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?

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- Sep 16th 2012, 08:05 AMtmanfind a real solution for a complex root
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? - Sep 16th 2012, 08:31 AMProve ItRe: find a real solution for a complex root
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. - Sep 16th 2012, 10:32 AMDevenoRe: find a real solution for a complex root
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}. - Sep 16th 2012, 12:01 PMtmanRe: find a real solution for a complex root
you mean to factor out a x to get x^2-4x+4

then (x-2)(x-2)/x-4? - Sep 16th 2012, 08:50 PMProve ItRe: find a real solution for a complex root
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...

$\displaystyle \displaystyle \begin{align*} x^3 - 4x^2 + x - 4 &= x^3 + x - 4x^2 - 4 \\ &= x\left( x^2 + 1 \right) - 4 \left( x^2 + 1 \right) \\ &= (x - 4) \left( x^2 + 1 \right) \\ &= (x - 4) (x - i)(x + i) \end{align*}$ - Sep 16th 2012, 08:51 PMProve ItRe: find a real solution for a complex root