How do you write a polynomial in standard form given the zeros #x= -3i# and #sqrt3#?

1 Answer
George C.
Jun 24, 2016

According to what you want, the simplest polynomials are:

#x^2+(3i-sqrt(3))x-3sqrt(3)i#

#x^3-sqrt(3)x^2+9x-9sqrt(3)#

#x^4+6x^2-27#

Explanation:

A polynomial of lowest degree with these zeros is:

#(x+3i)(x-sqrt(3)) = x^2+(3i-sqrt(3))x-3sqrt(3)i#

Typically we would be interested in the polynomial having Real coefficients. If so, then any non-Real Complex zeros occur in Complex conjugate pairs. Hence #x = 3i# would also be a zero and the simplest polynomial is:

#(x-3i)(x+3i)(x-sqrt(3))#

#= (x^2+9)(x-sqrt(3))#

#= x^3-sqrt(3)x^2+9x-9sqrt(3)#

If we also want the polynomial to have rational coefficients, then any irrational zeros of the form #a+sqrt(b)# occur in radical conjugate pairs. That is, if #a+sqrt(b)# is a zero then so is #a-sqrt(b)#.

Hence the simplest polynomial with the given zeros and rational coefficients is the quartic:

#(x-3i)(x+3i)(x-sqrt(3))(x+sqrt(3))#

#= (x^2+9)(x^2-3)#

#= x^4+6x^2-27#

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