What are the solutions to x^3+3x^2-4=0?

[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 :/

 

[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)

 

[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.

 

[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 :)

 

[Mark44]

x = 1 IS a solution of x³ + 3x² - 4 = 0, which means that x - 1 is a factor. There are two more solutions.