Complex Zeros & the Fundamental Theorem of Algebra

You’ll be able to

State how many zeros a polynomial of degree n has, counted with multiplicity
Use conjugate pairs to reconstruct a polynomial from partial information about its zeros
Explain why a real polynomial of odd degree must have a real zero

Every polynomial factors completely — over the complex numbers

x² + 1 has no real zeros; its graph never touches the axis. But it does have zeros i and −i, and with those the factorization is complete: x² + 1 = (x − i)(x + i). The Fundamental Theorem of Algebra promises this always works: a polynomial of degree n ≥ 1 has exactly n complex zeros when counted with multiplicity, and therefore factors into exactly n linear factors over the complex numbers. "No zeros" is only ever a statement about the reals.

Fundamental Theorem of Algebra

degree n ⟹ exactly n complex zeros counted with multiplicity

x² − 6x + 9 = (x − 3)² has "two" zeros: 3 twice. Multiplicity is what makes the count exact.

Non-real zeros arrive in pairs

When a polynomial has real coefficients, its non-real zeros come in complex conjugate pairs: if a + bi is a zero, so is a − bi. The reason is mechanical — conjugating the whole equation p(z) = 0 leaves the real coefficients alone and conjugates z. The consequence is powerful: non-real zeros are always even in number, so an odd-degree real polynomial cannot pair them all up and must have at least one real zero. That is why every real cubic crosses the x-axis somewhere.

Conjugate pair theorem

real coefficients and p(a + bi) = 0 ⟹ p(a − bi) = 0

A conjugate pair multiplies to the real quadratic (x − a)² + b², which is why real polynomials factor into real linear and real quadratic pieces.

Checkpoint

A polynomial with real coefficients has degree 4 and zeros 2 and 3i. What are its other zeros?

Worked example

Find a polynomial of least degree with real coefficients whose zeros include 1 and 2 − i.

1.Real coefficients force the conjugate 2 + i to be a zero as well, so there are at least three zeros.
2.Build the factors: (x − 1), (x − (2 − i)), (x − (2 + i)).
3.Multiply the conjugate pair first, since it must give a real quadratic: [(x − 2) + i][(x − 2) − i] = (x − 2)² − i² = (x − 2)² + 1.
4.Expand: x² − 4x + 4 + 1 = x² − 4x + 5.
5.Multiply by the remaining factor: (x − 1)(x² − 4x + 5) = x³ − 4x² + 5x − x² + 4x − 5.
6.Combine: x³ − 5x² + 9x − 5.

Answer: p(x) = x³ − 5x² + 9x − 5, of degree 3. Any nonzero real multiple of this also works — the zeros determine the polynomial only up to a constant factor.

Tip

Always multiply the conjugate pair together before bringing in the real factors. Grouping as [(x − 2) + i][(x − 2) − i] turns into a difference of squares and the imaginary units cancel in one step, with no i left to lose track of.

What the graph does and does not show

A graph shows real zeros and nothing else, so it can undercount. A quartic whose graph crosses the axis twice has two real zeros and one conjugate pair. A quartic whose graph never touches the axis has four non-real zeros in two pairs. Counting x-intercepts therefore gives a lower bound on the number of real zeros, and the missing ones — measured by degree minus (real zeros with multiplicity) — must be non-real and even in number.

Checkpoint

The graph of a degree-5 polynomial with real coefficients crosses the x-axis exactly once, with no flattening. How many non-real zeros does it have?

On the exam

Non-real zeros are examinable on AP Precalculus, but building polynomials from complex zeros shows up far more often than computing with i. Practice the reconstruction direction — given some zeros, produce the polynomial.

Answer the 2 checkpoints as you read.