Fundamental Theorem of Algebra
Lesson · Intermediate
Algebra: Polynomials
Every polynomial of degree n ≥ 1 has exactly n complex roots, counted with multiplicity.
Equivalently, if
then there exist complex numbers r₁, r₂, ..., rₙ such that
The roots rᵢ do not have to be distinct.
For example,
has degree 4 and roots 2, 2, 2, -1 when counted with multiplicity.
A polynomial of degree n has at most n distinct real roots.
If P and Q both have degree at most n, and
for n + 1 distinct values x₁, x₂, ..., xₙ₊₁, then
Consider
Then deg R ≤ n, but R has at least n + 1 distinct roots.
The only possibility is
Therefore,
A polynomial P(x) of degree at most 4 satisfies
Find P(100).
Consider
Since deg P ≤ 4, we have deg Q ≤ 4.
But
Thus Q has 5 distinct roots.
A nonzero polynomial of degree at most 4 cannot have 5 distinct roots.
Therefore,
so
Hence
Let P(x) be a polynomial of degree at most 5 such that
for k = 0, 1, 2, 3, 4, 5.
Find P(6).
We cannot simply compare P(x) with 2ˣ because P(x) - 2ˣ is not a polynomial, so the polynomial root bound cannot be applied directly.
Instead, construct a polynomial that agrees with 2ᵏ at those integer values.
Recall the binomial identity
Define
This is a polynomial of degree 5.
For each k = 0, 1, 2, 3, 4, 5, the terms after the kth binomial coefficient become 0, so
Hence
Therefore, P(k) = Q(k) for six distinct values k = 0, 1, 2, 3, 4, 5.
But both P and Q have degree at most 5.
Thus
has six distinct roots while having degree at most 5.
Therefore,
so
for every x.
Now evaluate at x = 6:
But
so