How do you write a polynomial function of least degree that has real coefficients, the following given zeros 8,-i, i and a leading coefficient of 1?

1 Answer
Nov 12, 2016

Answer:

#x^4-9x^3+9x^2-9x+8#

Explanation:

As the two zeros among four zeros, viz. #8#, #-i#, #i# and #1# have two complex numbers, which are complex conjugate of each other, it is possible to write a polynomial function of least degree (which is #4# as there are #4# zeros with no multiplicity, i.e. repeat zeros) that has real coefficients.

A function #f(x)# with zeros #a#, #b#, #c# and #d# is #(x-a)(x-b)(x-c)(x-d)# and hence the function with #8#, #-i#, #i# and #1#as zeros, will be

#(x-8)(x-(-i))(x+i)(x-1)#

= #(x-8)(x+i)(x-i)(x-1)#

= #(x-8)(x-1)(x^2+ix-ix+(-i^2))#

= #(x-8)(x-1)(x^2+1)# as #i^2=-1#

= #(x^2-9x+8)(x^2+1)#

= #x^4-9x^3+8x^2+x^2-9x+8#

= #x^4-9x^3+9x^2-9x+8#