How do you find a polynomial function that has zeros #2, 4+sqrt5, 4-sqrt5#?

1 Answer
Jul 22, 2017

Answer:

#f(x) = x^3-10x^2+27x-22#

Explanation:

If #x=a# is a zero then #(x-a)# is a factor.

So the simplest polynomial function of #x# that has these zeros is:

#f(x) = (x-2)(x-(4+sqrt(5)))(x-(4-sqrt(5)))#

#color(white)(f(x)) = (x-2)((x-4)+sqrt(5))((x-4)-sqrt(5))#

#color(white)(f(x)) = (x-2)((x-4)^2-5)#

#color(white)(f(x)) = (x-2)(x^2-8x+16-5)#

#color(white)(f(x)) = (x-2)(x^2-8x+11)#

#color(white)(f(x)) = x^3-10x^2+27x-22#

Any polynomial in #x# with these zeros will be a multiple (scalar or polynomial) of this #f(x)#.

#color(white)()#
Footnote

In this example, we were asked for a function with zeros including both #4+sqrt(5)# and its radical conjugate #4-sqrt(5)#. If only one had been specified then we would still have had to include the other if we wanted our function to have rational coefficients.