What are the zero(s) for $f(x) = 2x^6 +x^3+3$ ?

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Answer 1 · 2015-10-24T21:29:05

$f(x)$ has six Complex zeros which we can find by recognising that $f(x)$ is a quadratic in $x^3$ .

$f(x) = 2x^6+x^3+3 = 2(x^3)^2 + x^3+3$

Using the quadratic formula we find:

$x^3 = (-1+-sqrt(1^2-4xx2xx3))/(2*2)$

$=(-1+-sqrt(-23))/4 = (-1+-i sqrt(23))/4$

So $f(x)$ has zeros:

$x_(1,2) = root(3)((-1+-i sqrt(23))/4)$

$x_(3,4) = omega root(3)((-1+-i sqrt(23))/4)$

$x_(5,6) = omega^2 root(3)((-1+-i sqrt(23))/4)$

where $omega = -1/2+sqrt(3)/2i$ is the primitive Complex cube root of unity.

What are the zero(s) for f(x) = 2x^6 +x^3+3f(x) = 2x^6 +x^3+3?