1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Third degree polynomial

  1. Oct 28, 2013 #1
    Hi guys, somehow after a couple years of not doing math I got a bit rusty...
    How do I solve x^3+3x^2-4=0 ?

    I'm kinda stuck? I figured factorizing but I can't seem to find any good factors :/
     
  2. jcsd
  3. Oct 28, 2013 #2

    hilbert2

    User Avatar
    Science Advisor
    Gold Member

    You can factorize it as ##x^{3}+3x^{2}-4=(x-1)(x^{2}+4x+4)##. When looking for zeroes of a polynomial ##ax^{3}+bx^{2}+cx+d## with integer coefficients ##a,b,c,d##, try the integer divisors of constant term ##d##. (here the divisors of -4 are 1,-1,2,-2,4 and -4)
     
  4. Oct 28, 2013 #3
    To elaborate on the above:
    - When you're looking for zeroes of [itex]ax^3+bx^2+cx+d[/itex] with [itex]a,b,c,d[/itex] all integers, then try for everything of the form [itex]\frac{m}{n}[/itex] with [itex]m[/itex] a divisor of [itex]d[/itex] and [itex]n[/itex] a divisor of [itex]a[/itex]. Of course, when [itex]a=1[/itex] (as in your example), it's exactly as hilbert2 said.
    - If [itex]r[/itex] is a zero of your polynomial, then [itex](x-r)[/itex] is a factor. i.e. If [itex]p(r)=0[/itex], then [itex]p[/itex] is of the form [itex]p(x)=(x-r)q(x)[/itex] for some polynomial [itex]q[/itex].
     
  5. Oct 28, 2013 #4
    Thank you guys. I don't know where I went wrong, I got 1 as a solution, but I must've screwed up during polynomial divison.
    Oh well, it's been a long day.
    Thank you :)
     
  6. Oct 28, 2013 #5

    Mark44

    Staff: Mentor

    x = 1 IS a solution of x3 + 3x2 - 4 = 0, which means that x - 1 is a factor. There are two more solutions.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...