Can you find all quartic polynomials with real, rational coefficients having #2-isqrt(3)# and #sqrt2 +1# as two of the zeros?

1 Answer
George C.
Apr 2, 2017

#f(x) = k(x^4-6x^3+14x^2-10x-7)#

for any rational #k != 0#

Explanation:

Note that:

#(x-alpha)(x-beta) = x^2-(alpha+beta)x+alphabeta#

So given #alpha# of the form #a+bsqrt(c)# with #a, b, c# non-zero rational, we need #beta# to be a #a-bsqrt(c)# in order to make both #alpha+beta# and #alphabeta# rational.

In our example, that means that in addition to the given zeros:

#2-isqrt(3)" "# and #" "sqrt(2)+1#

we must have the conjugate zeros:

#2+isqrt(3)" "# and #" "-sqrt(2)+1#

So the simplest polynomial with rational coefficients and these zeros is:

#(x-2-isqrt(3))(x-2+isqrt(3))(x-1-sqrt(2))(x-1+sqrt(2))#

#= ((x-2)^2-(isqrt(3))^2)((x-1)^2-(sqrt(2))^2)#

#= ((x^2-4x+4)+3)((x^2-2x+1)-2)#

#= (x^2-4x+7)(x^2-2x-1)#

#= x^4-6x^3+14x^2-10x-7#

Any quartic with rational coefficients and these zeros will be a rational multiple of this one.

So the possible quartics are:

#f(x) = k(x^4-6x^3+14x^2-10x-7)#

for any rational #k != 0#

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