How do you find the polynomial function with roots #2 + 3sqrt3#?

1 Answer
Aug 1, 2017

Answer:

#f(x)=x^2-4x-23#

Explanation:

If a function has the root #a+sqrtb#, it must also have the root #a-sqrtb#.

So, this function actually has the roots #2+3sqrt3# and #2-3sqrt3#. Since these are roots, when #x# takes on these values, #f(x)=0#. Thus, each factor of the polynomial will be #x - #the root.

#f(x)=(x-(2+3sqrt3))(x-(2-3sqrt3))#

#f(x)=(x-2-3sqrt3)(x-2+3sqrt3)#

#f(x)=(x-2)^2+(-3sqrt3)(3sqrt3)#

#f(x)=x^2-4x+4-27#

#f(x)=x^2-4x-23#