Every characteristic polynomial, minimal polynomial, and invariant factor this course computes is, underneath the linear-algebra dressing, an ordinary single-variable polynomial — and every one of those objects depends on facts about division, roots, and factorization that a College Algebra course covers but rarely revisits at the depth this course needs. This section is that revisit: polynomial long division with a remainder, what a root's multiplicity means, and exactly how factoring over \mathbb C differs from factoring over \mathbb R.
Factoring a polynomial over \mathbb C is like taking apart a machine down to every last screw. Nothing more to disassemble. Factoring the same machine over \mathbb R stops one level short whenever two screws were actually a matched complex-conjugate pair. Those two stay bolted together as one irreducible real piece, a quadratic factor, because separating them would need a real number that doesn't exist. Multiplicity just counts how many identical copies of the same screw turned up. Long division is just checking whether a candidate screw actually unscrews cleanly: divide it out, and a leftover remainder of zero is what "clean" means.
A root of a polynomial p is a number r with p(r) = 0. Polynomial division is the process of writing p(x) = q(x)(x-r) + c for some quotient q and remainder c; r is a root exactly when that remainder c comes out zero, so (x-r) divides p evenly. The Fundamental Theorem of Algebra guarantees a degree-n polynomial has n roots counted with multiplicity, over \mathbb C — always. Over \mathbb R, the same polynomial factors only as far as its real roots allow; any complex root pairs with its conjugate to reconstitute an irreducible real quadratic factor, since a real polynomial's non-real roots always come in conjugate pairs (a direct consequence of conjugation distributing over arithmetic, reviewed in the previous section). A root r has multiplicity k when (x-r)^k divides the polynomial but (x-r)^{k+1} does not — the exact number that becomes an eigenvalue's algebraic multiplicity the moment this polynomial is a characteristic polynomial, in chapter 06.
Theorem
Fundamental Theorem of Algebra
Recalled from College Algebra · Complex numbers and the Fundamental Theorem — first taught there.
Every polynomial of degree n \geq 1 with complex (in particular, real) coefficients has at least one complex zero.
Every known proof leans on machinery outside this course — complex analysis (Liouville's theorem: a bounded entire function is constant, applied to 1/f), topology (a degree-n map wraps the plane around itself n times, so it must hit 0), or Galois theory. None of those tools are available yet, so the result is taken on faith here, the way it is in every first course that uses it. What this course can do with it: combined with the Factor Theorem, it guarantees a degree-n polynomial factors completely into n linear factors over the complex numbers — that consequence, the Linear Factorization Theorem, is used freely and does have its own proof in the registry.
Multiplicity controls bounce versus cross
Slide a root until it overlaps another: watch the curve switch from crossing the x-axis to merely touching it, exactly where multiplicity becomes even.
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A polynomial with no real root at all
Drag λ across the whole visible range — p(λ) never once touches zero, which is exactly what 'no real eigenvalue' looks like as a graph.
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Recipe Factor a polynomial completely and state each root's multiplicity Use when: you have a polynomial and need every root, over \mathbb R or over \mathbb C, along with how many times each one repeats Goal: the polynomial written as a product of irreducible factors, with each root's multiplicity read straight off
Quick recipe
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1
Try plugging in small integers, 0,\pm1,\pm2,\dots, until one makes the polynomial equal zero. The Rational Root Theorem narrows the search further when needed.
Example
For q(x)=x^3-3x^2+4: q(-1)=-1-3+4=0, so x=-1 is a root.
⚡ Shortcut
The Rational Root Theorem restricts candidates to \pm(a divisor of the constant term) over (a divisor of the leading coefficient), instead of testing every integer blindly.
Why: the Factor Theorem guarantees (x-r) divides the polynomial exactly when r is a root, so one successful test already hands over a whole factor.
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2
Divide the polynomial by (x-r) to get a quotient of one lower degree, and confirm the remainder is genuinely zero.
Example
Dividing x^3-3x^2+4 by (x+1) gives quotient x^2-4x+4 with remainder 0.
Why: long division peels off the known root exactly, leaving a lower-degree polynomial whose remaining roots are still unknown.
