# Find the quadratic polynomial whose one zero is 2+sqrt(3) Find a quadratic polynomial with rational coefficients with `(2+sqrt(3))` as a zero:

If `2+sqrt(3)` is a zero, so is the conjugate `2-sqrt(3)` .

Also, if a is a zero, then (x-a) is a factor, thus the factors of the quadratic are:

`(x-2+sqrt(3))(x-2-sqrt(3))`

Multiplying we get:

`x^2-2x-xsqrt(3)-2x+4+2sqrt(3)+xsqrt(3)-2sqrt(3)-3`

`x^2-4x+1`

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Find a quadratic polynomial with rational coefficients with `(2+sqrt(3))` as a zero:

If `2+sqrt(3)` is a zero, so is the conjugate `2-sqrt(3)` .

Also, if a is a zero, then (x-a) is a factor, thus the factors of the quadratic are:

`(x-2+sqrt(3))(x-2-sqrt(3))`

Multiplying we get:

`x^2-2x-xsqrt(3)-2x+4+2sqrt(3)+xsqrt(3)-2sqrt(3)-3`

`x^2-4x+1`

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The monic quadratic polynomial with rational coefficients is:

`x^2-4x+1`

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** Checking we can use the quadratic formula to find the roots(zeros):

`x=(4+-sqrt(16-4(1)(1)))/(2(1))`

`=(4+-sqrt(12))/2`

`=(4+-2sqrt(3))/2`

`=2+-sqrt(3)`

and we see that `2+sqrt(3)` is a zero.

*** Note that this is not the only such quadratic polynomial, even if we restrict the coefficients to be rational. If the leading coefficient need not be one (need not be a monic polynomial), then `2x^2-8x+2` works also; there are an infinite number of such polynomials.

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