I factored $x^3 - 8 = 0$ and I only got $x = 2$, but the answer said it's $x=2$ and $x=-1 \pm \sqrt{3}i$? How do you get $x=-1 \pm\sqrt{3}i$?
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$\begingroup$First of all, just a comment about the terminology. You can factor a polynomial, you can't factor an equation. What you are trying to do is solve the equation $x^3 - 8 = 0$.
As $2$ is a zero of the polynomial, $(x-2)$ is a factor by the factor theorem. So $x^3 - 8 = (x-2)p(x)$ for some polynomial. As the degree of a product of non-zero polynomials is the sum of their degrees, we see that $p(x)$ is a quadratic, so $p(x) = ax^2 + bx + c$. Comparing coefficients, we can then deduce the values of $a$, $b$, and $c$ which gives $x^3 - 2 = (x-2)(x^2+2x+4)$.
Now $x^3 - 8 = 0$ if $x - 2 = 0$ or $x^2 + 2x + 4 = 0$ by the null factor law. Solving the second equation (by using the quadratic formula for example), you find $x = -1\pm i\sqrt{3}$.
Alternatively, if you know some complex analysis (in particular, De Moivre's Theorem), you can obtain all three solutions at once.
$\endgroup$ 0 $\begingroup$Let $x = y-1$.
Then $x^3 - 8 = y^3 - 3y^2 + 3y - 9 = (y-3)(y^2 + 3)$.
Setting this last expression equal to zero, we find all three roots:
One is $y = 3$ and the other two are $y = \pm i\sqrt{3}$.
Since $x = y-1$, we conclude the roots are $x = 2$ and $x = \pm i\sqrt{3} - 1$.
"QED"
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