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Suppose that P(x) is a polynomial with rational coefficients that has a root at x = 8 + sqrt(3). Find another root of P(x).

 Dec 17, 2021
 #1
avatar+478 
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Somebody correct me if I am wrong, but another root would be 8 - sqrt(3). My reason is when using the quadratic formula, you get the root and its conjugate. 

 Dec 18, 2021
 #2
avatar+117466 
+1

I think the main point here is that the polynomial has rational coefficients.

 

is  [x-8  - sqrt3 ] is one factor then [x-8  + sqrt3] would have to be another. 

 

So when they are multiplied the factor will be [(x-8)^2 - 3] or  [x^2-16x+61]  only rational coefficients are left.

 Dec 18, 2021

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