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

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Answer 1 · 2017-08-14T02:53:53

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

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" ")|)#

Answer 2 · 2017-08-15T22:16:17

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

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)#.