When roots of polynomials occur in radical form, they occur as two conjugates. That is, The conjugate of (a + √b) is (a - √b) and vice versa. To show that the given conjugates come from a polynomial, we should create the polynomial from the given factors.
The first factor is x - (a + √b). The second factor is x - (a - √b).
The polynomial is f(x) = [x - (a + √b)]*[x - (a - √b)] = x² - x(a - √b) - x(a + √b) + (a + √b)(a - √b) = x² - 2ax + x√b - x√b + a² - b = x² - 2ax + a² - b
This is a quadratic polynomial, as expected.
If you solve the quadratic equation x² - 2ax + a² - b = 0 with the quadratic formula, it should yield the pair of conjugate radical roots. x = (1/2) [ 2a +/- √(4a² - 4(a² - b)] = a +/- (1/2)*√(4b) = a +/- √b x = a + √b, or x = a - √b, as expected.
