so according to factor theorem, x^4 + x^2 + 1 shouldn't be factorable since no value of x can make the equation equal to 0, and also, when graphed, it has no x-intercepts in the set of real elements.. but it is factorable... as (x^2 + x + 1)(x^2 - x + 1). so my question is what method was used to factor this? also, how can you tell when a polynomial that can't be factored using factor theorem is factorable?
here's how somebody else told me how to do it:
x^4 + x^2 + 1 = 0
let y = x^2
y^2 + y + 1 = 0
y^2 + y + 1 + 1/4 = 1/4
(y+1/2)^2 + 1 = 1/4
(y+1/2)^2 = -(3/4)
y = +-sqrt(-3/4) - 1/2
sub back in x
x^2 = +-sqrt(3/4) - 1/2
x = +-sqrt(+-sqrt(3/4) - 1/2)
x = 0.5 - sqrt(3)/2*i, -0.5 - sqrt(3)/2*i, -0.5 + sqrt(3)/2*i, 0.5 + sqrt(3)/2*i
then these are all the roots of the equation
so you have the factors as (x - 0.5 - sqrt(3)/2*i) (x - -0.5 - sqrt(3)/2*i)(x - -0.5 + sqrt(3)/2*i)(x - +0.5 + sqrt(3)/2*i)
then you multiply pairs with the opposite signs on the coefficent of the imaginary part of the factor and same sign on the coefficent of the real part.
so (x+(i sqrt(3))/2+1/2)( x-(i sqrt(3))/2+1/2) to get x^2 + x + 1
and (x+(i sqrt(3))/2-1/2)( x-(i sqrt(3))/2-1/2) to get x^2 - x + 1
then you have (x^2 + x + 1)(x^2 - x +1)
not sure what you're telling me though
Assuming is real:
and since your second example will have at least 4991174912 factors with real coefficients, I think I will give it a miss (but it is fairly simple to give the factors of this polynomial since they are all related, complex number theory will give all the roots of fairly easily, then the complex ones can be used in conjugate pairs to give the real quadratic factors and the real root will give you the real linear factor).
The thing about proving that something is the case in general is that it becomes unnecessary to demonstrate every particular instance. And if you read my previous post you will see that it contains a sketch of the proof of the result, and proof is how I know.i just don't get how you can know if it's factorable or not.
This one require some special techniques
. . based on observation and experience.
Add and subtract
We have a differece of squares: .
Therefore, we have:
The numerator is a difference of squares:
The sum and difference of cubes can also be factored.
The fraction become: .
. . which reduces to: .
A far as your is concerned, seeing that it has only even powers of x, I would let so the polynomial is the quadratic . Yes, "factoring" a polynomial is intimately related to solving the equation with the polynomial set equal to 0, and now we can use the quadratic formula, or complete the square, to get
In polar form, that would be and so the four values of x would be , , , and
I won't carry this any further, but, in rectangular form, those would be a+ bi, a- bi, c+ di, and c- di, for appropriate a, b, c, and d. The "linear factors", then, are (x- (a+bi))(x-(a-bi))(x- (c+di))(x- (c-di))= [(x-a)- bi][(x- a)+ bi][(x- c)- di][(x- c)- di]= [(x- a)^2+ b^2][(x-c)^2+ d^2], two quadratic factors with real, but not necessarily integer, coefficents. One can always factor a polynomial, with integer coefficents, into linear factors with complex coefficients, or at a most quadratic factors with real coefficients. The question of when one can factor a polynomial into factors (not necessarily linear) with integer coefficients is a very deep one- dealt with in "Galois theory", a part of abstract algebra.
(Galois, by the way, wrote out the details of his theory, virtually creating "abstract algebra" the night before his death- at the age of 18.)