How do you find a polynomial function that has zeros 2,4+5,45?

1 Answer
Jul 22, 2017

f(x)=x310x2+27x22

Explanation:

If x=a is a zero then (xa) is a factor.

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

f(x)=(x2)(x(4+5))(x(45))

f(x)=(x2)((x4)+5)((x4)5)

f(x)=(x2)((x4)25)

f(x)=(x2)(x28x+165)

f(x)=(x2)(x28x+11)

f(x)=x310x2+27x22

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


Footnote

In this example, we were asked for a function with zeros including both 4+5 and its radical conjugate 45. 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.