Hey guys I was having trouble with this question can someone plesase explain it to me in detail. thanks much appreciated.
$\displaystyle x^3 - 3^3 = (x - 3)(x^2 + 3x + 9)$.
Therefore the roots of $\displaystyle x^3 - 27$ are x = 3 and the solutions to $\displaystyle x^2 + 3x + 9 = 0$.
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A polynomial with roots $\displaystyle x = 3 \pm i$ and $\displaystyle x = -4$ is
$\displaystyle a (x - [-4])(x - [3 + i])(x - [3 - i]) = a (x + 4)([x - 3] - i)([x - 3] + i)$ $\displaystyle = a (x + 4)([x - 3]^2 + 1) = a(x + 4)(x^2 - 6x + 10)$
where a is any real number except zero.