Hello Experts,
I am given the polynomial f(x) = x^4 + p^2 where p is a prime number>2.
f(x) is a polynomial in the Z[x] ring (polynomials above integers).
Using the linear factorization of f(x) in the complex numbers field, how to factorize f to multiplication of 2 polynomials with degree 2 above the Real numbers.
How to show that f can't be factorized in the Q[x] ring (polynomials above rational numbers field).
Please give me a step by step answer.
I am given the polynomial f(x) = x^4 + p^2 where p is a prime number>2.
f(x) is a polynomial in the Z[x] ring (polynomials above integers).
Using the linear factorization of f(x) in the complex numbers field, how to factorize f to multiplication of 2 polynomials with degree 2 above the Real numbers.
How to show that f can't be factorized in the Q[x] ring (polynomials above rational numbers field).
Please give me a step by step answer.
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1) Write x^4 + p^2 = (X - sqrt(p)z)(X - sqrt(p)z^3)(X - sqrt(p)z^5)(X - sqrt(p)z^7), where z is a primitive eight root of 1, and z^2 = sqrt(-1) = i.
Since the complex conjugate of z^k = z^-k = z^8z^-k = z^(8-k) then z and z^7 and z^3 and z^5 are conjugate pairs.
Since z + z^7 = z + z^-1 = 2 cos(2PI/8) = 2cos(PI/4) = sqrt(2), and z^3 + z^5 = z^3 + z^-3 = 2cos(3*2PI/8) = 2cos(3PI/4) = -sqrt(2), then (X - sqrt(p)z)(X - sqrt(p)z^7) = X^2 - sqrt(2p)X + p and (X - sqrt(p)z^3)(X - sqrt(p)z^5) = X^2 + sqrt(2p)X + p.
2) Using 1) if f factors in Q[x] it cannot have a linear factor since all roots are complex. If f factors as two quadratics over Q then one must equal X^2 +/- sqrt(2p)X + p, which is impossible since sqrt(2p) is irrational.
Since the complex conjugate of z^k = z^-k = z^8z^-k = z^(8-k) then z and z^7 and z^3 and z^5 are conjugate pairs.
Since z + z^7 = z + z^-1 = 2 cos(2PI/8) = 2cos(PI/4) = sqrt(2), and z^3 + z^5 = z^3 + z^-3 = 2cos(3*2PI/8) = 2cos(3PI/4) = -sqrt(2), then (X - sqrt(p)z)(X - sqrt(p)z^7) = X^2 - sqrt(2p)X + p and (X - sqrt(p)z^3)(X - sqrt(p)z^5) = X^2 + sqrt(2p)X + p.
2) Using 1) if f factors in Q[x] it cannot have a linear factor since all roots are complex. If f factors as two quadratics over Q then one must equal X^2 +/- sqrt(2p)X + p, which is impossible since sqrt(2p) is irrational.