Question

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

Answer 1

I have absolutely no idea how to solve this!

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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?

Answer 6

do you know the quadratic formula?

Answer 7

Sorta.

Answer 8

that is what you need

Answer 9

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

Answer 10

actually in this case it is easier to complete the square

Answer 11

\[x^2+2x+4=0\]

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

Answer 12

Thanks so much! :)

Answer 13

yw