i am having problems understanding the roots
have the equation:
given that x = 4 is a real solution for f(x) = x^3 - 4x^2 + x - 4, find one of the complex roots?
and the steps to find a cube root of -8?
Since x = 4 is a solution, that means (x - 4) is a factor. Long divide to get the quadratic factor, and factorise that over the complex numbers to evaluate the other two solutions (they will be complex conjugates).
To find $\displaystyle \displaystyle \begin{align*} \sqrt[3]{-8} \end{align*}$...
$\displaystyle \displaystyle \begin{align*} z^3 &= -8 \\ z^3 + 8 &= 0 \\ (z + 2)\left( z^2 - 2z + 4 \right) &= 0 \\ z + 2 = 0 \textrm{ or } z^2 - 2z + 4 &= 0 \end{align*}$
Solve each of those two equations to find the three cube roots of -8.
an only slighter easier way:
suppose we already know how to find the 3 cube roots of 1 (by, say, using trigonometry), call these: 1, ω, and ω^{2} (= $\displaystyle \overline{\omega}$).
(it should be easy to see that if ω^{3} = 1, and ω ≠ 1, then (ω^{2})^{3} = (ω^{3})^{2} = (1)(1) = 1, as well).
(we can find these by factoring x^{3} - 1 = (x - 1)(x^{2} + x + 1), also).
then (x + 2)(x + 2ω)(x + 2ω^{2}) = (x + 2)(x^{2} + (2ω + 2ω^{2})x + 4ω^{3})
= (x + 2)(x^{2} + 2(ω + ω^{2})x + 4), and we know from above that: ω + ω^{2} = -1, so:
= (x + 2)(x^{2} - 2x^{2} + 4) = x^{3} + 8
so the three cube roots of -8 are -2,-2ω, and -2ω^{2}.
I don't know who you're speaking to, but no, there is no factoring of a single x. You are going to long divide $\displaystyle \displaystyle \begin{align*} \frac{x^3 - 4x^2 + x - 4}{x - 4} \end{align*}$.
The top is also able to be factorised by grouping...
$\displaystyle \displaystyle \begin{align*} x^3 - 4x^2 + x - 4 &= x^3 + x - 4x^2 - 4 \\ &= x\left( x^2 + 1 \right) - 4 \left( x^2 + 1 \right) \\ &= (x - 4) \left( x^2 + 1 \right) \\ &= (x - 4) (x - i)(x + i) \end{align*}$