That one is simple to factorize , though :
(1 - x^3)(2x + 1)
If you can't factorize directly , try to guess a root or two and then factorize.
You can also factorize this more. It's obvious by inspection that x=1 is a root of 1-x^3, so you can factorize it again.
(1-x)(1 + x + x^2)(2x+1)
To answer a more general question of "how do you find roots of polynomials that you can't factor", there are still some tricks you can use (feel free to ignore these if you feel they're too advanced, but if you want some awesome tricks that you won't find in your textbook, read on):
Rational Root Theorem: If p is a rational root of P(x) and can be written in lowest terms as a/b, then a is a factor of the constant coefficient, and b is a factor of the leading coefficient.
This drastically reduces the number of possibilities you have to check. Note that if every coefficient of the polynomial shares a common factor (for example 2x^2 + 4x + 24) then you should divide that common factor out first). For example, suppose I give you the following polynomial:
45x^4 - 276x^3 - 262x^2 - 76x - 7
and ask you to find the roots. The first thing to check is whether or not there's any rational roots. If there is a rational root then it can be written as a fraction a/b. By the theorem, a divides evenly into 7 and b divides evenly into 45.
Thus, we get the following choices:
Possibilities for a: ±1, ±7
Possibilities for b: ±1, ±3, ±5, ±9, ±15, ±45
If you actually test each of these, you'll find that the roots are -1/3, -1/5, and 7
But that's a lot of possibilities to check. You can limit your search space by using a couple of other theorems:
Descartes' Rule of Signs: Write the Polynomial in standard form (i.e. biggest power of x on the left, in decreasing order), then count then number of times the sign changes when moving to the next coefficient. Call that number N. The number of positive roots is either equal to N, or less than N by a positive multiple.
In the above example, the sign changes 1 time. Between the 45 coefficient, and the -276 coefficient. Thus, the number of positive roots is either 1, or -1, or -3, or -5, etc. Obviously the NUMBER of positive roots can't be negative, so we already know there is exactly 1 positive root.
Intermediate Value Theorem: The generalized version of this theorem you learn in Calculus, but for polynomials there's a simplified version. if P(a) < 0 and P(b) > 0, then there is a root between a and b.
To apply this theorem, what you want to do is just test different values from the possibilities you determined using the Rational Root Theorem, and try to limit the search space. Try 0 first because it's simple, and then try integers (even if ±1 is not a possibility for the denominator based on the results of the RRT) because they're also simple and they allow you to shorten the interval of possibilities quickly.
P(0) = -7
P(1) = -576
Sign still hasn't changed, so now we know that the positive root is strictly greater than 1 and as a result the numerator is exactly 7. (DRS told us that there is 1 positive root, IVT told us that it's not between 0 and 1, RRT told us that the only other choice for numerator is 7). You try x=7 first because it's an integer AND a possible root, and you find that it works.
At this point you can factor x=7 out of the polynomial to get a 3rd degree polynomial, and you get a new 3rd degree polynomial, Q(x) = 45x^3 + 39x^2 + 11x + 1.
Re-apply the RRT because your search space will be reduced further. Now you get:
Possibilities for a: ±1
Possibilities for b: ±1, ±3, ±5, ±9, ±15, ±45
Possibilities for x: -1, -1/3, -1/5, -1/9, -1/15, -1/45
There are other more advanced tests, but these are the polynomial root finder's primary toolbox.
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