [itex] (x^3-7x^2+14x-8) [/itex] instead, and reading off the dimensions of the box from the factorization. I think that is what is asked for. :smile:
    edit: Nope that doesn't make sense either. You were of course correct.
    Last edited: Sep 25, 2006
  12. Sep 25, 2006 #11
    He needs to factorise [itex]x^3-7x^2+14x-8[/itex] and solve [itex]x^3-7x^2+14x-8=12[/itex] (see my previous posts).

    You can just use the quadratic formula on [itex]x^2-2x+4=0[/itex] (it doesn't have any real roots though).

    All the roots of [itex]x^3-7x^2+14x-8[/itex] are small integers, so it should be easy for you to factor :smile:.
  13. Sep 25, 2006 #12
    I don't think so, because, if the dimensions are linear factors of (x^3-7x^2+14x-8), say, (x-a), (x-b), (x-c), so the Volume is (x-a)(x-b)(x-c), but this is equal to the first polynomial, so setting
    [itex](x-a)(x-b)(x-c) = x^3-7x^2+14x-8=12[/itex] was in my eyes the right thing to do.
    edit: And if this equations yield a complex solution, it doesn't help that the other linear factors are real - because the dimensions of the box you get by subtracting some real number from a complex one keep their imaginary part. This problem beats me ... :confused:
    Last edited: Sep 25, 2006
  14. Sep 25, 2006 #13
    yes, it is. You can use that equation to solve for [itex]x[/itex], and then the box has dimensions [itex](x-a) \times (x-b) \times (x-c)[/itex]. So you need to solve [itex]x^3-7x^2+14x-8=12[/itex] as well as factorise [itex]x^3-7x^2+14x-8[/itex], as I said.
  15. Sep 25, 2006 #14
    But this doesn't help in my eyes, see my edit.
  16. Sep 25, 2006 #15
    but w/e i can't even factorise that though the last part
  17. Sep 25, 2006 #16
    You have to find the real solution(s) of [itex]x^3-7x^2+14x-8=12[/itex], yes, because otherwise you'll have a box with complex side lengths (in this case, there's only one real solution).

    The roots of [itex]x^3-7x^2+14x-8[/itex] (ie. the a, b, c s.t. [itex](x-a)(x-b)(x-c) = x^3-7x^2+14x-8[/itex]) are all real, so everything's fine.
    Last edited: Sep 25, 2006
  18. Sep 25, 2006 #17
    Okay, data is right. You can insert your solution x=5 into the factorization of the polynomial you are told to do in the assignment to get the dimensions of the box. Sorry for causing confusion ...
  19. Sep 25, 2006 #18
    don't worry thanks for the help anyway, any input is good
  20. Sep 25, 2006 #19


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    5 is the only real solution to x3- 7x2+ 14x- 8= 12.
    You also need to know that x3- 7x2+ 14x- 8 factors as (x- 1)(x- 2)(x- 4). Put x= 5 into each of those factors to get the dimensions of the box.
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