what to do when factor theorem doesn't work?

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?

Re: what to do when factor theorem doesn't work?

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Originally Posted by

**iragequit** 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?

The factor theorem only works for linear factors.

Once you deduce that has no linear factors, assume . If you're able to solve for a, b, c and d, the expression can be factored into 2 quadratic factors.

Re: what to do when factor theorem doesn't work?

Quote:

Originally Posted by

**iragequit** 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?

A polynomial with real coefficients has roots which are either real or if complex occur in conjugate pairs. This implies that a polynomial of arbitrary degree can always be factored into a product of quadratic and linear factors with real coefficients.

CB

Re: what to do when factor theorem doesn't work?

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Originally Posted by

**alexmahone** The factor theorem only works for linear factors.

Once you deduce that

has no linear factors, assume

. If you're able to solve for a, b, c and d, the expression can be factored into 2 quadratic factors.

how would you solve for a b c and d?

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Originally Posted by

**CaptainBlack** A polynomial with real coefficients has roots which are either real or if complex occur in conjugate pairs. This implies that a polynomial of arbitrary degree can always be factored into a product of quadratic and linear factors with real coefficients.

CB

then why is x^4+y^4 not factorable? or x^9982349823 + 34234? i just don't get how you can know if it's factorable or not.. and if it IS factorable without the use of the factor theorem, how would you go about factoring it

Re: what to do when factor theorem doesn't work?

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Originally Posted by

**iragequit** how would you solve for a b c and d?

Equate coefficients of the different powers of x.

Re: what to do when factor theorem doesn't work?

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Originally Posted by

**alexmahone** Equate coefficients of the different powers of x.

it'd help if you actually showed what you meant lol

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

Re: what to do when factor theorem doesn't work?

Re: what to do when factor theorem doesn't work?

Quote:

Originally Posted by

**iragequit**

then why is x^4+y^4 not factorable? or x^9982349823 + 34234? i just don't get how you can know if it's factorable or not.. and if it IS factorable without the use of the factor theorem, how would you go about factoring it

That you cannot factor a polynomial does not mean it cannot be done:

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).

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i just don't get how you can know if it's factorable or not.

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.

CB

Re: what to do when factor theorem doesn't work?

Re: what to do when factor theorem doesn't work?

Quote:

Originally Posted by

**iragequit** how would you solve for a b c and d?

then why is x^4+y^4 not factorable? or x^9982349823 + 34234? i just don't get how you can know if it's factorable or not.. and if it IS factorable without the use of the factor theorem, how would you go about factoring it

You and the others are using different notions of "factoring". **Any** polynomial can be 'factored' if you allow all complex numbers as coefficients but elementary algebra uses 'factored' in the sense of "linear factors with integer coefficients", a much more limited concept.

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.)