College Algebra used the Fundamental Theorem of Algebra to show something. Every polynomial equation you were asked to solve had a full set of roots waiting to be found, real or not. Nothing about the statement changes here. But this course will later give you a different reason it is true. That reason comes from a piece of complex analysis called Liouville's theorem, not from algebra or counting. That proof is many chapters away. The statement itself is worth having solid starting now. The next two chapters lean constantly on the idea that "this equation has a root somewhere in \mathbb C."
A real polynomial's graph is allowed to never touch the x-axis. The graph of x^2+1 never does. The real content of the Fundamental Theorem is that this "missing" crossing was only ever missing because the search was stuck on one line, the real axis. Widen the hunt to the whole plane, and a root always turns up. It is guaranteed, with no exceptions, for any polynomial of degree one or more.
Say plainly what the theorem does not promise. It never claims the root is easy to compute. It only says that hunting for one is never a wasted search. A computer or a clever substitution may still be the only practical way to land on it.
Combine the Fundamental Theorem of Algebra with the Factor Theorem. The result upgrades "at least one root" to a full count. Factor out that one root. That leaves a polynomial of degree one lower. By the same theorem it has a root of its own. Continue down to degree zero.
The result is the Linear Factorization Theorem. A degree-n polynomial always factors completely into n linear factors over \mathbb C: p(z)=a(z-r_1)(z-r_2)\cdots(z-r_n). So it has exactly n roots, once each root is counted as many times as its factor repeats. That count is its multiplicity.
Suppose every coefficient of p is real. Then nonreal roots still arrive in conjugate pairs a\pm bi. This is the Conjugate Pair Theorem, already reviewed alongside this one. So the nonreal roots of a real polynomial always total an even number. That forces the real-root count to match the polynomial's degree in parity, not in any looser sense.
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
A polynomial with real coefficients has zeros 3, -2, and 1 + i (each multiplicity 1). What is the minimum possible degree?
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Why
Real coefficients force the nonreal zero 1+i to bring along its conjugate 1-i as well, even though it isn't listed. That's four distinct zeros total (3, -2, 1+i, 1-i), so the minimum degree is 4.
Correct answer
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Why
Remember that a nonreal zero of a real-coefficient polynomial can never appear alone — its conjugate must be a zero too, even if the problem doesn't say so.
From College Algebra
A cubic polynomial with real coefficients has zeros 2 and 1 + i, where the nonreal zero's real part is positive. What is the third zero?
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That's the negative of 1 + i, not its conjugate 1 - i.
The conjugate pairs with 1 + i, not with the real zero 2 — real zeros are their own mirror image.
There is: a cubic has exactly 3 zeros, and the conjugate-pair rule forces the third one.
Why
A cubic has exactly three zeros (FTA), and real coefficients force nonreal zeros into conjugate pairs. Since 1 + i is a zero, so is 1 - i — and that fills the third and final slot.
The conjugate pairing rule applies to the nonreal zero, not the real one — think about what must pair with 1 + i.
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.
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All three roots, and this time all of them real
Drag any root — for this particular cubic, every root the Fundamental Theorem promises happens to be a real number. That won't always be true; the total count of three, matching the degree, always is.
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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
Quick recipe
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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.
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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.
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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.
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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.
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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.
Worked example
Factoring completely over \mathbb C, given one real root
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1
Factor p(x)=x^3-3x^2+4x-2 completely over \mathbb C. You are given that x=1 is a root. This runs the recipe above start to finish. Steps 1-2 divide out the known root. Step 3 solves the quadratic that's left. Step 5 assembles the final answer. Divide out the known root: x^3-3x^2+4x-2 = (x-1)(x^2-2x+2).
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2
Solve the remaining quadratic with the quadratic formula: x = \frac{2\pm\sqrt{4-8}}{2} = \frac{2\pm\sqrt{-4}}{2} = 1\pm i
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3
A degree-3 polynomial has three roots, as the theorem guarantees. Here there is one real root and one conjugate pair. \boxed{p(x) = (x-1)\big(x-(1+i)\big)\big(x-(1-i)\big)}
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Worked example
Counting roots without finding a single one
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1
A degree-5 polynomial with real coefficients — could it have exactly two nonreal roots? The Fundamental Theorem guarantees 5 roots total, counting multiplicity.
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2
Nonreal roots of a real polynomial always arrive in conjugate pairs, so the nonreal count must be even: 0, 2, or 4, never 1, 3, or 5.
