Here is the kind of question to learn to recognise in under a second.
"A polynomial has roots 2, -1, 3. Find it."
Your hand should already be writing
before you finish reading the sentence. That is the polynomial — the factored form is a complete answer, and it is the only answer up to a constant multiple. This article builds the reflex: when roots are given and a polynomial is wanted, your first move is always the product of (x - r) factors.
The factor theorem, restated
The one fact doing all the work is this:
r is a root of p(x) if and only if (x - r) is a factor of p(x).
This is the factor theorem, and the "if and only if" is the important part. Roots and linear factors are the same information dressed in different clothes. When the question goes backwards — here are the roots, give me a polynomial — you run the factor theorem in reverse. Each root r contributes one factor (x - r). Multiply them and you are done.
Construction pipeline
The whole procedure has three steps:
-
Take the roots. Given r_1, r_2, \dots, r_k, write
p(x) = a \cdot (x - r_1)(x - r_2) \cdots (x - r_k). -
Pick the leading coefficient a. The roots do not determine a — any nonzero scalar multiple has the same roots. If the problem says "monic," pick a = 1. If it specifies a y-intercept or some other value, solve for a. If nothing is said, default to a = 1.
-
Expand if asked. Factored form is usually more informative, but many problems want standard form.
Worked example 1: roots 2, 3, 5
Monic polynomial with roots 2, 3, 5:
Expand two factors first: (x - 2)(x - 3) = x^2 - 5x + 6. Now multiply by the third:
Sanity check: the constant term -30 = (-1)^3 \cdot (2)(3)(5), matching Vieta's.
Worked example 2: roots -1 and 2
Watch the sign flip — root -1 gives (x + 1), not (x - 1). Expand:
Vieta's check: sum of roots = 1 = -b/a, product = -2 = c/a. Both match.
Worked example 3: root 0 and a repeated root 4
Roots 0, 4, 4. The root 4 appears twice, so its factor appears twice; (x - 0) collapses to x:
No constant term, because 0 is a root forces p(0) = 0. The graph passes through the origin.
Dealing with multiplicities
A root can repeat. "Double root at 3" or "multiplicity 2" means the factor appears twice. Multiplicity 3 means three copies, and so on.
- Root 3 with multiplicity 2 \longrightarrow (x - 3)^2.
- Roots \{3, 3, -1\} \longrightarrow (x - 3)^2 (x + 1), degree 3.
- Root -2 with multiplicity 4 \longrightarrow (x + 2)^4, degree 4.
The degree equals the total multiplicity. A degree-5 polynomial with roots \{1, 1, 1, 2, -3\} is (x - 1)^3 (x - 2)(x + 3) — five factors total, matching degree 5.
Irrational and complex roots come in pairs
A trap. If the problem says "one root is 1 + \sqrt{2}, write a polynomial with rational coefficients," you cannot use that root alone — (x - (1 + \sqrt{2})) has irrational coefficients. You need its partner.
For polynomials with real coefficients:
- Irrational pair. If a + b\sqrt{c} is a root, so is a - b\sqrt{c}. Combined: (x - a)^2 - b^2 c, rational if a, b, c are.
- Complex pair. If a + bi is a root, so is a - bi. Combined: (x - a)^2 + b^2, real.
Rule: whenever a root is irrational or complex, include its conjugate before writing factors, or the polynomial will not live in the number system the problem expects. Roots \{1 + \sqrt{2}, 1 - \sqrt{2}\} give (x - 1)^2 - 2 = x^2 - 2x - 1 — integer coefficients.
The leading coefficient a
Roots pin down a polynomial only up to scaling. (x - 2)(x - 3), 5(x - 2)(x - 3), and -\tfrac{1}{2}(x - 2)(x - 3) all have the same roots \{2, 3\}. a is a free parameter.
- No extra info: take a = 1 (monic).
- y-intercept given: plug in x = 0, solve for a. Roots \{2, 3\} and p(0) = 10: a(-2)(-3) = 6a = 10, so a = 5/3.
- Value at a specific x: same trick.
Given Vieta's data, reconstruct the polynomial
Sometimes the problem gives you the sum and product instead of the roots themselves. For a monic quadratic x^2 + bx + c with roots r_1, r_2, Vieta's says r_1 + r_2 = -b and r_1 r_2 = c. "Sum = 7, product = 10" gives
revealing roots 2 and 5. Same idea at higher degree: for a monic cubic x^3 + bx^2 + cx + d, the coefficients are the elementary symmetric functions with alternating signs. "Sum = 6, pairwise sum = 11, triple product = 6" gives x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3).
Recognition drill
Write the monic polynomial with the given roots. Under ten seconds each.
- \{1, 2, 3\}: (x - 1)(x - 2)(x - 3) = x^3 - 6x^2 + 11x - 6.
- \{-1, -2\}: (x + 1)(x + 2) = x^2 + 3x + 2.
- \{0, 4, 4\}: x(x - 4)^2 = x^3 - 8x^2 + 16x.
- \{1 + \sqrt{2}, 1 - \sqrt{2}\}: (x - 1)^2 - 2 = x^2 - 2x - 1.
- \{2 + 3i, 2 - 3i\}: (x - 2)^2 + 9 = x^2 - 4x + 13.
Common confusions
- "(x - 3) gives root -3." No. (x - 3) = 0 forces x = 3. The sign inside the factor is opposite to the sign of the root.
- "(x + 5) gives root +5." No. (x + 5) = 0 forces x = -5. Rewrite (x + 5) = (x - (-5)) to see it.
- "Different leading coefficients give different roots." No. Scaling by a nonzero constant does not move roots. 3(x - 2), (x - 2), and -100(x - 2) all have the single root x = 2.
- "Use just one of a complex conjugate pair." Only if you want complex coefficients. For a real-coefficient polynomial, complex roots come in pairs — use both.
The reflex, in one line
Roots and linear factors are the same information with the sign flipped. Given a list of roots, reach for (x - r_1)(x - r_2) \cdots (x - r_k) immediately, pick a from whatever the problem says (default 1), include conjugates for any irrational or complex root, and expand if asked. The degrees add, the coefficients fall out, and the factor theorem does all the thinking.