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Recipe

Factor a polynomial completely over \mathbb C

Use when: you're given a polynomial of degree n\ge3 and at least one of its roots, and asked for the complete factorization into n linear factors

Goal: the polynomial written as a product of exactly n linear factors, with every root the Fundamental Theorem promises actually accounted for

Taught in The Fundamental Theorem of Algebra, revisited

Quick recipe

  1. Count how many roots to expect
  2. Divide out the known root
  3. Solve what's left
  4. Check nonreal roots came in a conjugate pair
  5. Assemble the complete factorization
  1. 1

    Read the target root count off the degree n. The Fundamental Theorem of Algebra guarantees exactly n roots in \mathbb C, counting multiplicity. You know this before any work is done.

    Example

    p(x)=x^3-3x^2+4x-2 has degree 3. So exactly 3 roots, with multiplicity, are waiting to be found, real or not.

    Why: this is the target you are building toward. Without it there is no way to know when the factorization is finished.

  2. 2

    Use polynomial or synthetic division to divide by (x-r) for the known root r. This lowers the degree by exactly one.

    Example

    Dividing x^3-3x^2+4x-2 by (x-1) gives x^3-3x^2+4x-2=(x-1)(x^2-2x+2).

    Why: the Factor Theorem says (x-r) divides evenly, with no remainder, exactly when r is a genuine root.

  3. 3

    Solve the remaining factor for its own roots.

    Example

    x^2-2x+2=0 gives x=\frac{2\pm\sqrt{4-8}}{2}=\frac{2\pm\sqrt{-4}}{2}=1\pm i.

    • the remaining factor is quadratic — apply the quadratic formula directly.
    • the remaining factor still has degree 3 or higher — find or guess another root and repeat Step 2 on it.

    ⚡ Shortcut

    Say the original polynomial has degree exactly 3 and you already have one root. Then the leftover factor after Step 2 is automatically quadratic. You need not check the case fork above. Go straight to the quadratic formula.

    Why: every root of the leftover factor is also a root of the original polynomial. Dividing it out only removed the one root r.

  4. 4

    If every coefficient of the original polynomial is real, confirm that any nonreal roots arrived in a conjugate pair a\pm bi. They should not appear on their own.

    Example

    1+i and 1-i form a genuine conjugate pair, as a real polynomial requires.

    Why: the Conjugate Pair Theorem says a lone, unpaired nonreal root is impossible for a real polynomial. So this check catches an arithmetic slip before you write the final answer.

  5. 5

    Write the polynomial as the leading coefficient times all n linear factors, one per root. A repeated root gives repeated factors.

    Example

    p(x)=(x-1)\big(x-(1+i)\big)\big(x-(1-i)\big)

    This is three linear factors for a degree-3 polynomial, as Step 1 promised.

    Why: this shows that the polynomial is factored completely over \mathbb C. It does not just assume it.

Common mistakes

  • Dividing by (x+r) instead of (x-r) for a known root r.
  • Stopping after the real roots and never solving the leftover factor for its nonreal roots.
  • Assuming a negative discriminant means "no roots" instead of a genuine conjugate pair of complex roots.
  • Losing track of a repeated root's multiplicity when a factor divides out more than once.