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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


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    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


    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.
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