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3
Two nonreal roots is even, so it's allowed — it would leave 5-2=3 real roots, which is consistent. What's actually impossible is a real count of exactly 4: that would leave one nonreal root with no partner. \boxed{\text{Yes -- two nonreal roots is possible; exactly one nonreal root never is.}}
Where this goes next
A theorem this course eventually reproves, differently
Every proof available today reaches outside plain algebra for its machinery. That includes the one attached to the card above. Much later in this course, Liouville's theorem gives a proof that has nothing to do with counting roots. The theorem says a bounded function with no bad points anywhere in \mathbb C must be constant.
Here is how the proof goes. Suppose a nonconstant polynomial were never zero. Then 1/p(z) is defined and bounded everywhere. Liouville's theorem then forces it to be constant. That is a contradiction. Only complex analysis, not algebra, can produce it.
Check your understanding (counts toward your progress)
Could a real-coefficient degree-4 polynomial have exactly one real root, with the rest nonreal?
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There IS a constraint: nonreal roots of a real polynomial must pair up, which caps how the real and nonreal counts can split.
Degree alone doesn't force any minimum number of real roots — x^4+1, for instance, has zero real roots at all.
Multiplicity 4 on one real root means there is nothing left over — that contradicts 'the rest nonreal,' since there's no rest at all.
Why
Exactly one real root would leave three nonreal roots — an odd number — which is impossible, since nonreal roots always pair up.
A degree-4 polynomial's nonreal roots must still come in conjugate pairs.
A real-coefficient degree-6 polynomial has exactly 2 real roots. How many nonreal roots does it have?
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Why
6-2=4 nonreal roots, which is consistent since 4 is even (two conjugate pairs).
Correct answer
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Why
The total root count (with multiplicity) must equal the degree.
Which statement IS the Fundamental Theorem of Algebra?
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This is false in general — x^2+1 has no real root at all, only \pm i.
Rational factorization is a much stronger and generally false claim — most polynomials need irrational or nonreal coefficients to factor completely.
This is true, but it's a consequence of the theorem, not the theorem itself. The theorem's actual content is guaranteeing a COMPLEX root exists at all.
Why
The theorem guarantees at least one complex root for any polynomial of degree 1 or higher, real coefficients or not.
The theorem's whole point is widening the search from real numbers to complex numbers.
How many roots, counting multiplicity, does a degree-7 polynomial have over \mathbb C?
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Why
The Fundamental Theorem of Algebra, applied repeatedly through the Factor Theorem, guarantees exactly 7 roots for a degree-7 polynomial.
Correct answer
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Why
Practice: Counting and finding roots with the Fundamental Theorem (5 problems · doesn’t affect your progress)
The Fundamental Theorem of Algebra guarantees a degree-n polynomial has...
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The guarantee is for complex roots specifically — real roots aren't promised at all in general.
Repeated roots (multiplicity greater than 1) are explicitly allowed by the theorem, not ruled out.
The guarantee holds for every degree n\ge1, odd or even.
Why
The theorem guarantees at least one complex root for any polynomial of degree 1 or higher.
At least one complex root
A degree-9 polynomial over \mathbb C has how many roots in total, multiplicity included?
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Why
Repeated application of the Factor Theorem gives exactly 9 roots, counting multiplicity.
Correct answer
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Why
Given x=-1 is a root of p(x)=x^3+x^2+x+1, find the remaining quadratic quotient.
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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
Synthetic division by -1 gives quotient coefficients 1,0,1, i.e. x^2+1, remainder 0.
Correct answer
Also accepted:
Why
Use synthetic division by -1 on the coefficients 1,1,1,1.
Using the quotient x^2+1 from p(x)=x^3+x^2+x+1, find the nonreal root with negative imaginary part.
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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+1=0 gives x=\pm i; the one with negative imaginary part is -i.
Correct answer
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Why
A student concludes a real quadratic with no real roots has 'no solutions.' What's the more accurate statement?
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The Fundamental Theorem guarantees roots exist — 'no real roots' only rules out ONE kind of root, not every kind.
A repeated single root is what a perfect-square quadratic has — that's a different situation from having no real roots.
A quadratic never has infinitely many roots — it has exactly two, counting multiplicity.
Why
'No real roots' means both roots are nonreal — they still exist, as a conjugate pair a\pm bi.
It has two nonreal complex roots, forming a conjugate pair