There are some shortcuts that work in specific forms, but they usually aren't worthwhile (e.g., there is actually a cubic formula just like there is a quadratic formula, but it's more complicated and hardly anyone bothers).
For generally factoring polynomials, there are some things that can make life easier:
1) Rational root theorem. Whenever the coefficients are all integers (!), the only possible rational roots all have the form ±{divisor of constant term}/{divisor of leading coefficient}. So in your case,
X^6 - X^5 + X^4 + X^3 + X^2 + X + 14 = 0 --> possible roots: ±1,±2,±7,±14.
y^3 - 13 y^2 + 54 y - 72 = 0 --> possible roots: ±1,±2,±3,±4,±6,±8,±12,±24,±36,±72
So it's really useful whenever the first and last coefficients have a small number of factors, though unfortunately 72 has lots. But at least it tells you what roots to not even bother trying.
2)
Descartes rule of signs: the number of
positive root is the number of sign changes in the coefficients (written in standard order), or that number less an even number. Zero counts as no change.
First polynomial: 2 sign changes: So the number of positive roots is either 2 or 0.
Second polynomial: 3 sign changes: So the number of positive roots is either 3 or 1.
You can get the negative roots by plugging in (-X) for X and working counting sign changes in that polynomial instead (that just flips the sign of every odd-power term):
First polynomial: 4 sign changes: so the number of negative roots is 4,2,or 0.
Second polynomial: no sign changes. So there are no negative roots.
3)
Synthetic division. Say you want to know the value of your second polynomial at y = 2. You could just plug it in: 2^3 - 13*2^2 + 54*2 - 72 = -8.
You could evaluate it like this: y³ - 13y² + 54y - 72 = y(y(1y-13)+54)-72.
Or you could evaluate it like this, by writing the coefficients in order (i]including zeros, if any!):
2 | 1 -13 54 -72 <-- coefficients; do not skip missing terms (put 0 instead)
| 2 -22 64 <-- drop down first term, multiply by 2 and
+------------+------ write the result diagonally. Then add down.
1 -11 32 | -8 <-- the value is the last term.
Right now, that tells you that 2 is
not a root. But if you plug in 3, you get 0, so y = 3 is a root. The nice thing about synthetic division is that it also tells you the quotient after division by y-3. So:
3 | 1 -13 54 -72
| 3 -30 72
+------------+----
4 | 1 -10 24 | 0 <-- y-3 is a factor and the quotient is y² - 10y + 24.
| 4 -24 So we can try factoring that instead.
+--------+----
1 -6 | 0 <-- y-4 is a factor and the quotient is y - 6
So in factored form, your second polynomial is (y-3)(y-4)(y-6). Though after you get the quotient, you're free to try to factor it in any manner you like, e.g., what two numbers multiply to be 24 and add to be -10. -4 and -6.
(Common mistake: note that the number on the left is the root, i.e., what you're
subtracting when in factored form. So if you want to check whether, say, (y+6) is a factor, you should think of it as y-(-6).)
Your first polynomial has no real roots, and hence cannot be factored over the real numbers.
Reply
