5. What are the real or imaginary solutions of the polynomial equation? x3 – 8 =0 (1 point)

Question

5.   What are the real or imaginary solutions of the polynomial equation? x3 – 8 =0  (1 point)

anonymous · Answer

I have absolutely no idea how to solve this!

anonymous · Answer

$$(x-2)(x^2+2x+4)=0$$ so one solution is obvious, namely $x=2$ the other two via quadratic formula

anonymous · Answer

oh the first step was to factor $x^3-8$ it is the difference of two cubes

anonymous · Answer

$a^3-b^3=(a-b)(a^2+ab+b^2)$

anonymous · Answer

1 + isquare root 3, and 1 – isquare root 3
2, –1 + isquare root 3, and –1 – isquare root 3
2, 1 + 2isquare root 3, and 1 – 2i square root 3
2, 2 +2isquare root 3, and 2 – 2isquare root 3

These are my choices. Which one is it?

anonymous · Answer

do you know the quadratic formula?

anonymous · Answer

Sorta.

anonymous · Answer

that is what you need

anonymous · Answer

$$x^2+2x+4=0$$ use 
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ with 
$$a=1,b=2, c=4$$

anonymous · Answer

actually in this case it is easier to complete the square

anonymous · Answer

$$x^2+2x+4=0$$

$$x^2+2x=-4$$
$$(x+1)^2=-4+1=-3$$
$$x+1=\pm\sqrt{-3}$$
$$x=-1\pm\sqrt{-3}$$

anonymous · Answer

Thanks so much! :)

anonymous · Answer

yw