How do you write a polynomial function given the real zeroes 3-i,5i and coefficient 1?

1 Answer
Jan 10, 2016

Answer:

#y=x^4-6x^3+35x^2-150x+250#

Explanation:

Complex roots always come in pairs.

#3-i# will be a #0#, but so will #3+i#.

#5i# will be a #0#, but so will #-5i#.

This gives us the function

#y=(x-(3-i))(x-(3+i))(x-5i)(x-(-5i))#

#y=(x-3+i)(x-3-i)(x-5i)(x+5i)#

#y=((x-3)+i)((x-3)-i)(x^2-25i^2)#

#y=((x-3)^2-i^2)(x^2+25)#

#y=(x^2-6x+9+1)(x^2+25)#

#y=(x^2-6x+10)(x^2+25)#

#y=x^4-6x^3+35x^2-150x+250#

Notice how the function only has imaginary roots (never crosses the #x#-axis):

graph{x^4-6x^3+35x^2-150x+250 [-5, 8, -36, 300]}