How do you write a polynomial with zeros 4, 4, 2+i and leading coefficient 1?

1 Answer
iceman
Feb 18, 2017

#P(x)=x^4-12x^3+53x^2-104x+80#

Explanation:

#4,4,2+i#
Root #4# is multiplicity #2#
#2+i# Complex Conjugate Zero is #2-i#, we have:
#P(x)=(x-4)^2(x-2-i)(x-2+i)#
#P(x)=(x^2-8x+16)[(x-2)^2-i^2]#
#P(x)=(x^2-8x+16)(x^2-4x+4-(-1)]#
#P(x)=(x^2-8x+16)(x^2-4x+5)#
#P(x)=x^4-4x^3+5x^2-8x^3+32x^2-40x+16x^2-64x+80#
#P(x)=x^4-12x^3+53x^2-104x+80#

Related questions
See all questions in Complex Conjugate Zeros
Impact of this question
2492 views around the world
You can reuse this answer
Creative Commons License