How can you tell whether a polynomial is factorable and find whatever factors it has?
2 Answers
Answer:
Some thoughts...
Explanation:
First let's talk about what is meant by "factorable / not factorable".
If you are given a polynomial with integer coefficients then it may be factorable as a product of simpler polynomials also with integer coefficients.
For example:
#x^23x+2 = (x1)(x2)#
I think we would agree that that counts as factorable.
The next level of complexity would be irrational coefficients, for example:
#x^22x2 = (x1sqrt(3))(x1+sqrt(3))#
We could say that
The next level would be Complex coefficients. For example:
#x^22x+2 = (x1i)(x1+i)#
The quadratic
We can tell whether a quadratic
#Delta = b^24ac#

If
#Delta > 0# then#ax^2+bx+c# has two distinct Real zeros and is factorable over the Reals. In addition, if#Delta# is a perfect square (and#a, b, c# are rational) then it can be factored over the rationals. 
If
#Delta = 0# then#ax^2+bx+c# is a perfect square trinomial, expressible as#(sqrt(a)x+sqrt(c))^2# or as#(sqrt(a)xsqrt(c))^2# . 
If
#Delta < 0# then#ax^2+bx+c# has two distinct Complex zeros and is not factorable over the reals. It is factorable if you allow Complex coefficients.
You will probably get used to seeing
#x^4+4 = (x^22x+2)(x^2+2x+2)#
In theory any polynomial in one variable with real coefficients is factorable as a product of linear and quadratic factors with real coefficients. In practice, those coefficients are not necessarily expressible in terms of
An example of a polynomial whose factors are not expressible in terms of
#x^5+4x+2#
Answer:
Some more thoughts...
Explanation:
Descartes' Rule of Signs
Given any polynomial in one variable with Real coefficients:
#f(x) = a_nx^n+a_(n1)x^(n1)+...+a_1x+a_0#
Look at the pattern of the signs of the coefficients and write it as a sequence of pluses and minuses.
For example, given:
#f(x) = x^321x90#
we would write:
#+  #
Then count the number of changes of sign. In our example, there's one change of sign.
This count tells us the possible number of positive Real zeros, and if the polynomial has fewer positive zeros then it is less by a multiple of
Further, by examining the signs of the coefficients of
In our example:
#f(x) = x^3+21x90#
which has coefficients with signs
With
Rational roots theorem
Given any polynomial in one variable with integer coefficients:
#f(x) = a_nx^n+a_(n1)x^(n1)+...+a_1x+a_0#
the rational roots theorem tells us that any rational zeros of
Any such rational zero
If you can find such a factor, then you can divide your starting polynomial by it to get a simpler polynomial to factor.
For example, given:
#f(x) = x^321x90#
By the rational roots theorem, any rational zeros of
So the only possible rational zeros are:
#+1, +2, +3, +5, +6, +9, +10, +15, +18, +30, +45, +90#
We know from Descartes' Rule of Signs that this
We find:
#f(6) = 6^321*690 = 216  126  90 = 0#
So
#x^321x90 = (x6)(x^2+6x+15)#
Then examining the discriminant of the remaining quadratic factor, we can observe that it is negative, so has no Real zeros and no further linear factors with Real coefficients.