Now run the factor theorem backwards: from the zeros, build the polynomial. Remember from the The Fundamental Theorem of Algebra that complex zeros of real polynomials come in pairs.
Key idea
Each zero $r$ gives a factor $(x - r)$. Multiply the factors, multiply by a constant $a$, and use any extra point to find $a$. Multiply conjugate or $\pm\sqrt{m}$ pairs first: $(x - (p + qi))(x - (p - qi))$ $= (x - p)^2 + q^2$ and $(x - \sqrt{m})(x + \sqrt{m}) = x^2 - m$.
Why it works
By the factor theorem, $p(r) = 0$ exactly when $(x - r)$ divides $p(x)$, so a polynomial with all those zeros must contain all those factors. Any constant multiple has the same zeros, so a point off the axis is needed to choose the one you want. Pairing conjugates first keeps the arithmetic real: it is the difference of squares $(x - p)^2 - (qi)^2$, and $i^2 = -1$ turns the minus into a plus.
Worked examples
Worked example: Integer zeros and a point
Find the cubic with zeros $-2$, $1$, $3$ that passes through $(0, 12)$.
- $p(x) = a(x + 2)(x - 1)(x - 3)$.
- At $x = 0$: $a(2)(-1)(-3) = 6a = 12$, so $a = 2$.
- Expand: $2(x + 2)(x^2 - 4x + 3)$ $= 2(x^3 - 2x^2 - 5x + 6)$.
Answer: $p(x) = 2x^3 - 4x^2 - 10x + 12$
Worked example: A complex zero
Find the polynomial of lowest degree with real coefficients, leading coefficient 1, and zeros $2$ and $1 + 3i$.
- $1 - 3i$ is also a zero, so the degree is 3.
- Pair the conjugates: $(x - 1)^2 + 9 = x^2 - 2x + 10$.
- Multiply by $(x - 2)$: $x^3 - 2x^2 + 10x - 2x^2 + 4x - 20$.
Answer: $x^3 - 4x^2 + 14x - 20$
Common mistake
Writing $(x + 3)$ for the zero $3$. A zero $r$ needs a factor that is 0 at $r$, which is $(x - r)$. Check: $(3) - 3 = 0$. Also include the conjugate of every complex zero, or the coefficients will not be real.