How do you find a polynomial function that has zeros 1+sqrt3, 1-sqrt3?

2 Answers
Aug 14, 2017

f(x) = x^2 - 2x - 2

Explanation:

Since these are the zeros, we can make the following equation:

(x-(1+sqrt3))(x-(1-sqrt3)) = 0

Or

(x-1-sqrt3)(x-1+sqrt3) = 0

When we expand this, we get

x^2 - x + xsqrt3 - x + 1 - sqrt3 - xsqrt3 + sqrt3 - 3 = 0

Combining like terms:

color(blue)(ulbar(|stackrel(" ")(" "f(x) = x^2 - 2x - 2" ")|)

Aug 15, 2017

f(x) = x^2-2x-2

Explanation:

The simplest polynomial with distinct zeros alpha and beta is:

(x-alpha)(x-beta) = x^2-(alpha+beta)x+alphabeta

With alpha = 1+sqrt(3) and beta = 1-sqrt(3), we find:

{ (alpha+beta = (1+sqrt(3))+(1-sqrt(3)) = color(red)(2)), (alphabeta = (1+sqrt(3))(1-sqrt(3)) = 1^2-(sqrt(3))^2 = 1-3 = color(blue)(-2)) :}

So a suitable polynomial function would be:

f(x) = x^2-color(red)(2)xcolor(blue)(-2)

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