Fundamental Theorem of Algebra

The fundamental theorem of algebra states that every non-constant polynomial with complex (or real) coefficients has at least one complex zero. This sounds simple, but it is one of the most powerful facts in algebra: it guarantees that polynomial equations always have solutions, as long as you are willing to work with complex numbers instead of restricting yourself to real numbers.

Consider \(f(x) = x^2 + 1\). There is no real number that satisfies \(x^2 + 1 = 0\), since \(x^2\) can never equal \(-1\) for a real \(x\). But once complex numbers are allowed, \(x = i\) and \(x = -i\) both work, because \(i^2 = -1\). The fundamental theorem of algebra tells us that this will always happen: a polynomial can "run out" of real zeros, but it never runs out of complex ones.

The theorem becomes even more useful when combined with the factor theorem. If \(c\) is a zero of a polynomial function \(f(x)\), then \((x - c)\) is a factor of \(f(x)\). Applying the fundamental theorem of algebra repeatedly, dividing out one linear factor at a time, leads to a key corollary:

Every polynomial of degree \(n\) (with \(n \ge 1\)) can be written as a product of exactly \(n\) linear factors, and therefore has exactly \(n\) zeros, counted with multiplicity, over the complex numbers.

"Counted with multiplicity" matters. The polynomial \(f(x) = (x - 2)^2\) has degree 2, but it has only one distinct zero, \(x = 2\), which is repeated twice. The theorem still holds because that zero is counted twice when tallying to reach the total of 2.

When a polynomial's coefficients are all real numbers, any non-real complex zeros must occur in conjugate pairs. If \(a + bi\) (with \(b \ne 0\)) is a zero, then \(a - bi\) must also be a zero. This is why the quadratic \(x^2 + 1\) produced the pair \(i\) and \(-i\) rather than just one complex zero on its own; a real-coefficient polynomial can never have an odd number of non-real zeros left over.

This pairing rule is also why a real-coefficient polynomial of odd degree must have at least one real zero: the non-real zeros always come in pairs, so if the degree is odd, at least one zero has to be real.

Find all zeros of \(f(x) = x^3 - x^2 + x - 1\).

Group and factor by pairs: \(f(x) = x^2(x - 1) + 1(x - 1) = (x - 1)(x^2 + 1)\). Setting each factor equal to zero gives \(x = 1\) from the linear factor, and \(x = i\), \(x = -i\) from \(x^2 + 1 = 0\). That is three zeros for a degree-3 polynomial, exactly matching the fundamental theorem of algebra: one real zero and one conjugate pair of complex zeros.

Find all zeros of \(g(x) = (x - 3)^2(x + 2)\) and confirm the total matches the degree.

Expanding shows \(g(x)\) has degree 3. The zeros are \(x = 3\) (with multiplicity 2, from the squared factor) and \(x = -2\) (with multiplicity 1). Counted with multiplicity, that is \(2 + 1 = 3\) zeros total, matching the degree exactly as the theorem predicts.

Find all zeros of \(h(x) = x^2 - 4x + 13\).

Using the quadratic formula, \(x = \frac{4 \pm \sqrt{16 - 52}}{2} = \frac{4 \pm \sqrt{-36}}{2} = 2 \pm 3i\). The zeros are \(2 + 3i\) and \(2 - 3i\), a conjugate pair, giving two zeros for a degree-2 polynomial. The graph of \(h(x)\) never touches the x-axis, but the theorem still guarantees two zeros exist, just not among the real numbers.

The fundamental theorem of algebra is the reason techniques like the rational zero test are useful: they help you locate the real zeros among the finitely many guaranteed candidates, and once one is found, dividing it out reduces the degree so the remaining complex zeros are easier to uncover. Knowing in advance exactly how many zeros to expect, real and complex combined, keeps you from stopping a factoring problem too early or assuming a polynomial has no solutions just because it has no real ones.

Related lessons

• Algebra Help