How do you find the imaginary roots of a fourth degree polynomial that cannot be simplified?

Question

I started out with $f(x)=16x^6-1$, and I got it down to $64x^4+16x^2+4$ by synthetically dividing by roots $0.5$ and $-0.5$ How should I continue in order to find the other roots?

Answer 1

\begin{align} 64x^4 + 16 x^2 + 4 & = 0 \\[6pt] 64u^2 + 16 u + 4 & = 0 \qquad\text{where }u = x^2 \\[6pt] u & = \frac{-64\pm\sqrt{16^2 - 4\cdot64\cdot 4}}{128} \\[6pt] & = \frac{-16\left(4\mp\sqrt{1 - 4}\ \right)}{16\cdot8} \\[6pt] & = \frac{-4\mp\sqrt{-3}}{8}. \end{align} The problem now is how to find $x$ if $x^2=\dfrac{-4\mp i\sqrt 3}{8}$. That's a tractable but somewhat different question from what we've done above.

Answer 2

Hint: \begin{align}64x^4 + 16 x^2 + 4 &= (8x^2-2)^2 + 32x^2 + 16x^2 \\&= (8x^2-2)^2 + 48x^2 \\&= (8x^2-2)^2 - (4i\sqrt 3 x)^2 \end{align}

Answer 3

Your proposed roots $\pm1/2$ are not roots of $16x^6-1$, since the sixth power of $1/2$ is $1/64$, not $1/16$. Is there a typo in your formulation of $f$?

No matter, I’ll show you how to find the roots of $f(x)=16x^6-1$. This is a very special polynomial, and its factorization falls to special methods. First, let’s see what the roots of $x^6-1$ are. These are the six sixth roots of unity, and various methods, like DeMoivre’s Theorem, apply. These roots are $\cos(60k^\circ)+i\sin(60k^\circ)$ for $k$ running from $0$ to $5$ inclusive; in other words $\pm1$ and the four numbers $(\pm1\pm i\sqrt3)/2$. Now for your original polynomial $16x^6-1$. You can see that if $\rho$ is a root of this, then $2^{2/3}\rho$ is a sixth root of unity. In other words, the roots of $16x^6-1$ are $2^{-2/3}\zeta$, for $\zeta$ running through the six sixth roots of unity.

Answer 4

Note that $f\left(\pm \frac{1}{2}\right) = 16\left(\pm \frac{1}{2}\right)^6 - 1 = -\frac{3}{4}$, so $\pm \frac{1}{2}$ are not roots of $f$.

One way to proceed with the original problem is to write $v^6 := 16 x^6$, that is, $v = \sqrt[3]{4} x$, so that the original polynomial is $v^6 - 1$. This clearly has roots $v = \pm 1$, which (substituting) gives corresponds to $$x = \pm 4^{-1/3}.$$ Using synthetic division gives $$v^6 - 1 = (v - 1)(v + 1)(v^4 + v^2 + 1).$$ One can find the roots of the latter by writing $w = v^2$ and applying the quadratic formula to the resulting expression $w^2 + w + 1$; this only has complex roots, $w = -\frac{1}{2} \pm \frac{\sqrt{3}}{2} i,$ and we can then solve readily for $v$ and then $x$.

We can also attempt to factor $v^4 + v^2 + 1$. By the Rational Root Theorem it has no roots in $\mathbb{Q}$, so if it has a nontrivial factorization, it must be into a product of quadratics. One can set this up manually---writing down a product of general quadratics, distributing, and equating coefficients---and find the factorization $$(v^2 + v + 1)(v^2 - v + 1).$$

(On the other hand, if one knows about the cyclotomic polynomials $\Phi_n$, one can write down the factorization of any polynomial $v^m - 1$ as a product of such; in our case, $$v^6 - 1 = \Phi_1(v) \Phi_2(v) \Phi_3(v) \Phi_6(v) = (v - 1) (v + 1) (v^2 + v + 1) (v^2 - v + 1).)$$

Answer 5

$16x^6=1$

$(\sqrt[3]{4}x)^6=1$

$u=\sqrt[3]{4}x$

$u^6=1$

These roots are spaced at u=$\pm 1$ and $\pm \frac{1}{2} \pm \frac{i\sqrt{3}}{2}$ ($n^{th}$ roots of unity) Or $+1,-1, +\frac{1}{2} -\frac{i\sqrt{3}}{2}, +\frac{1}{2} -\frac{i\sqrt{3}}{2} , - \frac{1}{2} + \frac{i\sqrt{3}}{2}, - \frac{1}{2} - \frac{i\sqrt{3}}{2}$ So substituting x back into the solutions... and solving for x, you get the solutions