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3
Apply Steps 1 and 2 again to the quotient. Hitting the same root twice raises its multiplicity by one each time.
Example
x^2-4x+4=(x-2)^2: testing x=2 succeeds twice in a row, so its multiplicity is 2.
- the quotient factors further over the reals — repeat Steps 1 and 2 on it again.
- the quotient is an irreducible quadratic with a negative discriminant — stop dividing over \mathbb R; it only splits further over \mathbb C.
Why: multiplicity is nothing more than counting how many times the same linear factor divides out before the quotient stops containing it.
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4
If an irreducible real quadratic factor remains and complete factorization is required, solve it with the quadratic formula; its two roots always form a conjugate pair.
Example
\lambda^2+1 has no real root, but over \mathbb C it factors as (\lambda-i)(\lambda+i).
Why: the Fundamental Theorem of Algebra guarantees a degree-n polynomial has exactly n roots counted with multiplicity, but only once \mathbb C is available.
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5
Write the polynomial as a product of irreducible factors, then read each root's multiplicity off as the exponent on its linear factor.
Example
q(x)=(x+1)(x-2)^2, so x=-1 has multiplicity 1 and x=2 has multiplicity 2.
Why: this multiplicity is the exact number that becomes an eigenvalue's algebraic multiplicity once this polynomial is a characteristic polynomial.
Common mistakes
- Stopping after finding one root instead of dividing out and testing the quotient for repeats.
- Reporting a repeated root only once when its factor's exponent is greater than one.
- Treating an irreducible real quadratic as unfactorable instead of splitting it over \mathbb C.
- A sign slip while testing a candidate root, especially confusing p(-r) with -p(r).
Worked example
Long division with a remainder
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1set up polynomial long division exactly as College Algebra taught it, tracking the remainder explicitly
This is Step 2 of the recipe above, dividing out a known root by hand. Divide p(x) = x^3 - 2x^2 - x + 2 by (x-1), and interpret the remainder. x^3 - 2x^2 - x + 2 = (x-1)(x^2 + ax + b) + r
set up polynomial long division exactly as College Algebra taught it, tracking the remainder explicitly
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2
Carrying out the division: x^3-2x^2-x+2 = (x-1)(x^2-x-2)+0 — the remainder is exactly 0.
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3\boxed{p(x) = (x-1)(x^2-x-2),\ \text{remainder } 0 \implies x=1 \text{ is an exact root}}
A zero remainder is the Factor Theorem in action: (x-1) divides evenly exactly because p(1) = 1-2-1+2 = 0.
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Worked example
Multiplicity from a repeated factor
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1test small integer candidates first — the Rational Root Theorem narrows the search
This runs the full recipe above, Steps 1 through 3, on the polynomial the recipe's own examples used. Factor q(x) = x^3 - 3x^2 + 4 completely and state each root's multiplicity. q(-1) = -1-3+4 = 0 \implies (x+1) \text{ is a factor}
test small integer candidates first — the Rational Root Theorem narrows the search
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2
Dividing out (x+1): q(x) = (x+1)(x^2-4x+4) = (x+1)(x-2)^2.
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3\boxed{q(x) = (x+1)(x-2)^2,\ \text{root } x=-1 \text{ has multiplicity } 1,\ \text{root } x=2 \text{ has multiplicity } 2}
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Worked example
Factoring the same polynomial over R versus over C
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1group terms in pairs to spot the common factor, the same technique from College Algebra
Factor p(\lambda) = \lambda^3 - \lambda^2 + \lambda - 1 completely, first over \mathbb R, then over \mathbb C. p(\lambda) = \lambda^2(\lambda - 1) + (\lambda - 1) = (\lambda-1)(\lambda^2+1)
group terms in pairs to spot the common factor, the same technique from College Algebra
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2
Over \mathbb R, \lambda^2+1 has no real root (section 8's rotation matrix example, again) — so (\lambda-1)(\lambda^2+1) is already completely factored over \mathbb R.
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3\boxed{\text{Over } \mathbb C:\ p(\lambda) = (\lambda-1)(\lambda-i)(\lambda+i)}
The irreducible real quadratic \lambda^2+1 splits into the conjugate pair \pm i exactly once complex roots are allowed — the general pattern this course leans on every time a characteristic polynomial doesn't factor completely over \mathbb R.
