Does the Fundamental Theorem of Algebra really mean every polynomial has a solution?

Yes, but only in the complex numbers (which include real numbers). A real polynomial like x^2 + 1 = 0 has no real solutions, but it does have two complex solutions: x = \pm i. The theorem guarantees a solution always exists if you allow complex numbers.

How many roots does a polynomial really have?

A polynomial of degree n has exactly n roots when you count multiplicity and include complex roots. For example, (x-2)^3 = 0 has degree 3 and exactly 3 roots (all equal to 2, counted three times). Without multiplicity, it has one distinct root.

Can a polynomial of degree 5 have only 2 real roots?

Yes. A degree-5 polynomial has exactly 5 roots total (by the Fundamental Theorem), but some can be real and others complex. For instance, x^5 - x factors as x(x-1)(x+1)(x^2+1), giving 3 real roots (0, 1, -1) and 2 complex roots (\pm i).

What if I can't factor the polynomial—does the theorem still apply?

Yes, absolutely. The Fundamental Theorem applies to all polynomials, whether they factor nicely or not. Even if you can't find the roots by hand, they still exist in the complex numbers. You can use numerical methods, graphing, or formulas to approximate them.

Fundamental Theorem of Algebra

Every polynomial equation of degree $n$ has exactly $n$ complex roots (counting multiplicity) — this guarantees solutions always exist.

\deg P = n \implies P \text{ has } n \text{ complex roots}

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What each symbol means

What Fundamental Theorem of Algebra takes

Symbol	Meaning
$P$	A polynomial — a sum of terms like $x^2 + 3x - 5$ . Each term is a constant times a power of $x$ . If $P$ is written as $P(x) = 2x^3 + 5x - 1$ , then $P$ is the function, and $P(x)$ is its value at $x$ .
$n$	The degree of the polynomial — the highest power of $x$ that appears with a nonzero coefficient. For $x^3 - 6x^2 + 11x - 6$ , the degree is 3 because the highest power is $x^3$ . If you confuse degree with leading coefficient, you will miscount the number of roots.

When to use it

Reach for this when you need to know how many roots a polynomial must have, or when you're trying to factor and want to verify you've found them all.

Level

Usually taught in: Algebra II

Worked examples

1. Solve a cubic with three distinct real roots

Problem

Solve

x^3 - 6x^2 + 11x - 6 = 0

for

x

$P(1) = 1 - 6 + 11 - 6 = 0$
Test $x = 1$ by substituting into the polynomial. Since $P(1) = 0$ , we know $x = 1$ is a root.
$x^3 - 6x^2 + 11x - 6 = (x-1)(x^2-5x+6)$
Factor out $(x-1)$ using synthetic division or polynomial long division.
$x^2 - 5x + 6 = (x-2)(x-3)$
Factor the quadratic by finding two numbers that multiply to 6 and add to $-5$ : those are $-2$ and $-3$ .
$(x-1)(x-2)(x-3) = 0$
Write the complete factorization of the original polynomial.
$x = 1 \text{ or } x = 2 \text{ or } x = 3$
Set each factor equal to zero to find the three roots.

Answer: $x = 1 \text{ or } x = 2 \text{ or } x = 3$

A degree-3 polynomial must have exactly 3 roots (counting multiplicity) by the Fundamental Theorem. In this case, all three are distinct real roots. We found them by testing to identify one root, then factoring completely.

2. Solve a cubic with negative and fractional roots

Problem

Solve

2x^3 + 3x^2 - 8x - 12 = 0

for

x

$2x^3 + 3x^2 - 8x - 12 = (2x^3 + 3x^2) + (-8x - 12)$
Group the terms into two pairs to find a common factor.
$= x^2(2x + 3) - 4(2x + 3)$
Factor $x^2$ from the first group and $-4$ from the second group.
$= (x^2 - 4)(2x + 3)$
Factor out the common binomial $(2x + 3)$ from both terms.
$= (x-2)(x+2)(2x+3)$
Recognize that $x^2 - 4$ is a difference of squares and factor it as $(x-2)(x+2)$ .
$x = 2 \text{ or } x = -2 \text{ or } x = -\frac{3}{2}$
Set each factor equal to zero: $x - 2 = 0$ gives $x = 2$ , $x + 2 = 0$ gives $x = -2$ , and $2x + 3 = 0$ gives $x = -\frac{3}{2}$ .

Answer: $x = 2 \text{ or } x = -2 \text{ or } x = -\frac{3}{2}$

This cubic has degree 3, so the Fundamental Theorem guarantees exactly 3 roots. By factoring using grouping, we found all three: two integers and one fraction. The presence of negative and fractional roots shows that the theorem applies regardless of the type of root.

3. Phone plan: find data usage from the bill

Problem

A phone plan charges a $20 monthly fee plus $4 per gigabyte of data. Customers who use limited data receive a loyalty discount of

x^2

dollars, where

x

is the number of gigabytes used. If your total bill for the month is $23, how many gigabytes of data did you use?

