Maple's factor command, which is behind the Context Panel option Factor, factors a polynomial over the field implied by the coefficients. Since the coefficients of the given polynomial are integers, the polynomial is factored over the integers. Since the two quadratic factors can't be factored to linear factors over the integers, this is as far as the unaided factor command can go.

By the fundamental theorem of algebra, a polynomial of degree n factors to linear factors over the complex numbers, and each such linear factor is of the form $x-r$, where r is a zero of the polynomial. Hence, to find all linear factors, the task is really to find all zeros. Hence, solve is as appropriate as factor in a problem such as this.

The Context Panel option Solve≻Obtain Solutions for≻$x$ returns the sequence of solutions

$1\,-1\,\frac{1}{2}\sqrt{-2-2\mathrm{I}\sqrt{3}}\,-\frac{1}{2}\sqrt{-2-2\mathrm{I}\sqrt{3}}\,\frac{1}{2}\sqrt{-2\+2\mathrm{I}\sqrt{3}}\,-\frac{1}{2}\sqrt{-2\+2\mathrm{I}\sqrt{3}}$

The Context Panel option Conversions≻To List changes this sequence to a list. The Context Panel option Map Command Onto≻evalc (where evalc must be typed into the dialog box) returns the solutions in the form

$\left[1\,-1\,\frac{1}{2}-\frac{1}{2}\mathrm{I}\sqrt{3}\,-\frac{1}{2}\+\frac{1}{2}\mathrm{I}\sqrt{3}\,\frac{1}{2}\+\frac{1}{2}\mathrm{I}\sqrt{3}\,-\frac{1}{2}-\frac{1}{2}\mathrm{I}\sqrt{3}\right]$

Use the Context Panel option Assign to a Name to assign this list a name such as s. Then, the product template from the Expression palette can be invoked to yield  

$\prod _{k\=1}^6\left(x-s_k\right)\=\left(x-1\right)\left(x\+1\right)\left(x-\frac{1\+i \sqrt{3}}{2}\right)\left(x-\frac{1-i \sqrt{3}}{2}\right)\left(x-\frac{-1\+i \sqrt{3}}{2}\right)\left(x-\frac{-1-i \sqrt{3}}{2}\right)$

The field over which factoring takes place in the factor command is modified by adding field extensions as a second parameter.