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Lecture 14

# MATH135 Lecture 14: Conjugate Roots Theorem and FTA

Department
Mathematics
Course Code
MATH135
Professor
Roxane Itier Itier
Lecture
14

This preview shows page 1. to view the full 5 pages of the document. MATH 135, Winter 2015
Conjugate Roots Theorem.
Fundamental Theorem of Algebra.
1 Conjugate Roots Theorem
Theorem 1. Conjugate Roots Theorem (CJRT)
Suppose fpxq P Rrxs. If cPCis a root of fpxq, then cis a root of fpxq.
Proof. Write fpxq  a0a1xa2x2    anxnwhere a0, a1, . . . , anPR. For any zPC, note
that by the properties of conjugation,
fpzq  a0a1za2z2    anzn
a0a1za2z2    anzn
a0a1za2z2    anzn
a0a1za2z2    anzn
fpzq.
(We used the fact that akaksince akPR.) In particular, if cPCsatisﬁes fpcq  0, then
fpcq  fpcq  0. l
Observation. For any cPC, the following quadratic polynomial has cand cas roots:
gcpxqpxcqpxcq
x2 pccqxcc
x22Repcqx |c|2.
The key point is that gcpxqhas real coeﬃcients.
Example 1.
(a) If c12i, then gcpxq  x22x5.
(b) If crpcos θisin θqwhere r¡0 and θPR, then gcpxq  x2 p2rcos θqxr2.
1
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Only page 1 are available for preview. Some parts have been intentionally blurred. Corollary 2. Real Quadratic Factors (RQF)
Suppose fpxq P Rrxs. If cPCis a non-real root of fpxq, then there exists a real quadratic
polynomial gpxqsuch that gpxq | fpxqin Rrxsand gpcq  0.
Proof. Since cis a root of the real polynomial fpxq, by CJRT, cis also a root of fpxq. Since c
is non-real, i.e., cc, it follows that cand care distinct roots of fpxq. By the Factor Theorem,
gpxq  pxcqpxcqis a factor of fpxqin Crxs. So, fpxq  gpxqhpxqfor some hpxq P Crxs. By
the preceding observation, gpxq P Rrxs. However, the argument via the Factor Theorem does not
guarantee that the quotient hpxqmust have real coeﬃcients.
(For easy reference, we recall that argument. Since fpaq  0, fpxqpxcqppxqfor some ppxq P
Crxs. Next, 0 fpcqpccqppcq. Then ppcq  0, and so ppxqpxcqhpxqfor some hpxq P Crxs.
Thus, fpxqpxcqpxcqhpxq  gpxqhpxq.)
We resolve this problem using the Division Algorithm for Polynomials (DAP). Since both fpxq
and gpxqare in Rrxs, there exist qpxq, rpxq P Rrxssuch that fpxq  gpxqqpxq  rpxq, where either
rpxq  0 or degprq  degpgq. Note that
gpxqhpxq  0fpxq  gpxqqpxq  rpxq.
If we consider the occurring polynomials fpxq, gpxq, hpxq, qpxq, rpxqas complex polynomials, then
by the uniqueness clause of DAP, rpxq  0 and hpxq  qpxq. In particular, hpxq P Rrxs, and we are
done. l
Example 2. Factor the polynomial fpxq  10x522x43x34x23x2 over C. Use the
fact that 1i
2is a root of fpxq.
Solution. First note that fpxqhas real coeﬃcients. Since c1i
2is a root of fpxq, by CJRT,
c1i
2is a second root of fpxq. By the proof of RQF, fpxqhas the real quadratic factor
gpxq  x22Repcqx |c|2x2x1
2.
Equivalently, fpxqhas the real quadratic factor
2gpxq  2x22x12x1i
2
x1i
2
.(1)
2
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