Page 79 - Linear Algebra Done Right

Degree
for all z ∈ F. Because p(λ) = 0, we have
2
0 = a 0 + a 1 λ + a 2 λ + ··· + a m λ .
Subtracting the last two equations, we get m 65
2
m
2
p(z) = a 1 (z − λ) + a 2 (z − λ ) +· · ·+ a m (z m − λ )
for all z ∈ F. For each j = 2,...,m, we can write
j
j
z − λ = (z − λ)q j−1 (z)
for all z ∈ F, where q j−1 is a polynomial with degree j − 1 (specifically,
take q j−1 (z) = z j−1 + z j−2 λ +· · ·+ zλ j−2 + λ j−1 ). Thus

p(z) = (z − λ) (a 1 + a 2 q 2 (z) + ··· + a m q m−1 (z))

q(z)

for all z ∈ F. Clearly q is a polynomial with degree m − 1, as desired.

Now we can prove that polynomials do not have too many roots.

4.3 Corollary: Suppose p ∈P(F) is a polynomial with degree m ≥ 0.
Then p has at most m distinct roots in F.

Proof: If m = 0, then p(z) = a 0 = 0 and so p has no roots. If
m = 1, then p(z) = a 0 + a 1 z, with a 1 = 0, and p has exactly one
root, namely, −a 0 /a 1 . Now suppose m> 1. We use induction on m,
assuming that every polynomial with degree m − 1 has at most m − 1
distinct roots. If p has no roots in F, then we are done. If p has a root
λ ∈ F, then by 4.1 there is a polynomial q with degree m − 1 such that

p(z) = (z − λ)q(z)

for all z ∈ F. The equation above shows that if p(z) = 0, then either
z = λ or q(z) = 0. In other words, the roots of p consist of λ and the
roots of q. By our induction hypothesis, q has at most m − 1 distinct
roots in F. Thus p has at most m distinct roots in F.

The next result states that if a polynomial is identically 0, then all
its coefficients must be 0.