Let F be a field and let f(x) = a_n x^n + a_n-1 x^n -1 +. .. + a_0 elementof F[x

Question

Let F be a field and let f(x) = a_n x^n + a_n-1 x^n -1 +. .. + a_0 elementof F[x]. The derivative, D_x f(x)), of f(x) is defined by D_x(f(x)) = na_n x^n-1 + (n-1) a_n-1 x^n-2 +. .. + a_1 where, as usual, na = a +a + ... +a (n times). Note that D_x (f(x)) is again a polynomial with coefficients in F. The polynomial f(x) is said to have a multiple root if there is some field E containing F and some alpha elementof E such that (x - alpha)^2 divides f(x) in E[x]. For example, the polynomial f(x) = (x - 1)^2 (x - 2) elementof Q[x] has alpha = 1 as a multiple root and the polynomial f(x) = x^4 + 2x^2 + 1 = (x^2 + 1)^2 elementof R[x] has alpha = plusminus i elementof C as multiple roots. We shall prove in Section 13.5 that a nonconstant polynomial f(x) has a multiple root if and only if f(x) is not relatively prime to its derivative (which can be detected by the Euclidean Algorithm in F[x]). Use this criterion to determine whether the following polynomials have multiple roots: (a) x^3 - 3x - 2 elementof Q[x]

Explanation / Answer

(a) x3 - 3x -2 = f(x) = (x-1)(x2+x+2)

f'(x) = 3x2 - 3 = 3(x2-1) = 3(x-1)(x+1)

We can see here that factor (x-1) is common to both the polynomials f(x) and f'(x)

Hence f(x) is not relatively prime to f'(x)

So f(x) = x3-3x-2 has multiple roots

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