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Irreducible polynomial; algebraic over R

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1. Prove that the polynomial x^4 - 16*(x^2) + 4 is irreducible in Q[x] (the ring of all polynomials with rational coefficients).

2. Show that every element of C (i.e., every complex number) is algebraic over R (the ring of all real numbers).

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Solution Summary

A detailed proof that the polynomial x^4 - 16*(x^2) + 4 is irreducible in Q[x] is given, as is a detailed proof that every complex number is algebraic over R.

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1. Proof that the polynomial f(x) = x^4 - 16*(x^2) + 4 (where a caret, "^", indicates raising to a power, hence x^4 means "x to the 4th power" and x^2 means "x squared") is irreducible in Q[x]:

To say that f(x) is irreducible in Q[x] means that f(x) cannot be factored into polynomials f1(x), f2(x) such that (a) f1 and f2 are of smaller degree than f and (b) f1 and f2 have rational coefficients.

So suppose, by way of contradiction, that f(x) is NOT irreducible in f(x).

Since f(x) is a 4th-degree polynomial, one possibility is to factor f(x) into a pair of quadratic polynomials, each of which has rational coefficients. To find those polynomials, we can proceed as follows:

Since the only terms of f(x) with nonzero coefficients are the x^4 and x^2 terms, we can use the quadratic formula (with a = 1, b = -16, and c = 4) to solve the polynomial f(x) = 0 for x^2:

x^2 = [-(-16) +- sqrt{(-16)^2 - (4*1*4)} ]/(2*1)

= [16 +- sqrt(256 - 16)]/2

= [16 +- sqrt(240)]/2

= {16 +- [4*sqrt(15)]}/2

= 8 +- [2*sqrt(15)]

Thus if we factor f(x) into a pair of quadratic polynomials, we get

f(x) = [x^2 - {8 + [2*sqrt(15)]}] * [x^2 - {8 - [2*sqrt(15)]}]

Since both 8 + [2*sqrt(15)] and 8 - [2*sqrt(15)] are irrational numbers (and since a ...

Solution provided by:
  • AB, Hood College
  • PhD, The Catholic University of America
  • PhD, The University of Maryland at College Park
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