# Third degree polynomial

1. Oct 28, 2013

### Elpinetos

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. Oct 28, 2013

### hilbert2

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)

3. Oct 28, 2013

### economicsnerd

To elaborate on the above:
- When you're looking for zeroes of $ax^3+bx^2+cx+d$ with $a,b,c,d$ all integers, then try for everything of the form $\frac{m}{n}$ with $m$ a divisor of $d$ and $n$ a divisor of $a$. Of course, when $a=1$ (as in your example), it's exactly as hilbert2 said.
- If $r$ is a zero of your polynomial, then $(x-r)$ is a factor. i.e. If $p(r)=0$, then $p$ is of the form $p(x)=(x-r)q(x)$ for some polynomial $q$.

4. Oct 28, 2013

### Elpinetos

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 :)

5. Oct 28, 2013

### 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.