Complete factorization of $64y^6 - 1$

Question

If we use the difference of two cubes identity, then we have $$64y^6 - 1 = (4y^2-1)(16y^4+4y^2+1) = (2y+1)(2y-1)(16y^4+4y^2+1)$$ which is the suggested solution of the textbook.

But if we use the difference of two squares identity, we then have $$64y^6 - 1 = (8y^3+1)(8y^3-1) = (2y+1)(2y-1)(4y^2-2y+1)(4y^2+2y+1).$$

I then came up with the question: what is complete factorization? Of course we can do this trick: $$16y^4+4y^2+1 = 16y^4+8y^2+1 - 4y^2 = (4y^2+1 - 2y)(4y^2+1+2y).$$ How do we know when to stop factorizing?

Answer 1

The answer depends on the base ring/field you consider:

The complete factorisation over $\mathbf Z, \mathbf Q$ or $\mathbf R$ is indeed $$(2x-1)(2x+1)(4x^2+2x+1)(4x^2-2x+1)$$

But on $\mathbf C$, you obtain a product of linear factors involving the sixth roots of unity: \begin{align} 64x^6-1&=\prod_{k=1}^6\Bigl(2x-\mathrm e^{\tfrac{2ik\pi}{6}}\Bigr)\\ &=\underbrace{\Bigl(2x-\mathrm e^{\tfrac{i\pi}{3}}\Bigr)\Bigl(2x-\mathrm e^{\tfrac{5i\pi}{3}}\Bigr)}_{\textstyle4x^2+2x+1}\,\underbrace{\Bigl(2x-\mathrm e^{\tfrac{2i\pi}{3}}\Bigr)\Bigl(2x-\mathrm e^{\tfrac{4i\pi}{3}}\Bigr)}_{\textstyle4x^2-2x+1}\,\Bigl(\underbrace{2x-\mathrm e^{\tfrac{3i\pi}{3}}\rule[-1.5ex]{0pt}{1.5ex}}_{\textstyle2x+1}\Bigr)\,\Bigl(\underbrace{2x-\mathrm e^{\tfrac{6i\pi}{3}}\rule[-1.5ex]{0pt}{1.5ex}}_{\textstyle2x-1}\Bigr) \end{align}\,

Answer 2

Are you looking for the factorization in $\mathbb{R}[x]$? Note that for all real values $y$, $$4y^2+1 - 2y=3y^2+(y-1)^2>0\quad 4y^2+1+2y=3y^2+(y+1)^2>0$$ so they cannot be factored as $a(y-y_1)(y-y_2)$ with $a,y_1,y_2\in \mathbb{R}$, otherwise they should be zero for $y_1$ and $y_2$.

Answer 3

The polynomials $$p(y)=4y^2-2y+1$$ $$q(y)=4y^2+2y+1$$

have negative discriminants so they cannot be factorized over the field of real numbers.

$q(y),p(y)>0,\forall y$

Now if you want to include complex numbers then just solve the quadratics.

Then you will have a complete factorization over the complex field.

Answer 4

This is a classic problem in complex analysis. Note first that $(1/2)^6 = 1/64$. Then you need to look for the six roots of unity. They are: $$ e^{n \pi i },~~n=0, 1/3, ...,5/3 $$ So all six roots are $$ (1/2)e^{n \pi i },~~n=0, 1/3, ...,5/3 $$ The n-th roots of unity form a group with the operation multiplication. They're an important early example in abstract algebra.

Answer 5

A simplistic answer for "when to stop factorizing" real polynomials is when every factor is either a first-degree real polynomial or an irreducible second-degree real polynomial.

A second-degree polynomial $ax^2 + bx + c$ is irreducible when $b^2 - 4ac < 0.$

I say this answer is "simplistic" because it ignores the problem of how you would actually obtain such a factorization from an arbitrary polynomial.