Factorising of x^4+1. Read the details before asking, please?

Question

Yes, it can be done.

Check this: (x^2+sqrt2 x + 1)(x^2- sqrt2 x + 1)

This is obtained finding the complex root and multiplying 2 of the factors

If you have some remark, please, post them

Ana

Thanks for this remark. I actually posted this question because someone else asked it and many persons answered that it couldnt be done.

Check that these are really the factors this way:

(x^2-sqrt2 x+1)(x^+ sqrt2 x + 1) =
(x^2 +1 - sqrt 2 x)(x^2 + 1 +sqrt2 x) =
((x^2+1)^2 - 2 x^2) =
x^4 + 2x + 1 - 2x^2 = x^4 + 1

2x^2

This was a typo mistake. Sorry

Answer 1

We are getting some incorrect responses here, so let's work this out.

First of all, x^4 + 1 is a palindromic polynomial. That is, if you reverse the order of the coefficients (which would turn 3x^2+5x+4 into 4x^2+5x+3), you get the same polynomial back again. If you substitute 1/x for x in a polynomial equation, multiplying by x^n (n is the degree) will get you the polynomial with the coefficients in reverse order. Since x^4+1 is palindromic, its reverse is the same as the original polynomial. Therefore, if r is a root, then so is 1/r, and so this polynomial is a factor:

x^2 - (r+1/r)x + 1

[provided r <> 1! But one can easily check that 1 is not a root of x^4+1=0.]

So set up this factorization:

(*) x^4+1 = (x^2+Ax+1)(x^2+Bx+1)
= x^4 + (A+B)x^3 + (2+AB)x^2 + (A+B)x+1

Therefore,

A+B = 0
2+AB = 0, or AB = -2

So A and B must be roots of the equation x^2 - 0x -2 = 0. In other words, A and B are roots of x^2 = 2, and so A and B are +sqrt(2) and -sqrt(2). Substituting these for A and B in (*) above gives you the factorization that you give us, which shows that it is correct.

Answer 2

This is a polynomial of order 4 with real coefficeints

Basis:
So this has 0 or 2 or 4 complex roots.

if a+ib is a complex root so is a- ib is a root so (x-a)^2+b^2 is a factor
So we can use a shorter method

solution:
it can be factored as (ax^2+bx+c)(dx^2+ex+f)

short method

can it be converted to a^2-b^2 form

Yes it can be becuase no x^3 , no x term

now we go
x^4+1
= (x^2)^1 + 1^2
add and subtract 2x^2

(x^2)^2+2x^2+1-2x^2
=(x^2+1)^2 - (xsqrt(2))^2
= (x^2+1+xsqrt(2))
(x^2+1-xsqrt(2))

same as yours after rearranging

Answer 3

have you ever easily examine a number of the further information on the following? many situations the question turned right into a lot extra powerful without them. Then too, further information seem as you're answering or afterwards... there have been some situations that i have neglected component to the files or did not examine them.

Answer 4

Why'd you post?

I don't think that you have it right. Where're the two terms x^2 times 1?

Doing what you said:

x^4 = -1 = exp(i*pi + 2k*pi)
x^2 = exp(i*pi/2 + k*pi)

again:

x = exp(i*pi/4 + k*pi/2)

for k = 0,1,2,3 you get the 4 roots of -1. They come in complex conjugates.

Answer 5

you need to be more specific when requesting a factorization.
namely, you need to say whether you want it factored by degree 1 polynomials, or degree 2.

so, if it is over the reals, then the polynomial can be factored by 2 polynomials of degree 2, but it cannot be factored by polynomials of degree 1.
If it is over the complex numbers, then the polynomial can be factored by 4 polynomials of degree 1.


'

Answer 6

ok
yes your proof is correct
it is factorised over real numbers

Answer 7

Look here: x^4=-1; -1=exp(-pi*i) = exp(-3pi*i) = exp(+3pi*i) = exp(pi*i)
Thus x1=exp(-pi*i/4)= cos(pi/4)-i*sin(pi/4), x2= exp(-3pi*i/4) =cos(3pi/4)-i*sin(3pi/4),
And x3=exp(3pi*i/4)= cos(3pi/4)+i*sin(3pi/4), x4= exp(pi*i/4) =cos(pi/4)+i*sin(pi/4).
Now simplify: cos(pi/4)=sin(pi/4)=1/sqrt(2). – check it!!!

Answer 8

this can't be factorized because this equation has no real root