$20 + 4x - x^2 = 23$
Write the equation for the total bill: base fee ($20) plus data charges ($4x) minus discount ($x^2$) equals the total ($23).
$-x^2 + 4x + 20 - 23 = 0$
Subtract 23 from both sides to set the equation equal to zero.
$-x^2 + 4x - 3 = 0$
Simplify: $20 - 23 = -3$ .
$x^2 - 4x + 3 = 0$
Multiply both sides by $-1$ to make the leading coefficient positive.
$(x-1)(x-3) = 0$
Factor the quadratic by finding two numbers that multiply to 3 and add to $-4$ : those are $-1$ and $-3$ .
$x = 1 \text{ or } x = 3$
Set each factor equal to zero: $x - 1 = 0$ gives $x = 1$ GB, and $x - 3 = 0$ gives $x = 3$ GB.

Answer: $x = 1 \text{ gigabyte or } x = 3 \text{ gigabytes}$

The quadratic equation has degree 2, so by the Fundamental Theorem, it has exactly 2 roots. Both roots are real and positive, so both represent valid data usage. A customer could have used either 1 GB or 3 GB to generate a $23 bill.

Common mistakes

Where Fundamental Theorem of Algebra usually goes wrong

The mistake	Why it is wrong	The fix
The polynomial $x^3 - 8$ has only one root, which is 2.	This counts only real roots and ignores the Fundamental Theorem, which guarantees that a degree-3 polynomial has exactly 3 roots total, including complex ones.	Factor completely: $x^3 - 8 = (x-2)(x^2+2x+4)$ . The cubic formula or the quadratic formula on $x^2 + 2x + 4 = 0$ gives two complex roots: $x = -1 \pm i\sqrt{3}$ . So there are 3 roots total.
The polynomial $x^2 + 4$ has no roots.	This assumes only real roots exist. The Fundamental Theorem guarantees that every polynomial has roots in the complex numbers.	Solve $x^2 + 4 = 0$ by rearranging: $x^2 = -4$ , so $x = \pm 2i$ . The polynomial has exactly 2 complex roots.
The polynomial $(x-3)^2(x+1) = 0$ has 2 roots.	This counts only distinct root values, not multiplicity. The Fundamental Theorem counts each occurrence of a repeated root separately.	This polynomial has degree 3 (when expanded), so it has exactly 3 roots counting multiplicity: $x = 3$ (with multiplicity 2) and $x = -1$ (with multiplicity 1).

Tips and when to use something else

The Fundamental Theorem guarantees $n$ roots exist, but doesn't tell you what they are. Use the Rational Root Theorem or Factor Theorem to find them.
A repeated root counts as multiple roots: the factor $(x-3)^2$ means $x = 3$ appears twice, so it counts as 2 roots, not 1.
If a real polynomial of degree $n$ has fewer than $n$ real roots, the remaining roots must be complex conjugate pairs.
For polynomial equations you can't factor by hand, use the Remainder Theorem to test candidate rational roots, or switch to the quadratic formula for degree 2.

Related formulas

Rational Root TheoremThe Rational Root Theorem tells you which rational numbers could possibly be roots of a polynomial with integer coefficients.Factor TheoremThe Factor Theorem tells you when a linear binomial divides a polynomial: (x - c) is a factor of P(x) exactly when P(c) = 0.Complex NumbersSolve equations with no real solutions and multiply combinations of real and imaginary numbers using the complex number system.Multiplicity of a RootMultiplicity describes how many times a root appears as a factor, determining whether a polynomial graph touches or crosses the axis.Remainder TheoremThe Remainder Theorem lets you find the remainder when a polynomial is divided by (x - c) by simply evaluating the polynomial at c.

Frequently asked questions

Does the Fundamental Theorem of Algebra really mean every polynomial has a solution?: Yes, but only in the complex numbers (which include real numbers). A real polynomial like $x^2 + 1 = 0$ has no real solutions, but it does have two complex solutions: $x = \pm i$ . The theorem guarantees a solution always exists if you allow complex numbers.
How many roots does a polynomial really have?: A polynomial of degree $n$ has exactly $n$ roots when you count multiplicity and include complex roots. For example, $(x-2)^3 = 0$ has degree 3 and exactly 3 roots (all equal to 2, counted three times). Without multiplicity, it has one distinct root.
Can a polynomial of degree 5 have only 2 real roots?: Yes. A degree-5 polynomial has exactly 5 roots total (by the Fundamental Theorem), but some can be real and others complex. For instance, $x^5 - x$ factors as $x(x-1)(x+1)(x^2+1)$ , giving 3 real roots (0, 1, $-1$ ) and 2 complex roots ( $\pm i$ ).
What if I can't factor the polynomial—does the theorem still apply?: Yes, absolutely. The Fundamental Theorem applies to all polynomials, whether they factor nicely or not. Even if you can't find the roots by hand, they still exist in the complex numbers. You can use numerical methods, graphing, or formulas to approximate them.

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Reviewed 2026-09-18