# 1-√3,1 + √3 - 2 find polynomial function of degree with given numbers as zeros

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### 3 Answers

let x1 and x2 be the roots such that:

x1= 1- sqrt3

x2= 1+sqrt3

Then ( x-x1) and (x-x2) are factors of the function :

==> f(x) = [x - (1-sqrt3)] [ x - (1+sqrt3)]

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

Now let us expand the brackets:

= x^2 - x - sqrt3*x - x + 1 + sqrt3 + sqrt3*x - sqrt3 - 3

Reduce and group similar terms:

= x^2 - 2x - 2

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

Since the polynomial has 3 zeroes, that means that the polynomial has 3 roots, then the polynomial is of 3rd degree.

f(x) = ax^3 + bx^2 + cx + d

According to the rule, a polynomial could be written as product of linear factors, also.

Since the roots of polynomial are:

x1 = 1 - sqrt3

x2 = 1 + sqrt3

x3 = -2

f(x) = (x - 1 + sqrt3)(x - 1 - sqrt3)(x + 2)

f(x) = [(x-1)^2 - 3](x+2)

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

f(x) = (x^2 - 2x - 1)(x+2)

We'll remove the brackets and we'll get:

f(x) = x^3 + 2x^2 - 2x^2 - 4x - x - 2

We'll eliminate and combine like terms:

f(x) = x^3 - 5x - 2

**The function that has the 3 given roots is:**

**f(x) = x^3 - 5x - 2**

A value of x = x1 is said to be the zero of a polynomial f(x), if we substitute x1 for x in the polynomial f(x), then f(x) vanishes. So if f(x) is a polynomial, and if f(x1) = 0, then x1 is a zero of the polynomial f(x).

We know that if x1,x2 and x3 are the zeros of a polynomial f(x), then the polynomial could be written as:

f(x) = (x-x1)(x-x2)(x-x3), as putting x= x1 or x= x2 and x=x3 makes th polynomial vanish.

Given are the zeros : 1-sqrt3, 1+sqrt3 and -2 are the 3 zero.

Then f(x) = {x - (1-sqr3)}{x-(1+sqrt3)}{x-(-2)}.

f(x) = {x^2 +(-1+sqrt3-1-sqrt3)x + (1-sqrt3)(1+sqrt3)}{x+2}.

f(x) = {x^2 -2x + [1^2-(sqrt3)^2]}(x+2).

f(x) = {x^2-2x+1-3}(x+2).

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

f(x) = {x^3-2x^2-2x +2x^2-4x-4}.

f(x) =x^3 -6x -4 .

Therefore x^3-6x-4 is the polynomial with zeros 1-sqrt3 , 1+sqrt3 , -2.