Factoring a polynomial #p(x)# means to find its roots #x_i# and divide the polynomial by the term #(x-x_i)#, writing #p(x)# as #(x-x_i)q(x)#, and then iterating the process searching the roots of #q(x)# (which of course are roots of #p(x)#, too. The fundamental theorem of algebra states that you can always write a polynomial of degree #k# as the product of #k# linear factors #(x-x_i)#, i.e., every polynomial of degree #k# has #k# complex roots.
These roots can be complex but not real, and in that case, you cannot simplify the polynomial in #\mathbb{R}#.
In your case, the solutions are easy to find, without doing any calculation: you can remember the rule which states that, if you have a quadratic of the form #x^2-sx+p#, the product of the solutions equals #p# and the sum of the solutions equals #s#.
So, we have to find two numbers which sum up to #-6#, and that multiplied one for the other give #-16#. You can easily convince yourself that these two numbers are #-2# and #8#
If #-2# and #8# are roots of our polynomial, for what we said above the following must hold:
#x^2−6x−16=(x+2)(x-8)#
