Find a polynomial function of lowest degree with rational coefficients that has the given numbers as some of its zeros.

Question

2-i,sqrt2

f(x)= ?

Thanks!

Answer 1

since the coefficients are real, complex roots occur in conjugate pairs. So, 2+i is another root. Similarly, since the coefficients are rational, we need to balance that √2 with -√2.

f(x) = (x-(2-i))(x-(2+i))(x-√2)(x-(-√2))
= ((x-2)+i)((x-2)-i)(x-√2)(x+√2)
= ((x-2)^2+1)(x^2-2)
Now all the i's and √2's are gone
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