What must the third zero be? Why? Similarly, for any √ ?polynomial with integer coefficients, if it has one irrational zero of the form a + b c (with a, b, c integers, and c > 0), what can we conclude about some of its other zeros?

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

x^{3} - 2x^{2} - 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.

## Comments

Do you know what the polynomial is?

NO.