Exercise 14.1.6

By Example 8.5.5, f = x 5 6 x + 3 is not solvable by radicals over . Give a new proof of this fact using the previous exercise together with the irreducibility of f and part (b) of Exercise 6 from Section 6.4.

Answers

Proof. The given polynomial f has prime degree 5 and only three real roots, according to part (b) of Exercise 6.4.6. Since f has more than one but less than 5 real roots, it is not solvable by radicals by Exercise 14.1.5. □

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2022-07-19 00:00
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