# Finding Imaginary Values of a Degree 6 Polynomial

Here's the problem: $0 = x^6 - 65x^3 + 64$.

I understand to insert "z" for $x^3$, which gets roots 4 and 1. That leaves you with a $4^{th}$ degree polynomial (which I have found). I know how to synthetically divide for 4th degree polynomials, and how to solve for most special cases. However, the answer guide I have says that the remaining 4 answers are all imaginary.

How do you find the roots of a degree 4 polynomial if all 4 are imaginary, and the problem is not easily reducible?

Alternately, is there a different/faster way to solve a degree 6 polynomial?

If you can factor $$p(x) = x^6 - 65x^3 + 64 = (x^3 - 64)(x^3 - 1),$$

then you can use the formula for a difference of cubes, $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ to factor $p(x)$ into a product of two linear factors and two irreducible quadratic factors. For example, the first quadratic factor is $x^2 + 4x + 16$.

You can use the quadratic formula on each of these quadratics to get your four complex roots.

The complex solutions of $z^3=a$ for $a>0$ are given by

$z_1=a^{1/3}$, $z_{2,3}=a^{1/3}(-\frac{1}{2}\pm \frac{\sqrt{3}}{2}i)$.

In your case, you will need to find them for $z^3=1$ and $z^3=64$

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Regarding solving a degree $6$ polynomial, your example is not typical because it is in fact a "disguised" quadratic.