How do you find a fourth degree polynomial given roots #3i# and #sqrt6#?

2 Answers
Aug 28, 2017

Answer:

#x^4+3x^2-54 = 0#

Explanation:

The simplest polynomial with these zeros is the quadratic:

#(x-3i)(x-sqrt(6)) = x^2-(sqrt(6)+3i)x+3sqrt(6)i#

If we want integer coefficients, then #-3i# and #-sqrt(6)# are also zeros and the simplest polynomial, which is a quartic, is:

#(x-3i)(x+3i)(x-sqrt(6))(x+sqrt(6)) = (x^2-(3i)^2)(x^2-(sqrt(6))^2)#

#color(white)((x-3i)(x+3i)(x-sqrt(6))(x+sqrt(6))) = (x^2+9)(x^2-6)#

#color(white)((x-3i)(x+3i)(x-sqrt(6))(x+sqrt(6))) = x^4+3x^2-54#

Aug 28, 2017

Answer:

See below.

Explanation:

The complex conjugate root theorem states that if a polynomial has a root #a+bi#, then it also has the root #a-bi#. There are also radical conjugate roots: if #a+sqrtb# is a root, then #a-sqrtb# must also be a root.

Thus, if #3i# and #sqrt6# are roots, then #-3i# and #-sqrt6# must also be roots. Now that we have the four roots, we can establish the factors that go along with them.

#(x-3i)(x-(-3i))(x-sqrt6)(x-(-sqrt6))#

#(x-3i)(x+3i)(x-sqrt6)(x+sqrt6)#

We can now expand to find the fourth-degree polynomial. Since #(a+b)(a-b) = a^2-b^2#, we can rewrite the above expression as

#(x^2-(3i)^2)(x^2-6)#

#(x^2-9i^2)(x^2-6)#

#(x^2+9)(x^2-6)#

Expanding this, we get

#x^4-6x^2+9x^2-54#

#x^4+3x^2-54#