Minimal polynomial
Let α = 5 − 3 , and let f ( x ) be the minimal polynomial of α . That is, f ( x ) is a monic polynomial with rational coefficients of the smallest possible nonzero degree such that f ( α ) = 0 .
Find f ( 1 ) .
The answer is -11.
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2 solutions
Moderator note:
As explained in the comments, one has to be careful to check that the minimal degree is actually 4. In general, all that we know is the minimal degree is less than the product of the individual degrees in such a summation.
Yes indeed. Is it immediately clear that the polynomial cannot factor as the product of two quadratics? (Or, for that matter, that none of the roots are rational?)
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Indeed it is. The four roots are irrational since their squares are (given that the root of an integer n is irrational if n isn't a perfect square, with n = 1 5 in our case). If, hypothetically, f ( x ) were to factor into two quadratic polynomials over Q , then the sums and the products of two of the roots would have to be rational (in pairs), which isn't the case.
Do you find my solution satisfactory?
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Yes, very nice.
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@Patrick Corn – Just checking... it has been a while since I have done this kind of stuff. I hope you will find the time to solve the problem inspired by this ;)
An alternative solution which gives the same result:
The minimal polynomial will have the factor ( x − 5 + 3 ) . Pair this factor with its conjugate:
( x − 5 + 3 ) ( x − 5 − 3 ) x 2 − 2 5 x + 2
Now pair this factor with its conjugate:
( x 2 + 2 5 x + 2 ) ( x 2 − 2 5 x + 2 ) x 4 − 1 6 x 2 + 4
Note that the 1st degree factors of this polynomial cannot be paired in such a way as to produce a polynomial with rational coefficients. Thus, this is the minimal polynomial. If f ( x ) = x 4 − 1 6 x 2 + 4 , then f ( 1 ) = − 1 1 .
Squaring gives α 2 = 8 − 2 1 5 so α 2 − 8 = − 2 1 5 . Squaring again gives α 4 − 1 6 α 2 + 6 4 = 6 0 and α 4 − 1 6 α 2 + 4 = 0 . Since the roots of f ( x ) = x 4 − 1 6 x 2 + 4 are ± 5 ± 3 , the polynomial f ( x ) is irreducible over Q , so that f ( x ) is the minimal polynomial of its roots. Now f ( 1 ) = − 1 1 .