How do you write a polynomial function of least degree given the zeros 2i, -2i, 2+2i?

1 Answer
Douglas K.
Oct 14, 2016

Please see the explanation for the process.

#y = k(x^4 - 4x^3 + 12x^2 -16x + 32)#

Explanation:

You cannot have an odd number of complex or imaginary roots, because they always exist is conjugate pairs, therefore, another root must be 2 - 2i and the polynomial is of the form:

#y = k(x - 2i)(x + 2i)(x - 2 -2i)(x - 2 + 2i)#

#y = k(x^2 - 4i^2)(x - 2 -2i)(x - 2 + 2i)#

#y = k(x^2 + 4)(x - 2 -2i)(x - 2 + 2i)#

#y = k(x^2 + 4)(x^2 - 2x + 2ix -2x + 4 - 4i - 2ix +4i - 4i^2)#

#y = k(x^2 + 4)(x^2 - 4x + 4 - 4i^2)#

#y = k(x^2 + 4)(x^2 - 4x + 4 + 4)#

#y = k(x^2 + 4)(x^2 - 4x + 8)#

#y = k(x^4 - 4x^3 + 8x^2 + 4x^2 -16x + 32)#

#y = k(x^4 - 4x^3 + 12x^2 -16x + 32)#

k exists to allow the polynomial to pass through a specified point.

Related questions
See all questions in Complex Conjugate Zeros
Impact of this question
2425 views around the world
You can reuse this answer
Creative Commons License