Page 85 - Linear Algebra Done Right

Real Coefficients
2
First suppose that α
< 4β. Then clearly the right side of the
equation above is positive for every x ∈ R, and hence the polynomial
2
x + αx + β has no real roots. Thus no factorization of the form 4.12, 71
with λ 1 ,λ 2 ∈ R, can exist.
2
Conversely, now suppose that α ≥ 4β. Thus there is a real number
2 α 2
c such that c = − β. From 4.13, we have
4
α
2 2 2
x + αx + β = (x + ) − c
2
α α
= (x + + c)(x + − c),
2 2
which gives the desired factorization.
2
In the following theorem, each term of the form x + α j x + β j , with
2
α j < 4β j , cannot be factored into the product of two polynomials with
real coefficients and degree 1 (by 4.11). Note that in the factorization
below, the numbers λ 1 ,...,λ m are precisely the real roots of p, for these
are the only real values of x for which the right side of the equation
below equals 0.
4.14 Theorem: If p ∈P(R) is a nonconstant polynomial, then p
has a unique factorization (except for the order of the factors) of the
form

2
2
p(x) = c(x − λ 1 )...(x − λ m )(x + α 1 x + β 1 )...(x + α M x + β M ),
2 2
where c, λ 1 ,...,λ m ∈ R and (α 1 ,β 1 ),...,(α M ,β M ) ∈ R with α j < 4β j Here either m or M
for each j. may equal 0.

Proof: Let p ∈P(R) be a nonconstant polynomial. We can think
of p as an element of P(C) (because every real number is a complex
number). The idea of the proof is to use the factorization 4.8 of p as a
polynomial with complex coefficients. Complex but nonreal roots of p
come in pairs; see 4.10. Thus if the factorization of p as an element
of P(C) includes terms of the form (x − λ) with λ a nonreal complex
¯
number, then (x − λ) is also a term in the factorization. Combining
these two terms, we get a quadratic term of the required form.
The idea sketched in the paragraph above almost provides a proof
of the existence of our desired factorization. However, we need to
be careful about one point. Suppose λ is a nonreal complex number