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Check your understanding (counts toward your progress)
The Factor Theorem says (x-r) divides p(x) evenly exactly when
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This reverses the roles — the theorem is about plugging r INTO p, not reading r off from p(0).
A polynomial's degree is a fixed count of terms, unrelated to any particular root's value.
The sign of p(r) is irrelevant — the Factor Theorem is specifically about p(r) landing on exactly zero.
Why
Dividing p(x) by (x-r) always leaves a remainder equal to p(r) (the Remainder Theorem) — that remainder is zero, meaning exact division, precisely when r is a root.
Recall the Remainder Theorem: dividing p(x) by (x-r) always leaves remainder exactly p(r).
Over \mathbb R, an irreducible quadratic factor like x^2+1 corresponds, over \mathbb C, to
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A repeated real root would make the quadratic a perfect square like (x-r)^2, which x^2+1 is not.
Two distinct real roots would mean the quadratic factors over R already — that's exactly what "irreducible over R" rules out.
The Fundamental Theorem of Algebra guarantees every nonconstant polynomial has a root over C — "no roots at all" never happens there.
Why
x^2+1 = (x-i)(x+i) over \mathbb C — the conjugate pair \pm i is exactly what an irreducible real quadratic splits into.
Solve x^2+1=0 over \mathbb C directly and see what kind of roots come out.
Factor x^3+x completely over \mathbb C.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
x^3+x = x(x^2+1) = x(x-i)(x+i) — the real root 0 stays as is, and the irreducible real quadratic splits into a conjugate pair over C.
Correct answer
Also accepted:
Why
Factor out the common x first, then factor the remaining quadratic over \mathbb C.
A degree-5 real polynomial has roots 2 (multiplicity 2), -1, and one conjugate pair. How many DISTINCT roots does it have?
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Why
Distinct roots: 2, -1, and the conjugate pair contributes 2 more distinct values — 4 total, even though multiplicities sum to 5 (2+1+1+1).
Correct answer
Also accepted:
Why
Count distinct values, not multiplicities: 2, -1, and the two members of the conjugate pair.
Practice: Polynomial factoring practice (4 problems · doesn’t affect your progress)
Divide x^2-4 by (x-2).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
(x-2)(x+2) = x^2-4 — a difference of squares, exact division with remainder 0.
Correct answer
Also accepted:
Why
Given p(x) = (x+2)^3(x-1), determine the multiplicity of its root x=-2.
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Why
The factor (x+2) appears raised to the third power — multiplicity 3.
Correct answer
Also accepted:
Why
Write out the complete factorization of x^2+4 into linear factors over \mathbb C.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
x^2+4 = x^2-(2i)^2 = (x-2i)(x+2i) — the conjugate pair \pm 2i.
Correct answer
Also accepted:
Why
A degree-6 real polynomial has 2 distinct real roots and the rest non-real. How many non-real roots does it have (counted with multiplicity), assuming no repeated roots at all?
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Why
6 - 2 = 4 non-real roots, forming 2 conjugate pairs — consistent with the conjugate-pairing rule.
Correct answer
Also accepted:
Why
6 total roots minus 2 real roots leaves the non-real count, and non-real roots always pair up.
Where this shows up: the characteristic and minimal polynomials
Chapter 06 opens by turning \det(A-\lambda I) into a genuine polynomial in \lambda and asking exactly the questions this section rehearsed: what are its roots, what are their multiplicities, and does it factor completely over \mathbb R or only over \mathbb C? Chapter 07's Jordan form is, underneath its geometric description, a statement about how a characteristic polynomial's repeated roots interact with the operator's actual eigenvectors — multiplicity is the thread connecting every chapter from here to the end of canonical forms.
Where this goes next
Numerical root-finding and control theory
A control system's stability check reduces to finding whether a polynomial's roots (the system's poles) all have negative real part — the Routh–Hurwitz criterion answers this without ever finding the roots explicitly, using only the polynomial's coefficients, precisely because factoring a high-degree polynomial exactly is often infeasible even though its coefficients are known exactly.