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Suppose that p(x) is a polynomial with integer coefficients of degree 3 and x = 2 is ??a zero, and x = v 5 is a zero.

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1 Answer

My inclination would be to say that the third zero would be x=  -√5. This would then make the factors 

(x - 2)(x - √5)(x + √5) = 0. Multiplying out the last two and rewriting, we get

(x - 2)(x2 - 5) = 0. Multiplying out remaining factors, we get for our equation

x3 - 2x2 - 5x + 10 = 0. 

Any root but -√5 would leave us with an irrational coefficient somewhere.

If it has integer coefficients, then we have either 0 or 2 irrational roots. 

If it was a 3rd degree polynomial with irrational coefficients, then we would have only one irrational root with two rational roots, or three irrational roots and 0 rational roots.