Every polynomial has a root
Over the real numbers, some polynomials have no roots at all — x^2 + 1 = 0 stops you cold. The fundamental theorem of algebra says this never happens once you allow complex numbers: every non-constant polynomial with complex (hence also real) coefficients has at least one complex root. Apply it repeatedly and a degree-n polynomial factors completely into n linear factors, so it has exactly n roots when you count multiplicity.
Conjugate pairs and the discriminant
When a polynomial has real coefficients, its non-real roots arrive in conjugate pairs: if a + bi is a root, so is a - bi. That is why a real cubic always has at least one real root — three roots, but non-real ones come two at a time, so one must be left over and real. For a quadratic ax^2 + bx + c = 0, the discriminant b^2 - 4ac tells the tale: negative means two conjugate complex roots.
solve x^2 - 4x + 13 = 0 by the quadratic formula
x = ( -b ± sqrt(b^2 - 4ac) ) / (2a), a=1, b=-4, c=13
discriminant: (-4)^2 - 4·1·13 = 16 - 52 = -36 (< 0)
x = ( 4 ± sqrt(-36) ) / 2
= ( 4 ± 6i ) / 2
= 2 ± 3i
roots: 2 + 3i and 2 - 3i — a conjugate pair ✓Notice the quadratic formula never failed — it just produced sqrt(-36), which we now read as 6i. Complex numbers don't change the formula; they finally let it always finish.
Building a polynomial from its roots
Run the logic in reverse. Because roots determine the factors, you can build a real polynomial from prescribed roots — and conjugate pairs multiply back to a real quadratic, so no i is left behind.
build a polynomial with roots 2 + 3i and 2 - 3i:
(x - (2 + 3i))(x - (2 - 3i))
= ( (x - 2) - 3i )( (x - 2) + 3i )
= (x - 2)^2 - (3i)^2 [difference of squares]
= (x^2 - 4x + 4) - (9)(-1)
= x^2 - 4x + 4 + 9
= x^2 - 4x + 13 ← real coefficients, no i