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, 09:05 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, 09:31 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 ...

Solve each of those two equations to find the three cube roots of -8. - Sep 16th 2012, 11:32 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}(= ).

(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, 01:01 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, 09:50 PMProve ItRe: find a real solution for a complex root
- Sep 16th 2012, 09:51 PMProve ItRe: find a real solution for a complex root