Polynomial Zeros & Conjugate Pairs

Fundamental Observations on Quadratics

  • If x2=0x^2 = 0 then the only solution is x=0x = 0.
  • If x2+1=0x^2 + 1 = 0, then x2=−1x^2 = -1 and the solutions are the purely imaginary numbers x=ix = i and x=−ix = -i.
  • Illustration: a degree–4 polynomial in the morning’s example had the zeros 0, i, −i0,\, i,\, -i (multiplicity explains why only three distinct values appeared even though four zeros—counted with multiplicity—are required by the Fundamental Theorem of Algebra).

The Complex Conjugate‐Root (Pair) Theorem

  • Main fact for this section: If a polynomial has real coefficients and a complex zero z=a+biz = a + bi (with b≠0b \neq 0), then its conjugate zˉ=a−bi\bar z = a - bi is automatically a zero as well.
  • Complex roots therefore appear only in pairs. Permissible counts of non-real zeros for a real polynomial: 0,2,4,6,…0, 2, 4, 6, \dots. You can never have exactly one or three non-real roots.
  • Practical consequence for multiple-choice exams: once a single non-real root is supplied, you can immediately record its conjugate as another answer choice.

Worked Example: Constructing a Degree-4 Polynomial

Given zeros: 2,  −3,  4i2,\; -3,\; 4i.

  • By the conjugate‐root theorem, −4i-4i must also be a zero.
  • Factors corresponding to the zeros:
    • 2  ⟹  (x−2)2 \implies (x - 2)
    • −3  ⟹  (x+3)-3 \implies (x + 3)
    • 4i  ⟹  (x−4i)4i \implies (x - 4i)
    • −4i  ⟹  (x+4i)-4i \implies (x + 4i)
  • Initial (monic) polynomial:
    f(x)=(x−2)(x+3)(x−4i)(x+4i)f(x) = (x - 2)(x + 3)(x - 4i)(x + 4i)
Step 1: Eliminate the imaginaries first

Multiply the conjugate pair factors:
(x−4i)(x+4i)=x2−(4i)2=x2−16i2=x2+16(x - 4i)(x + 4i) = x^2 - (4i)^2 = x^2 -16 i^2 = x^2 + 16
(because i2=−1i^2 = -1).

Step 2: Multiply the real linear factors

(x−2)(x+3)=x2+x−6(x - 2)(x + 3) = x^2 + x - 6

Step 3: Final product

\begin{aligned}
f(x) &= (x^2 + x - 6)(x^2 + 16)\[4pt]
&= x^2(x^2 + 16) + x(x^2 + 16) - 6(x^2 + 16)\[4pt]
&= x^4 + 16x^2 + x^3 + 16x - 6x^2 - 96\[4pt]
&= x^4 + x^3 + 10x^2 + 16x - 96.
\end{aligned}

  • Resulting monic degree-4 polynomial: f(x)=x4+x3+10x2+16x−96f(x) = x^4 + x^3 + 10x^2 + 16x - 96.
Adjusting for a Specified Leading Coefficient
  • If the problem additionally states “leading coefficient 33,” multiply every term by 33:
    fnew(x)=3x4+3x3+30x2+48x−288.f_{\text{new}}(x) = 3x^4 + 3x^3 + 30x^2 + 48x - 288.

Algebraic Identities Used & Re-derived

  • Conjugate pair product: (a−b)(a+b)=a2−b2(a - b)(a + b) = a^2 - b^2 (specifically a=x,  b=4ia = x,\; b = 4i).
  • Square of an imaginary number: i2=−1;  (4i)2=−16i^2 = -1;\; (4i)^2 = -16.
  • FOIL or distributive law for multiplying binomials and polynomials.

Practical/Exam-Related Remarks

  • Expect multiple-choice questions where identifying the missing conjugate root is enough to select the right answer.
  • When writing final answers do not leave the symbol ii inside factored pairs; expand the conjugate factors to remove imaginary units.
  • Homework reference (page 381):
    • Practice problems on zeros: 9–19 odd, plus 21 & 23.
    • Some ask only “what is the other zero?”; others require full polynomial construction.
  • Chapter 3 is complete; a practice test (Test 3) will be distributed tomorrow afternoon. Test 2 results were good, but Test 3 will be more challenging.

Ethical & Pedagogical Emphases

  • Demonstrating the conjugate‐root property avoids incorrect assumptions about lone complex roots.
  • Emphasis on exact arithmetic and expansion ensures answers remain within the real‐coefficient polynomial family, which aligns with standard conventions in algebra courses.