what are the real or imaginary solutions of the polynomial equation x^3-8

Question

.sam. · Answer

Let $p(x)=x^3-8$
find zeros when $p(x)=0$
$$x^3-8=0 \ \ x^3=8 \ \ x=2$$
1 real solution, can you see it now?

raden · Answer

hint :
x^3 - 8 = (x-2)(x^2+2x+4)

raden · Answer

then make all factors be equal zero, solve for x

anonymous · Answer

@.Sam. : It is x^3 NOT x^2

.sam. · Answer

The other 2 are imaginary

anonymous · Answer

@.Sam. : What do you mean? The solution shall be x=-2 and x=+2 (with +2 being repeated roots). So yeah, it has 2 solutions +2 and -2, but your's isn't correct.

.sam. · Answer

The real solution of p(x) is x=2,
(x-2) $is $ a factor of p(x), Then you'll use (x-2) do synthetic division and get $x^2+2x+4$

$$=(x-2)(x^2+2x+4)$$
Then just find the imaginary solution on $x^2+2x+4$

reemii · Answer

you must solve (in the complex plane) the equation $z^3 = 8$, that is, in polar coordinates $r^3e^{i3\theta} = 2$. you'll get 3 solutions, and one of those is $z=2$.

anonymous · Answer

x^2+2x+4= (x+2)^2... the roots of this is x=-2. -2 is a real number. There is no complex solution to it.

@reemii - Yeah, one can do it as well, given all real numbers can be represented on the complex plane.

anonymous · Answer

umm... so it's not 2?

raden · Answer

well to get roots of x^2 + 2x + 4 = 0, use the quadratic formula :
x = (-b +- sqrt(b^2 - 4ac))/(2a)

here, given a = 1, b=2 and c = 4

anonymous · Answer

alright, thank you very much.

raden · Answer

and actually the solutions above be complex, because the value of discriminant (D) < 0
with D = b^2 - 4ac